Abstract Algebra

Group Theory

Introduction

Def. A binary operation \(*\) on a set \(G\) is a function \(*:G\times G\to G\). For any \(a,b\in G\), we shall write \(a*b\) for \(*(a,b)\).

A binary operation \(*\) on a set \(G\) is associative if for all \(a,b,c\in G\) we have \(a*(b*c)=(a*b)*c\).
If \(*\) is a binary operation on a set \(G\) we say elements \(a\) and \(b\) of \(G\) commute if \(a*b=b*a\). We say \(*\) (or \(G\)) is commutative if for all \(a,b\in G\), \(a*b=b*a\).

Def. A group is an ordered pair \((G,*)\) satisfying:

\(\exist\) an identity \(e\) such that \(e*a=a*e=a\) for \(\forall a\in G\);
For \(\forall a\in G\), \(\exist a^{-1}\in G\), called an inverse of \(a\), such that \(a*a^{-1}=a^{-1}*a=e\);
For \(\forall a,b,c\in G\), \(a*(b*c)=(a*b)*c\).

\((G,*)\) is called an Abelian (or commutative) if \(a*b=b*a\) for \(\forall a,b\in G\).

Prop. Let \((G,*)\) be a group, then

\(e\) is unique;
\(a^{-1}\) is unique for \(\forall a\in G\);
\((a^{-1})^{-1}=a\) for \(\forall a\in G\);
\((a*b)^{-1}=(b^{-1})*(a^{-1})\) for \(\forall a,b\in G\);
Generalized associative law: \(a_1*a_2*\cdots*a_n\) is independent of how the expression is bracketed, for \(\forall a_1,a_2,\cdots,a_n\in G\).

Def. Let \(G\) be a group and \(x\in G\). The order of \(x\) is defined to be the smallest positive integer \(n\) such that \(x^n=e\), denoted as \(|x|\). If no such integer exists, then \(|x|=\infty\).

Note that the order of a finite set is the number of its element.

Prop. (Cancellation law) \(\forall a,b,u,v\in G\), \(au=av\Rightarrow u=v\), \(ub=vb\Rightarrow u=v\).

Prove by inverse.

Dihedral Group

\(D_{2n}=\langle r,s\mid r^n=s^2=1,rs=sr^{-1}\rangle\)

Symmetric Group

Let \(\Omega=\{1,2,\cdots,n\}\), \(S_n\) denote all permutations of \(\Omega\). Obviously, \(S_n\) forms a group under function composition and \(|S_n|=n!\).

A cycle is a string of integers which represents the element of \(S_n\) that cyclically permutes these integers (fix all other integers). That is, \((a_1~a_2~\cdots~a_m)\) means \(a_1\to a_2,a_2\to a_3,\cdots,a_m\to a_1\). \(\forall \sigma\in S_n\), it can be rearranged and grouped into some cycles of the form

\[\sigma=(a_1~a_2\cdots a_{m_1})(a_{m_1+1}~a_{m_1+2}\cdots~a_{m_2})\cdots(a_{m_{k-1}+1}~a_{m_{k-1}+2}\cdots a_{m_k}) \]

Example.

\[\begin{aligned}\sigma&=\begin{pmatrix}1&2&3&4&5&6&7&8&9&10&11&12&13\\12&13&3&1&11&9&5&10&6&4&7&8&2\end{pmatrix}\\&=(1~~12~~8~~10~~4)(2~~13)(3)(5~~11~~7)(6~~9)\\&=(1~~12~~8~~10~~4)(2~~13)(5~~11~~7)(6~~9)\end{aligned} \]

A cycle of length \(t\) is called \(t\)-cycle.

Computing the products in \(S_n\) is straightforward.

Example. \((1~~2)\circ (1~~3)=(1~~3~~2)\), \((1~~3)\circ (1~~2)=(1~~2~~3)\). By the way we find that \(S_n\) is non-Abelian.

对于给定的排列求循环表示：不断增加新的循环，每次从还未加入循环的最小数开始，不断将它的后继加入当前循环。

求若干循环表示的复合：在上述“找后继”部分，从右往左看每个循环。例如，在 \((1~~2)\circ (1~~3)\) 中找 \(1\) 的后继，先在 \((1~~3)\) 中让 \(1\to 3\)，再在 \((1~ ~2)\) 中让 \(3\to 3\).

The quaternion group

\[Q_8=\{1,-1,i,-i,j,-j,k,-k\} \]

The product \(\cdot\) is defined as

\[\begin{aligned} &1\cdot a=a\cdot 1=a, \forall a\in Q_8\\ &(-1)\cdot (-1)=1,(-1)\cdot a=a\cdot (-1)=-a,\forall a\in Q_8\\ &i\cdot i=j\cdot j=k\cdot k=-1\\ &i\cdot j=k,~~~~j\cdot i=-k,~~~~j\cdot k=i,\\ &k\cdot j=-i,~~~~k\cdot i=j,~~~~i\cdot k=-j \end{aligned} \]

Then \(|Q_8|=8\) and \(Q_8\) is non-Abelian.

顺着 \(i,j,k\) 的顺序就是正的，逆着 \(i,j,k\) 的顺序就是负的。

Homomorphisms & Isomorhpisms

Def. Let \((G,*)\) and \((H,\diamond)\) be groups. A map \(\varphi:G\to H\) such that \(\varphi(x*y)=\varphi(x)\diamond\varphi(y),\forall x,y\in G\) is called a homomorphism（同态）. If \(\varphi\) is bijective（双射的）, then we call \(\varphi\) an isomorphism（同构） and \(G\) and \(H\) are said to be isomorphic, written \(G\cong H\).

Intutively, homomorphisms respects the group structures.

Example.

\(G\cong G\), \(\cong\) is an equivalence relation.
The exponential map \(\exp:\R\to \R^{+}\) defined by \(\exp(x)=e^x\) is an isomorphism from \((\R,+)\) to \((\R^{+},\times)\).
The symmetric groups \(S_{\Delta}\) and \(S_{\Omega}\) are isomorphic iff \(|S_{\Delta}|=|S_{\Omega}|\).

Prop. If \(\varphi:G\to H\) is an isomorphism, then

\(|G|=|H|\) (the order of \(G\) and \(H\) are same).
\(G\) is Abelian if and only if \(H\) is Abelian.
For \(\forall x\in G\), \(|x|=|\varphi(x)|\) (the order of \(x\) and \(\varphi(x)\) are same).

Group actions

Def. A group action（群作用） of a group \(G\) on a set \(A\) (not necessarily a group) is a map from \(G\times A\to A\) (written as \(g\cdot a\), for \(\forall g\in G,a\in A\)) satisfying:

\(g_1\cdot (g_2\cdot a)=(g_1*g_2)\cdot a\), for \(\forall g_1,g_2\in G,a\in A\).
\(1\cdot a=a\) for \(\forall a\in A\).

\(\cdot\) is NOT a binary operation. Informally, we say \(G\) is a group acting on set \(A\), and omit \(*\) and \(\cdot\).

Let \(G\) act on \(A\), for \(\forall g\in G\), we define a map \(\sigma_g:A\to A\) by \(\sigma_g(a)=g\cdot a\). Then \(\sigma_g\) is a permutation of \(A\), and the map from \(G\) to \(S_A\) defined by \(g\to \sigma_g\) is a homomorphism.

By \((\sigma_{g^{-1}}\circ \sigma_g)(a)=a\) for \(\forall a\in A\), we know that \(\sigma_g\) is a permutation. Let \(\varphi:G\to S_A\) be defined by \(\varphi(g)=\sigma_g\) for \(\forall g\in G\). Then \(\forall a\in A\), we have \(\varphi(g_1g_2)(a)=\sigma_{g_1g_2}(a)=(g_1g_2)a=g_1(g_2a)=\sigma_{g_1}(\sigma_{g_2}(a))=(\varphi(g_1)\circ\varphi(g_2))(a)\), so \(\varphi\) is a homomorphism.

The homomorphism given above is called the permutation representation associated to the given action.

Subgroups

Def. Let \(G\) be a group. A nonempty subset \(H\) of \(G\) is a subgroup of \(G\) if \(H\) is closed under products and inverses, written \(H\le G\).

Subgroups of \(G\) are just subsets of \(G\) which are themselves groups.

If \(H\le G\) and \(H\neq G\), we shall write \(H\lt G\) to emphasize that \(H\) is a proper subset.

Prop. A subset \(H\) of a group \(G\) is a subgroup iff \(H\neq \empty\) and \(\forall x,y\in H\), \(xy^{-1}\in H\). Furthermore, if \(H\) is finite, then it suffices to check that \(H\) is nonempty and closed under multiplication.

注意不要忽略无限群的情况，这和有限群可能完全不同。验证无限群的子群，需要验证非空和 \(xy^{-1}\in H\)；验证有限群的子群，只要验证非空和 \(xy\in H\)。

Let \(G\) be a group and \(A\) be any nonempty subset of \(G\). We define centralizer of \(A\) in \(G\): \(C_{G}(A)=\{g\in G\mid gag^{-1}=a,\forall a\in A\}\). Note that \(gag^{-1}=a\Leftrightarrow ga=ag\).

Prop. \(C_G(A)\le G\).

First, \(e\in C_G(A)\), so \(C_G(A)\neq \emptyset\). Second, \(\forall x,y\in C_G(A),a\in A\), we have \((xy^{-1})a(xy^{-1})^{-1}=x(y^{-1}ay)x^{-1}=xax^{-1}=a\), so \(xy^{-1}\in C_G(A)\). Thus \(xy^{-1}\in C_G(A)\), \(C_G(A)\le G\).

center of \(G\): \(Z(G)=\{g\in G\mid gx=xg,\forall x\in G\}\). \(Z(G)=C_G(G)\le G\).

normalizer of \(A\) in \(G\): \(N_G(A)=\{g\in G\mid gAg^{-1}=A\}\), where \(gAg^{-1}=\{gag^{-1}\mid a\in A\}\). We have \(N_G(A)\le G\) and \(C_G(A)\le N_G(A)\).

If \(G\) is Abelian then \(Z(G)=C_G(A)=N_G(A)=G\).

stabilizers of group actions: Let \(G\) act on \(S\) and \(s\in S\), the stabilizer of \(s\) in \(G\) is the set \(G_s=\{g\in G\mid gs=s\}\). We have \(G_s\le G\).

Cyclic Group

Def. A group \(H\) is cyclic if \(H\) can be generated by a single element, i.e. \(\exist x\in H\) such that \(H=\{x^n\mid n\in \Z\}\), written \(H=<x>\).

Example. \(\Z=<1>\) is a cyclic group.

Prop. Let \(G\) be an arbitrary group, \(x\in G\) and \(m,n\in \Z\). If \(x^n=x^m=1\), then \(x^d=1\) where \(d=\gcd(m,n)\). In particular, if \(x^m=1\), then \(|x|\) divides \(m\).

Prop. Any two cyclic groups of the same order are isomorphic.

Prop. Let \(G\) be a group, \(x\in G\) and \(a\neq 0\in \Z\).

If \(|x|=\infty\), then \(|x^a|=\infty\).
If \(|x|=n\lt \infty\), then \(|x^a|=\frac{n}{(n,a)}\).

Prop. Let \(H=<x>\).

Assume \(|x|=\infty\), then \(H=<x^a>\) iff \(a=\pm 1\).
Assume \(|x|=n\lt \infty\), then \(H=<x^a>\) iff \((a,n)=1\). In particular, the number of generators of \(H\) is \(\varphi(n)\) (Euler function).

再次提醒，注意无穷群和有限群的区别。

Subgroup generated by subsets

Defined as \(<A>=\{a_1^{\alpha_1}a_2^{\alpha_2}\cdots a_n^{\alpha_n}\mid a_i\in A,\alpha_i\in \Z,a_i\neq a_{i+1},n\in \Z^{+}\}\).

If \(G\) is Abelian, then \(|<A>|\le |a_1|\cdot|a_2|\cdot\cdots\cdots|a_n|\). Otherwise, this formula doesn't hold.

Example:

\(G=D_8=<r,s>\) and \(A=\{s,rs\}\). Then \(|a|=|b|=2\), but \(|<A>|=|G|=8\).
\(G=GL_2(\R)\) and \(A=\{a=\begin{pmatrix}0&1\\1&0\end{pmatrix},b=\begin{pmatrix}0&2\\\frac12&0\end{pmatrix}\}\). Then \(|a|=|b|=2\), but \(|<A>|=\infty\) (since \(|ab|=\infty\)).

The Lattice of subgroups of a group

列出所有的子群，根据包含关系连边。

Quotient Groups

Let \(\varphi\) be a homomorphism from a group \(G\) to a group \(H\), recall that the fibers of \(\varphi\) are the sets of elements of \(G\) projecting to single element of \(H\). We denote the fiber above \(a\) as \(\varphi^{-1}(a)\).

Multiplication of fibers

Let \(X_g\) denote the fibers above \(g\). For \(a,b\in H\), the product of \(X_a\) and \(X_b\) id defined to be the fiber \(X_{ab}\) above \(ab\). This makes the set of fibers into a group, called a quotient group of \(G\).

Example. Let \(G=\Z\), \(H=\Z_n=<x>\) be the cyclic group of order \(n\), and define \(\varphi: G\to H\) by \(\varphi(a)=x^a\). Then \(\varphi\) is a homomorphism. The fibers \(\varphi^{-1}(x^a)=\{m\in\Z\mid m\equiv a\pmod n\}=\bar a\).

Def. If \(\varphi\) is a homomorphism \(\varphi:G\to H\), then the kernel of \(\varphi\) is the set \(\ker \varphi=\{g\in G\mid \varphi(g)=1\}\).

Def. Let \(\varphi:G\to H\) be a homomorphism with kernel \(K\). The quotient group or factor group \(G/K\), is the group whose elements are fibers of \(\varphi\) with group operation defined by \(X_aX_b=X_{ab}\).

Prop. Let \(X\in G/K\) be the fiber above \(a\), i.e., \(X=\varphi^{-1}(a)\). Then for \(\forall u\in X\), we have \(X=uK=Ku\) or \(X=\{uk\mid k\in K\}=\{ku\mid k\in K\}\).

Def. For any \(N\le G\) and any \(g\in G\), the left coset and the right coset are \(gN=\{gn\mid n\in N\}\) and \(Ng=\{ng\mid n\in N\}\), respectively. Any element of a coset is called a representative for the coset.

Prop. The set of left/right cosets of \(K\) with operation \(uK\circ vK=(uv)K\)/\(Ku\circ Kv=K(uv)\) forms the group \(G/K\).

这一命题是商群最常用的定义，即所有 coset 组成的群。

Prop. Let \(N\le G\), then the set of left cosets of \(N\) in \(G\) forms a partition of \(G\). Furthermore, for \(u,v\in G\), \(uN=vN\) iff \(v^{-1}u\in N\), and in particular, \(uN=vN\) iff \(u\) and \(v\) are representatives of the same coset.

注意这个定理是对所有 \(N\le G\) 成立的，没有要求 \(N\trianglelefteq G\)，但此时所有 coset 组成的集合就不一定是群。

Prop. Let \(N\le G\).

\(uN \cdot vN = (uv)N\) is well defined iff \(gng^{-1}\in N\) for \(\forall g\in G,n\in N\).
If the above operation is well defined, then it makes the set of left cosets of \(N\) in \(G\) into a group.

Let \(u_1=un\) and \(v_1=vm\) for some \(n,m\in N\), then \(u,u_1\in uN,v,v_1\in vN\). Note that \(u_1v_1=unvm=u(vv^{-1})nvm=(uv)(v^{-1}nv)m\), so \(uvN=u_1v_1N\Leftrightarrow u_1v_1\in uvN\Leftrightarrow (v^{-1}nv)m\in N,\forall m\in N\Leftrightarrow v^{-1}nv\in N\).

Def. Let \(N\le G\). The element \(gng^{-1}\) is called the conjugate of \(n\) by \(g\). The set \(gNg^{-1}\) is called the conjugate of \(N\) by \(g\). The element \(g\) is said to normalize \(N\) if \(gNg^{-1}=N\). A subgroup \(N\) of \(G\) is called normal if \(gNg^{-1}=N\) for \(\forall g\in G\), written \(N\trianglelefteq G\).

Prop. A subgroup \(N\) of \(G\) is normal iff it is the kernel of some homomorphism.

Define \(\pi:G\to G/N\) by \(\pi(g)=gN\), then \(\ker\pi=N\).

正规子群一定是某个同态的核，也正因此只有对正规子群才能定义商群。而有了 coset 的定义，不需要同态也可以定义商群。

Def. The homomorphism \(\pi:G\to G/N\) defined by \(\pi(g)=gN\) is called the natural projection (homomorphism) of \(G\) onto \(G/N\). If \(\bar H\le G/N\), the complete preimage of \(\bar H\) in \(G\) is the preimage of \(\bar H\) under \(\pi\).

preimage 的中文意思是“原像”。一个集合在映射下的原像是已经定义的，这里定义的是一个群在另一个群下的原像。

Lagrange Theorem

If \(G\) is a finite group and \(H\le G\), then \(|H|\mid |G|\) and the number of left cosets of \(H\) in \(G\) equals \(\dfrac{|G|}{|H|}\).

定理的前半部分是说 \(H\) 的元素个数是 \(G\) 的元素个数的因数。因此 Lagrange Theorem 只对有限群有效。

定理的证明，注意到对所有 \(g\)，映射 \(f:H\to gH, f(h)=gh\) 是双射，加上 \(G\) 可以被划分成 \(k\) 个 \(H\) 的 left coset，就得到 \(|G|=k|H|\).

Def. The number of left cosets of \(H\) in \(G\) is called the index（指数） of \(H\) in \(G\), written as \(|G:H|\).

The full converse to Lagrange's Theorem is NOT true. Namely, if \(G\) is a finite group and \(n\mid |G|\), then \(G\) may not have a subgroup of order \(n\).

Cauchy Theorem. If \(|G|\lt \infty\) and \(p\) is a prime dividing \(|G|\), then \(\exist x\in G\) such that \(|x|=p\).

Sylow's Theorem. If \(|G|=p^\alpha m\), where \(p\) is a prime and \(p\) does not divide \(m\), then \(G\) has a subgroup of order \(p^\alpha\).

Def. Let \(H\) and \(K\) be subgroups of a group, and define

\[HK=\{hk\mid h\in H,k\in K\} \]

Note that \(HK\) may not be a group.

Prop. If \(H\) and \(K\) are finite subgroups of a group then

\[|HK|=\frac{|H||K|}{|H\cap K|} \]

Note that for \(h_1,h_2\in H\), we have \(h_1K=h_2K\Leftrightarrow h_1(H\cap K)=h_2(H\cap K)\). Thus the number of distinct cosets of the form \(hK\) for \(h\in H\) is the number of distinct cosets \(h(H\cap K)\), for \(h\in H\), which is, by Lagrange Theorem, \(\dfrac{|H|}{|H\cap K|}\). Thus, \(|HK|=\dfrac{|H||K|}{|H\cap K|}\).

Prop. If \(H,K\le G\), then \(HK=KH\) if and only if \(HK\le G\).

Corollary. If \(H,K\le G\) and \(H\le N_G(K)\), then \(HK\le G\). In particular, if \(K\trianglelefteq G\), then \(HK\le G\) for any \(H\le G\).

The Isomorphism Theorems

Thm. If \(\varphi:G\to H\) is a homomorphism of groups, then \(\ker \varphi \trianglelefteq G\) and \(G/\ker\varphi\cong \varphi(G)\).

This theorem is called The First Isomorphism Theorem or The Fundamental Theorem of Homomorphism.

Corollary. Let \(\varphi:G\to H\) be a homomorphism of groups.

\(\varphi\) is injective if and only if \(\ker \varphi = \{1\}\).
\(|G:\ker \varphi|=|\varphi(G)|\).

Thm. Let \(A,B\le G\) and \(A\le N_G(B)\), then \(AB\le G\), \(B\trianglelefteq AB\), \(A\cap B\trianglelefteq A\) and \(AB/B\cong A/(A\cap B)\).

\(A,B\le G, A\le N_G(B) \Rightarrow AB\le G\)

\(A,B\le N_G(B)\Rightarrow AB\le N_G(B)\Rightarrow B\trianglelefteq AB\)

Define \(\varphi:A\to AB/B\) by \(\varphi(a)=aB\), then \(\varphi\) is a homomorphism. Also \(\varphi\) is surjective.

Note that \(\ker \varphi=\{a\in A\mid aB=1\cdot B\}=A\cap B\), so \(A\cap B\trianglelefteq A\) and \(AB/B\cong A/(A\cap B)\) by The First Isomorphism Theorem.

This theorem is called The Second Isomorphism Theorem or The Diamond Isomorphism Theorem. The lattice of subgroups of \(G\) is as follows.

\[G\\ |\\ AB\\ \diagup ~~~~\diagdown\\ A~~~~~~~~~~~B\\ \diagdown~~~~\diagup\\ A\cap B\\ |\\ 1 \]

Thm. Let \(G\) be a group and let \(H\) and \(K\) be normal subgroups of \(G\) with \(H\le K\). Then \(K/H\trianglelefteq G/H\) and

\[(G/H)/(K/H)\cong G/K \]

First we have \(N_K(H)=K\Leftrightarrow H\trianglelefteq K\), so \(K/H\) is well defined. Obviously \(K/H\le G/H\) since \(K\trianglelefteq G\).

Consider \(aH\in G/H\), \(\forall a\in G\) and \(bH\in K/H\), \(\forall b\in K\). Then \((aH)(bH)(a^{-1}H)=(aba^{-1})H\in K/H\Rightarrow K/H\trianglelefteq G/H\).

Define \(\varphi:G/H\to G/K\) by \(\varphi(gH)=gK\). By \(H\le K\) it's easy to show that \(\varphi\) is well defined and \(\varphi\) is surjective.

Finally, \(\ker \varphi=\{gH\in G/H\mid gK=K\}=\{gH\mid g\in K\}=K/H\). By the First Isomorphism Theorem, we have \((G/H)/(K/H)\cong G/K\).

If we denote the quotient by \(H\) with a bar, this can be written \(\bar G/\bar K\cong G/K\). This theorem is called The Third Isomorphism Theorem.

Thm. Let \(N\trianglelefteq G\), then there exists a bijection from the set of subgroups \(A\) of \(G\) which contain \(N\) onto the set of subgroups \(\bar A=A/N\) of \(G/N\).

In particular, every subgroup of \(\bar G=G/N\) is of the form \(A/N\) for some \(A\le G\) and \(N\subseteq A\). This bijection has the following properties: for \(\forall A,B\le G\) with \(N\le A\) and \(N\le B\),

\(A\le B\) if and only if \(\bar A\le \bar B\),
if \(A\le B\) then \(|B:A|=|\bar B:\bar A|\)
\(\overline{\langle A,B\rangle}=\langle\bar A,\bar B\rangle\)
\(\overline{A\cap B}=\bar A\cap \bar B\)
\(A\trianglelefteq G\) if and only if \(\bar A\trianglelefteq \bar G\).

定理是在说，\(G/N\) 的子群一一对应于某个 \(A/N\)，其中 \(A\) 是 \(G\) 的子群，且 \(A\) 包含 \(N\)。这个一一对应用 \(\bar A\) 来表示。

This theorem is called The Fourth Isomorphism Theorem or The Lattice Isomorphism Theorem.

Composition series and the Holder program

Prop. If \(G\) is a finite Abelian group and \(p\) is a prime dividing \(|G|\), then \(G\) contains an element of order \(p\). This is a special case of Cauchy's theorem.

The proof proceeds by induction on \(|G|\). Namely, we assume the result is valid for every group whose order \(\lt |G|\), and then prove the result valid for \(|G|\).

Since \(|G|\gt 1\), \(\exist x\in G,x\neq 1\).

If \(|G|=p\), then \(|x|=p\), done. We therefore assume \(|G|\gt p\).

Suppose \(p\mid |x|\) and \(x=pn\) for some \(n\in \Z^{+}\), then \(|x^n|=p\), done.

Thus we may assume \(p\nmid |x|\). Let \(N=\langle x\rangle\), since \(G\) is Abelian we know \(N\trianglelefteq G\). By Lagrange's Theorem, \(|G/N|=\frac{|G|}{|N|}\lt |G|\) (note \(|N|>1\)).

Moreover, since \(p\nmid |N|\) and \(p\mid |G|\), we must have \(p\mid |G/N|\). Now we can apply the induction hypothesis on \(G/N\) to conclude it contains and element \(\bar y=yN\) of order \(p\).

Since \(|\bar y|=p, y\not\in N\) but \(\bar y ^p =\bar 1 \Rightarrow y^p\in N\), we must have \(\langle y^p\rangle \neq \langle y \rangle\). That is, \(|y^p|\lt |y|\). Thus \(p\mid |y|\). Let \(z=y^{|y|/p}\), then \(z\in G\) and \(|z|=p\). This completes the induction.

Def. A group \(G\) is called simple if \(|G|\gt 1\) and the only normal subgroups of \(G\) are \(1\) and \(G\).

中文“单群”。单群 \(G\) 不能分解成 \(N\) 和 \(G/N\) 的形式，它们类似于 \(\Z\) 中的质数。

Def. In a group \(G\), a sequence of subgroups

\[1=N_0\le N_1\le N_2\le\cdots\le N_{k-1}\le N_k=G \]

is called a composition series if \(N_i\trianglelefteq N_{i+1}\) and \(N_{i+1}/N_i\) is a simple group. If the above sequence is a composition series, the quotient groups \(N_{i+1}/N_i\) are called composition factors of \(G\).

Example. \(1\trianglelefteq \langle s\rangle \trianglelefteq \langle s,r^2\rangle \trianglelefteq D_8\), \(1\trianglelefteq\langle r^2\rangle \trianglelefteq\langle r\rangle \trianglelefteq D_8\).

Thm.(Jordan-Holder) Let \(G\) be a finite group with \(G\neq 1\). Then

\(G\) has a composition series and
The composition factors in a composition series are unique, namely, if
\(1=N_0\le N_1\le \cdots\le N_r=G\) and \(1=M_0\le M_1\le \cdots\le M_s=G\) are two composition series for \(G\), then \(r=s\) and there is some permutation \(\pi\) of \(\{1,2,\cdots,r=s\}\) such that \(M_{\pi(i)}/M_{\pi(i)-1}\cong N_i/N_{i-1}\) for \(1\le i\le r\).

The Holder Program is a two part program for classfying all finite groups up to isomorphism:

Classify all finite simple groups.
Find all ways of "putting simple groups together" to form other groups.

Thm. There is a list consisting of \(18\) (infinite) families of simple groups and \(26\) simple groups not belonging to these families (the sporadic simple groups) such that every finite simple group is isomorphic to one of the groups in this list.

Thm.(Feit-Thompson) If \(G\) is a simple group of odd order, then \(G\cong \Z_p\) for some prime \(p\).

Def. A group \(G\) is solvable if there is a chain of subgroups

\[1=G_0\trianglelefteq G_1\trianglelefteq G_2\trianglelefteq\cdots\trianglelefteq G_s=G \]

such that \(G_{i+1}/G_i\) is Abelian for \(i=0,1,\cdots,s-1\).

Thm. The finite group \(G\) is solvable if and only if for every divisor \(n\) of \(|G|\) such that \((n,\dfrac{|G|}{n})=1\), \(G\) has a subgroup of order \(n\).

Prop. Let \(N\trianglelefteq G\), then if \(N\) and \(G/N\) are solvable, then \(G\) is solvable.

Assume \(1=N_0\trianglelefteq N_1\trianglelefteq \cdots\trianglelefteq N_n=N\) and \(\bar 1=\bar G_0\trianglelefteq \bar G_1\trianglelefteq \cdots \trianglelefteq \bar G_m = \bar G\).

By Lattice (Fourth) Isomorphism Theorem, there exists subgroup \(G_i\) of \(G\) with \(N\le G_i\) such that \(G_i/N=\bar G_i\) and \(G_i\trianglelefteq G_{i+1},0\le i\le m\).

By the Third Isomorphism Theorem, \(\bar G_{i+1}/\bar G_i = (\bar G_{i+1}/N)/(\bar G_{i}/N)\cong G_{i+1}/G_i\).

Thus \(1=N_0\trianglelefteq N_1\trianglelefteq\cdots\trianglelefteq N_n=N=G_0\trianglelefteq G_1\trianglelefteq\cdots\trianglelefteq G_m=G\) is a chain of subgroups of \(G\) whose successive quotient groups are Abelian. Thus \(G\) is solvable.

Transposition and the alternating group

Def. A 2-cycle is called a transposition.

Prop. Every element of symmetric group \(S_n\) can be written as a product of transpositions.

Let \(x_1,\cdots,x_n\) be independent variables, we define \(\Delta\) be the polynomial

\[\Delta =\prod_{1\le i\lt j \le n}(x_i-x_j) \]

For each \(\sigma \in S_n\), let \(\sigma\) act on \(\Delta\)

\[\sigma(\Delta)=\prod_{1\le i\lt j\le n}(x_{\sigma(i)}-x_{\sigma(j)}) \]

For \(\forall \sigma\in S_n\), define \(\epsilon:S_n\to \Z_2\cong (\{\pm 1\},\times)\) by

\[\epsilon(\sigma)=\begin{cases}+1,\text{if }\sigma(\Delta)=\Delta\\-1,\text{if }\sigma(\Delta)=-\Delta\end{cases} \]

Def. \(\epsilon(\sigma)\) is called the sign of \(\sigma\). \(\sigma\) is called an even permutation if \(\epsilon(\sigma)=1\) and an odd permutation if \(\epsilon(\sigma)=-1\).

Prop. \(\epsilon\) is a homomorphism.

Prop. Transpositions are all odd permutations and \(\epsilon\) is surjective homo.

Def. The alternating group of degree \(n\), denoted by \(A_n\), is the kernel of the homomorphism \(\epsilon\) (i.e. the set of even permutations).

Prop. The permutation \(\sigma\) is odd if and only if the number of cycles of even length in its cycle decomposition is odd.

Prop. \(A_n\) is solvable if and only if \(n\le 4\).

When \(n=1,2\), \(A_n\cong\{1\}\) is trivial.

When \(n=3\), \(A_3\cong \Z_3\) is Abelian, so \(A_3\) is solvable.

When \(n=4\), \(1\trianglelefteq \langle(1~4)(2~3)\rangle\trianglelefteq\langle(1~2)(3~4),(1~3)(2~4)\rangle\trianglelefteq A_4\).

Now we are going to prove that \(A_n\) is simple for all \(n\ge 5\), and thus \(A_n\) is not solvable.

Lemma 1. \(A_n=\langle(1~2~3),(1~2~4), \cdots,(1~2~n)\rangle\) (for \(n\ge 3\)).

This lemma can be proven immediately by \(S_n=\langle(1~2),(1~3),\cdots,(1~n)\rangle\) (for \(n\ge 2\)) and \((1~i)(1~j)=(1~2~i)(1~2~j)^2\).

Now we return to the simplicity of \(A_n\). Suppose there exists some \(N\trianglelefteq A_n\) and \(N\neq 1\), we need to prove \(N=A_n\). In fact, we only need to show \(N\) contains a \(3\)-cycle. This suffices because for \(n\ge 5\), one can always choose an even permutation \(\gamma=\begin{pmatrix}1&2&3&4&5&\cdots\\i&j&k&l&m&\cdots \end{pmatrix}\). WLOG, we assuume \(A\) contains \((1~2~3)\), then any \(3\)-cycle \((i~j~k)=\gamma^{-1}(1~2~3)\), and thus \(N=A_n\).

Pick \(\sigma\in N,\sigma\neq 1\), write \(\sigma=\pi_1\pi_2\cdots\pi_k\), where \(\pi_j\)'s are disjoint cycles.

Case 1: some \(\pi_i\) has length at least \(4\). By relabelling we can assume \(\pi_1=(1~2~\cdots~r)\) for some \(r\ge 4\). Let \(\varphi=(1~2~3)\), then \(\varphi\sigma\varphi^{-1}\in N\), and \(\varphi\sigma\varphi^{-1}=\varphi\pi_1\varphi^{-1}\pi_2\cdots\pi_k=\varphi\pi_1\varphi^{-1}\pi_1^{-1}\sigma=(1~2~4)\sigma\), so \((1~2~4)=\varphi\sigma\varphi^{-1}\sigma^{-1}\in N\).

Case 2: Each \(\pi_i\) has length \(\le 3\), and at least two have length \(3\) (so \(n\ge 6\)). We assume \(\pi_1=(1~2~3)\) and \(\pi_2=(4~5~6)\). Let \(\varphi=(1~2~4)\), then \(\varphi\sigma\varphi^{-1}=\varphi\pi_1\pi_2\varphi^{-1}\pi_2^{-1}\pi_1^{-1}\sigma=(1~2~5~3~4)\sigma\), so \(5\)-cycle \((1~2~5~3~4)=\varphi\sigma\varphi^{-1}\sigma^{-1}\in N\). By Case 1, \(N\) contains a \(3\)-cycle.

Case 3: Each \(\pi_i\) has length \(\le 3\), and exactly one has length \(3\). In this case, \(\sigma^2\in N\) is a \(3\)-cycle.

Case 4: All \(\pi_i\)'s are \(2\)-cycle (so \(k\) is even; especially, \(k > 1\)). We assume \(\pi_1=(1~2)\) and \(\pi_2=(3~4)\). Let \(\varphi=(1~2~3)\), then \(\varphi\sigma\varphi^{-1}=\varphi\pi_1\pi_2\varphi^{-1}\pi_2^{-1}\pi_1^{-1}\sigma=(1~3)(2~4)\sigma\in N\), so \((1~3)(2~4)=\varphi\sigma\varphi^{-1}\sigma^{-1}\in N\). Let \(\psi=(1~3~5)\), then \((1~3)(2~4)\psi(1~3)(2~4)\psi^{-1}=(1~3~5)\in N\).

In summary, this proposition is proven.

Group Action

Recall that a group action of \(G\) on \(A\) is a map from \(G\times A\) to \(A\) (written as \(g\cdot a\)) such that (1) \(g_1\cdot (g_2\cdot a)=(g_1\cdot g_2)\cdot a,~\forall g_1,g_2\in G, a\in A\) and (2) \(1\cdot a=a,~\forall a\in A\).

For \(\forall g\in G\), the map \(\sigma_g:A\to A\) defined by \(\sigma_g(a)=g\cdot a\) is a permutation of \(A\). The homomorphism associated to an action of \(G\) on \(A\) \(\varphi:G\to S_A\) defined by \(\varphi(g)=\sigma_g\) is called the permutation representation associated to the given action.

Def.

The kernel of the action: \(\{g\in G\mid g\cdot a=a\text{ for } \forall a\in A\}\).
The stabilizer of \(a\in A\) in \(G\): \(G_a=\{g\in G\mid g\cdot a = a\}\).
An action is faithful if its kernel is the identity (distinct element of \(G\) induce distinct permutations).

Two group elements induce the same permutation on \(A\) iff they are in the same coset of the kernel.

Given any non-empty \(A\) and any homomorphism \(\varphi:G\to S_A\). We obtain an action of \(G\) on \(A\) by defining \(g\cdot a=\varphi(g)(a)\) for \(\forall g\in G,a\in A\). Hence we know for any group \(G\) and any nonempty set \(A\), there exists a bijection between the actions of \(G\) on \(A\) and the homomorphism of \(G\) into \(S_A\).

由一个 group action，可以对每个 \(g\in G\) 定义一个排列 \(\sigma(g)\)，这所有 \(\sigma(g)\) 又构成了一个同态 \(\sigma:G\to S_A\)，所有的 group action 又可以一一对应到这样的同态。~~套娃说是~~

Prop. Let \(G\) be a group acting on the nonempty set \(A\). The relation on \(A\) defined by \(a\sim b\) iff \(a=g\cdot b\) for some \(g\in G\), is an equivalence relation. For each \(a\in A\), the number of elements in \(\tilde a\) is \(|G:G_a|\), the index of stabilizer of \(a\).

Construct a bijection between the left cosets of \(G_a\) in \(G\) and the elements of \(\tilde a\). Suppose \(b=g\cdot a\), then \(gG_a\) is a left coset of \(G_a\) in \(G\). The map \(b=g\cdot a\mapsto gG_a\) is what we need.

这个定理就是说，一个元素所在的轨道大小等于它的 stabilizer 的 index。

Def. Let \(G\) be a group acting on the nonempty set \(A\).

The equivalence class \(\{g\cdot a\mid g\in G\}\) is called the orbit of \(G\) containing \(a\).
The action of \(G\) on \(A\) is called transitive if there is only one orbit, i.e. given any two elements \(a,b\in A\), \(\exist g\in G\) s.t. \(a=g\cdot b\).

Group acting on themselves by left multiplication and conjugation

Let \(H\le G\) and \(A\) be the set of all left cosets of \(H\) in \(G\). Define an action of \(G\) on \(A\) by \(g\cdot aH=(ga)H\) for \(\forall g\in G, aH\in A\).

Thm. Let \(G\) be a group, \(H\le G\) and \(G\) act by left multiplication on the set \(A\) of left cosets of \(H\) in \(G\). Let \(\pi_H\) be the associated permutation representation afforded by this action. Then

G acts transitively on \(A\).
The stabilizer in \(G\) of the point \(1H\in A\) is the subgroup \(H\).
The kernel of the action (i.e. the kernel of \(\pi_H\)) is \(\cap_{x\in G}xHx^{-1}\), and \(\ker \pi_H\) is the largest normal subgroup of \(G\) contained in \(H\).

前两个结论的证明是显然的，对于第三个：\(\ker \pi_H=\{g\in G\mid gxH=xH, \forall x\in G\}=\{g\in G\mid (x^{-1}gx)H=H, \forall x\in G\}=\{g\in G\mid x^{-1}gx\in H,\forall x\in G\}=\{g\in G\mid g\in xHx^{-1}, \forall x\in G\}=\cap_{x\in G}xHx^{-1}\). 由此还得到 \(\ker\pi_H\trianglelefteq G\) 和 \(\ker\pi_H\le H\). 如果 \(N\) 是 \(G\) 的某个包含于 \(H\) 的正规子集，那么 \(N=xNx^{-1}\le xHx^{-1},\forall x\in G\)，于是 \(N\le \cap_{x\in G}xHx^{-1}=\ker\pi_H\).

Corollary. (Cayley's Theorem) Every group is isomorphic to a subgroup of some symmetric group. If \(G\) is a group of order \(n\), then \(G\) is isomorphic to a subgroup of \(S_n\).

凯莱定理：所有群 \(G\) 同构于在 \(G\) 上的对称群的某个子群。

证明：令 \(H=1\)，然后应用上面的定理，得到一个 \(G\) 到 \(S_G\) 的同态。这个同态的核是 \(1\)（包含于 \(H\)），所以 \(G\) 和它在这个同态下的像同构。

Corollary. If \(G\) is a finite group of order \(n\), and \(p\) is the smallest prime dividing \(|G|\), then any subgroup of index \(p\) is normal.

Suppose \(H\le G\) and \(|G:H|=p\), and let \(\pi_H\) be the permutation representation afforded by multiplication on the set of left cosets of \(H\) in \(G\). Let \(K=\ker \pi_H\) and \(|H:K|=k\), then \(|G:K|=|G:H||H:K|=pk\). Since \(H\) has \(p\) left cosets, \(G/K\) is isomorphic to a subgroup of \(S_p\) (namely, the image of \(G\) under \(\pi_H\)) by the First Isomorphism Theorem.

By Lagrange's Theorem, \(pk=|G/K|\) divides \(p!=|S_p|\). Thus \(k\mid\frac{p!}{p}=(p-1)!\). But all prime divisors of \((p-1)!\) are less than \(p\) and by the minimality of \(p\), every prime divisor of \(k\) is greater than or equal to \(p\). This forces \(k=1\), so \(H=K\trianglelefteq G\).

Now we consider \(G\) acting on itself by conjugation: \(g\cdot a=gag^{-1}\) for \(\forall g\in G,a\in G\).

Def. Two elements \(a\) and \(b\) of \(G\) are said to be conjugate in \(G\) if there is some \(g\in G\) s.t. \(b=gag^{-1}\). The orbits of \(G\) acting on itself by conjugation are called the conjugacy classes of \(G\).

Prop. The number of conjugates of a subset \(S\) in subgroup \(G\) is the index of the normalizer of \(S\), \(|G:N_G(S)|\).

Thm. (The Class Equation) Let \(G\) be a finite group and let \(g_1,\cdots,g_r\) be representations of the distinct conjugacy classes of \(G\) not contained in the center \(Z(G)\) of \(G\). Then \(|G|=|Z(G)|+\sum_{i=1}^r|G:C_G(g_i)|\).

\(Z(G)\) 中的每个元素单独构成一个轨道（共轭类）。

Automorphism

Def. Let \(G\) be a group. An isomorphism from \(G\) to \(G\) itself is called an automorphism. The set of all automorphisms of \(G\) is denoted by \(Aut(G)\). Then \(Aut(G)\) is a group under composition, and \(Aut(G)\le S_G\).

Prop. Let \(H\trianglelefteq G\), then \(G\) acts by conjugation on \(H\) as automorphisms of \(H\). More specifically, the action of \(G\) on \(H\) by conjugation is defined for each \(g\in G\) by

\[h\mapsto ghg^{-1}\text{ for each }h\in H \]

For each \(g\in G\), conjugation by \(g\) is an automorphism of \(H\). The permutation representation afforded by this action is a homomorphism of \(G\) into \(Aut(H)\) with kernel \(C_G(H)\). In particular, \(G/C_G(H)\) is isomorphic to a subgroup of \(Aut(H)\).

Corollary. If \(K\le G\) and \(g\in G\), then \(K\cong gKg^{-1}\).

Let \(G=H\) in the propsition.

Corollary. For \(\forall H\le G\), \(N_G(H)/C_G(H)\) is isomorphic to a subgroup of \(Aut(H)\). In particular, \(G/Z(G)\) is isomorphic to a subgroup of \(Aut(G)\).

Apply the proposition with \(N_G(H)\) playing the role of \(G\).

Def. Let \(G\) be a group and let \(g\in G\). Conjugation by \(g\) is called an inner automorphism of \(G\) and the subgroup of \(Aut(G)\) consisting of all inner automorphisms is denoted by \(Inn(G)\). Then we have \(Inn(G)\cong G/Z(G)\).

Def. A subgroup \(H\) of a group \(G\) is called characteristic in \(G\), denoted \(H\) char \(G\), if every automorphism of \(G\) maps \(H\) to itself, i.e., \(\sigma(H)=H\) for all \(\sigma\in Aut(G)\).

Prop.

If \(H\) char \(G\), then \(H\trianglelefteq G\).
If \(H\) is the unique subgroup of \(G\) of a given order, then \(H\) char \(G\).
If \(K\) char \(H\) and \(H\trianglelefteq G\), then \(K\trianglelefteq G\) (Note "normality" itself is not transitive).

Thus we may think of characteristic subgroups as "strongly normal" subgroups.

Prop. The automorphism group of the cyclic group of order \(n\) is isomorphic to \((\Z/n\Z)^{\times}\), an Abelian group of order \(\varphi(n)\).

Let \(x\) be a generator of the cyclic group \(Z_n\). \(\psi\in Aut(Z_n)\Rightarrow \psi(x)=x^a\). Show that \(\psi\) exists if and only if \((a,n)=1\).

Sylow's Theorem

Def. Let \(G\) be a group and let \(p\) be a prime.

A group of order \(p^\alpha\) for some \(\alpha\ge 1\) is called a \(p\)-group. Subgroups of \(G\) which are \(p\)-groups are called \(p\)-subgroups.
If \(G\) is a group of order \(p^\alpha m\) where \(p\nmid m\), then a subgroup of order \(p^\alpha\) is called a Sylow \(p\)-subgroup of \(G\).
The set of Sylow \(p\)-subgroups of \(G\) will be denoted by \(Syl_p(G)\) and the number of Sylow \(p\)-subgroups of \(G\) will be denoted by \(n_p(G)\) (or just \(n_p\) when \(G\) is clear from the context).

Thm. (Sylow's Theorem) Let \(G\) be a group of order \(p^\alpha m\) where \(p\) is a prime not divding \(m\).

Sylow \(p\)-subgroups of \(G\) exist, i.e., \(Syl_p(G)\neq \emptyset\).
If \(P\) is a Sylow \(p\)-subgroup of \(G\) and \(Q\) is any \(p\)-subgroup \(G\), then there exists \(g\in G\) such that \(Q\le gPg^{-1}\), i.e., \(Q\) is contained in some conjugate of \(P\). In particular, any two Sylow \(p\)-subgroups of \(G\) are conjugate in \(G\).
The number of Sylow \(p\)-subgroups of \(G\) is of the form \(1+kp\), i.e., \(n_p\equiv 1 \pmod p\). Further, \(n_p\) is the index in \(G\) of the normalizer \(N_G(P)\) for any Sylow \(p\)-subgroup \(P\), hence \(n_p\) divids \(m\).

Corollary. Let \(P\) be a Sylow \(p\)-subgroup of \(G\). Then the following are equivalent:

\(P\) is the unique Sylow \(p\)-subgroup of \(G\), i.e. \(n_p=1\).
\(P\) is normal in \(G\).
\(P\) is characteristic in \(G\).
All subgroups generated by elements of \(p\)-power order are \(p\)-groups, i.e., if \(X\) is any subset of \(G\) such that \(|x|\) is a power of \(p\) for all \(x\in X\), then \(\langle X\rangle\) is a \(p\)-group.

(1) \(\Leftrightarrow gPg^{-1}=P,\forall g\in G\Leftrightarrow\) (2)

(3) implies (2) since characteristic subgroups are normal; (1),(2) implies (3) since normal subgroup of unique order is characteristic.

(1) \(\Rightarrow\) (4) since \(x\in gPg^{-1}\) for all \(x\in X\) and some \(g\in G\) by the conjugacy part of Sylow's theorem, and it follows that \(X\subseteq P\Rightarrow \langle X\rangle \le P\Rightarrow\) \(\langle X\rangle\) is a \(p\)-group.

(4) \(\Rightarrow\) (1) by letting \(X\) be the union of all Sylow \(p\)-subgroups of \(G\) and we have \(P\le \langle X\rangle \Rightarrow P=\langle X\rangle\).

Sylow 的定义：
\(p\)-group 指的是阶是 \(p\) 的幂的群；
\(p\)-subgroup 指的是阶是 \(p\) 的幂的子群；
Sylow \(p\)-subgroup 指的是阶是 \(p\) 的最高幂次的子群。
即，如果 \(G\) 的阶是 \(p^\alpha m\)，则 \(G\) 的 Sylow \(p\)-subgroup 的阶是 \(p^\alpha\)。

Sylow 定理是在说：
任何群都存在 Sylow \(p\)-subgroup；
如果 \(P\) 是 Sylow \(p\)-subgroup，\(Q\) 是 \(p\)-subgroup，则 \(Q\) 是 \(P\) 的某个 conjugate 的子群；
Sylow \(p\)-subgroup 的个数模 \(p\) 余 \(1\)，并且等于任何 Sylow \(p\)-subgroup 的 normalizer 的 index，所以是 \(m\) 的因数。

Sylow 定理的重要推论：
\(P\) 是 \(G\) 的唯一 Sylow \(p\)-subgroup，等价于 \(P\trianglelefteq G\)，等价于 \(P\) char \(G\)，等价于所有由“某些阶是 \(p\) 的幂次的元素”生成的子群是 \(p\)-subgroup。

Direct and Semidirect Products and Abelian Groups

Direct product 中文“直积”，顾名思义，是一种非常直接的乘积。

Note that sometimes \(G_i\) means the subgroup \(\{(1,1,\cdots,1,g_i,1,\cdots,1)\mid g_i\in G_i\}\) of \(G=G_1\times G_2\times \cdots \times G_i\times \cdots \times G_n\).

The fundamental theorem of finitely generated Abelian groups

Def. A group \(G\) is finitely generated if there is a finite subset \(A\) of \(G\) such that \(G=\langle A\rangle\) (note that \(G\) may be infinite).

Thm. (Fundamental theorem of finitely generated Abelian groups) Let \(G\) be a finitely generated Abelian group, then

\(G\cong \Z^r\times Z_{n_1}\times Z_{n_2}\times\cdots\times Z_{n_s}\) for some integers \(r,n_1,n_2,\cdots,n_s\) satisfying the following conditions: (a) \(r\ge 0,n_j\ge 2\) (b) \(n_{i+1}\mid n_i\).
The expression in 1. is unique.

Recall that \(Z_n\) is the cyclic group of order \(n\).

Def. The integer \(r\) in the theorem above is called the free rank or Betti number of \(G\) and the integers \(n_1,n_2,\cdots,n_s\) are called the invariant factors of \(G\). The description of \(G\) is called the invariant factor decomposition in \(G\).

By the fundamental theorem, we can calculate the number of finite Abelian groups of a specific order.

Prop. Let \(m,n\in \Z^{+}\).

\(Z_m\times Z_n \cong Z_{mn}\) if and only if \((m,n)=1\).
If \(n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}\) then \(Z_n\cong Z_{p_1^{\alpha_1}}\times Z_{p_2^{\alpha_2}}\times\cdots\times Z_{p_n^{\alpha_n}}\).

Recognizing Direct Products

Def. Let \(G\) be a group, let \(x,y\in G\) and let \(A,B\) be nonempty subsets of \(G\).
(1) Define \([x,y]=x^{-1}y^{-1}xy\), called the commutator of \(x\) and \(y\).
(2) Define \([A,B]=\langle[a,b]\mid a\in A,b\in B\rangle\), the group generated by commutators of elements from \(A\) and from \(B\).
(3) Define \(G'=\langle[x,y]\mid x,y\in G\rangle\), the subgroup of \(G\) generated by commutators of elements from \(G\), called the commutator subgroup of \(G\).

Prop. Let \(G\) be a group, let \(x,y\in G\) and let \(H\le G\). Then
(1) \(xy=yx[x,y]\) (in particular, \(xy=yx\) if and only if \([x,y]=1\)).
(2) \(H\trianglelefteq G\) if and only if \([H,G]\le H\).
(3) \(\sigma[x,y]=[\sigma(x),\sigma(y)]\) for any automorphism \(\sigma\) for \(G\), \(G'\) char \(G\) and \(G/G'\) is Abelian.
(4) \(G/G'\) is the largest Abelian quotient of \(G\) in the sense that if \(H\trianglelefteq G\) and \(G/H\) is Abelian, then \(G'\le H\). Conversely, if \(G'\le H\), then \(H\trianglelefteq G\) and \(G/H\) is Abelian.

Prop. Let \(H\) and \(K\) be subgroups of the group \(G\). The number of distinct ways of writing each element of the set \(HK\) in the form \(hk\), for some \(h\in H\) and \(k\in K\) is \(|H\cap K|\). In particular, if \(H\cap K=1\), then each element of \(HK\) can be written uniquely as a product \(hk\), for some \(h\in H\) and \(k\in K\).

\(x=h_1k_1=h_2k_2\Rightarrow k_1k_2^{-1}=h_1h_2^{-1}=r\in H\cap K\)

\(x=h_2k_2=(h_1r)(r^{-1}k_1)\)，每组 \((h_2,k_2)\) 唯一对应一个 \(r\)

Thm. (Recognition Theorem) Suppose \(G\) is a group with subgroups \(H\) and \(K\) such that \(H\) and \(K\) are normal in \(G\), and \(H\cap K=1\), then \(HK\cong H\times K\).

注意 \(HK=\{hk\mid h\in H,k\in K\}\)，\(H\times K=\{(h,k)\mid h\in H,k\in K\}\)

用上一个命题构造 isomorphism 即可

Semidirect products

Thm. Let \(H\) and \(K\) be groups and let \(\varphi\) be a homomorphism from \(K\) into \(Aut(H)\). Let \(\cdot\) denote the (left) action of \(K\) on \(H\) determined by \(\varphi\). Let \(G\) be the set of ordered pairs \((h,k)\) with \(h\in H\) and \(k\in K\) and define the following multiplication on \(G\):

\[(h_1,k_1)(h_2,k_2)=(h_1~k_1\cdot h_2,k_1k_2) \]

(1) This multiplication makes \(G\) into a group of order \(|G|=|H||K|\).
(2) The sets \(\{(h,1)\mid h\in H\}\) and \(\{(1,k)\mid k\in K\}\) are subgroups of \(G\) and the maps \(h\mapsto (h,1)\) for \(h\in H\) and \(k\mapsto (1,k)\) for \(k\in K\) are isomorphisms of these subgroups with the groups \(H\) and \(K\) respectively:

\[H\cong \{(h,1)\mid h\in H\}\text{ and }K\cong\{(1,k)\mid k\in K\} \]

(3) \(H\trianglelefteq G\)
(4) \(H\cap K = 1\)
(5) for all \(h\in H\) and \(k\in K\), \(khk^{-1}=k\cdot h=\varphi(k)(h)\).

对于 (5)，只是换了一个记号，等价于 \((1,k)(h,1)(1,k^{-1})=(kh,1)\)。

Def. Let \(H\) and \(K\) be the groups and let \(\varphi\) be a homomorphism from \(K\) into \(Aut(H)\). The group described in last Theorem is called the semidirect product of \(H\) and \(K\) with respect to \(\varphi\) and will be denoted by \(H\rtimes_\varphi K\) (when there is no danger of confusion we shall simply write \(H\rtimes K\)).

注意 \(H\times K\) 和 \(H\rtimes K\) 的集合是一样的，但是它们的运算不同，所以作为群不同

Prop. Let \(H\) and \(K\) be groups and let \(\varphi:K\to Aut(H)\) be a homomorphism. Then the following are equivalent:
(1) the identity (set) map between \(H\rtimes K\) and \(H\times K\) is a group homomorphism (hence an isomorphism).
(2) \(\varphi\) is the trivial homomorphism from \(K\) into \(Aut(H)\).
(3) \(K\trianglelefteq H\rtimes K\).

Ring Theory

Introduction to Rings

Def.
(1) A ring \(R\) is a set together with two binary operations \(+\) and \(\times\) (called addition and multiplication) satisfying the following axioms:

\((R,+)\) is an Abelian group.
\(\times\) is associative: \((a\times b)\times c=a\times(b\times c)\) for all \(a,b,c\in R\).
The distributive laws hold in R: for all \(a,b,c\in R\),
\[(a+b)\times c=(a\times c)+(b\times c)\text{ and } a\times (b+c)=(a\times b)+(a\times c) \]
(2) The ring \(R\) is commutative if multiplication is commutative.
(3) The ring \(R\) is said to have an identity (or contain a 1) if there is an element \(1\in R\) with \(1\times a=a\times 1=a\) for all \(a\in R\).

Note that \((R,\times)\) may be not a group, and the identity (for multiplication) may not exist (for example, \(2\mathbb Z\)).

We shall usually write simply \(ab\) rather than \(a\times b\). The additive identity of \(R\) will always be denoted by \(0\), and the additive inverse of the ring element \(a\) will be denoted by \(-a\).

From distributive law we can deduce that \((R,+)\) is Abelian.

Def. A ring \(R\) with identity \(1\), where \(1\neq 0\), is called a division ring (or skew field) if every nonzero element \(a\in R\) has a multiplicative inverse, i.e., there exists \(b\in R\) s.t. \(ab=ba=1\). A commutative division ring is called a field.

Def. Let \(R\) be a ring.
(1) A nonzero element \(a\) of \(R\) is called a zero divisor if there is a nonzero element \(b\in R\) such that either \(ab=0\) or \(ba=0\).
(2) Assume \(R\) has an identity \(1\neq 0\). An element \(u\) of \(R\) is called a unit in \(R\) if there is some \(v\in R\) such that \(uv=vu=1\). The set of units in \(R\) is denoted \(R^{\times}\).

Note that \((R^{\times},\times)\) is a group under multiplication, so it will be referred to as the group of units of \(R\). For a field \(F\), we have \(F^{\times}=F-\{0\}\), and \(F\) contains no zero divisors.

Def. A commutative ring with identity \(1\neq 0\) is called an integral domain if it has no zero divisors.

整环。如果一个交换环不存在两个非零的元素乘积为 \(0\)，则它是一个整数环。

Prop. Assume \(a,b\) and \(c\) are elements of any ring with \(a\) not a zero divisor. If \(ab=ac\), then either \(a=0\) or \(b=c\). In particular, if \(a,b\) and \(c\) are elements of an integral domain and \(ab=ac\), then either \(a=0\) or \(b=c\).

This is cancellation law for rings.

Corollary. Any finite integral domain is a field.

Let \(R\) be a finite integral domain and let \(a\) be a nonzero element of \(R\). By the cancellation law, the map \(x\mapsto ax\) is an injective function. Since \(R\) is finite, this map is also surjective. In particular, there is some \(b\in R\) such that \(ab=1\), i.e. \(a\) is a unit in \(R\). Since \(a\) was an arbitrary nonzero element, \(R\) is a field.

Def. A subring of the ring \(R\) is a subgroup of \(R\) that is closed under multiplication.

Let \(D\) be a squarefree integer. It is immediate from the addition and multiplication that the subset \(\Z(\sqrt D)=\{a+b\sqrt D\mid a,b\in \Z\}\) froms a subring of the quadratic field \(\mathbb Q(\sqrt D)=\{a+b\sqrt D\mid a,b\in \mathbb Q\}\). If \(D\equiv 1\pmod 4\) then the slightly larger subset \(\Z(\frac{1+\sqrt D}{2})=\{a+b\frac{1+\sqrt D}{2}\mid a,b\in \Z\}\) is also a subring.

Polynomial Rings

Let \(x\) be an indeterminate and let \(R\) be a commutative ring with identity. Then \(P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\) with \(n\in \Z,n\ge 0\) annd \(a_i\in R\) is called a polynomial in \(x\) with coefficients \(a_i\) in \(R\). If \(a_n\neq 0\), then the polynomial is said to be of degree \(n\). \(a_nx^n\) is called the leading term, and \(a_n\) is called the leading coefficient. The polynomial is monic if \(a_n=1\). The set of all such polynomials is called the ring of polynomials in variable \(x\) with coefficients in \(R\) (simply, polynomial rings) and will be denoted \(R[x]\).

The ring in which the coefficients are taken makes a substantial difference in the behavior of polynomials. For example, the polynomial \(x^2+1\) is not a perfect square in \(\Z[x]\), but is a perfect square in \(\Z/2\Z[x]\) since \((x+1)^2=x^2+1\).

Prop. Let \(R\) be an integral domain and let \(p(x),q(x)\) be nonzero elements of \(R[x]\). Then
(1) \(\deg p(x)q(x)=\deg p(x)+\deg q(x)\),
(2) the units of \(R[x]\) are just the units of \(R\),
(3) \(R[x]\) is an integral domain.

Group Rings

Fix a commutative ring \(R\) with identity \(1\neq 0\) and let \(G=\{g_1,g_2,\cdots,g_n\}\) be any finite group with group operation written multiplicatively. Define the group ring, \(RG\), of \(G\) with coefficients in \(R\) to be the set of all formal sums \(a_1g_1+a_2g_2+\cdots+a_ng_n\) with \(a_i\in R\) for \(1\le i\le n\).

Addition is defined componentwise, and multiplication is performed by first defining \((ag_i)(bg_j)=(ab)(g_ig_j)\) and then extending to all formal sums by distributive laws. The ring \(RG\) is commutative if and only if \(G\) is a Abelian group.

Ring homomorphisms and quotient rings

Def. Let \(R\) and \(S\) be rings.
(1) A ring homomorphism is a map \(\varphi:R\to S\) satisfying \(\varphi(a+b)=\varphi(a)+\varphi(b)\) and \(\varphi(ab)=\varphi(a)\varphi(b)\) for all \(a,b\in R\).
(2) The kernel of the ring homomorphism \(\varphi\), denoted \(\ker \varphi\), is the set of elements of \(R\) that map to \(0\) in \(S\).
(3) A bijective ring homomorphism is called an isomorphism.

Prop. Let \(R\) and \(S\) be rings and let \(\varphi:R\to S\) be a homomorphism.
(1) The image of \(\varphi\) is a subring of \(S\).
(2) The kernel of \(\varphi\) is a subring of \(R\). Furthermore, is \(\alpha\in \ker\varphi\) then \(r\alpha,\alpha r\in \ker\varphi\) for every \(r\in R\).

Def. Let \(R\) be a ring, let \(I\) be a subset of \(R\) and let \(r\in R\).
(1) \(rI=\{ra\mid a\in I\}\) and \(Ir=\{ar\mid a\in I\}\).
(2) A subset \(I\) of \(R\) is a left ideal of \(R\) if \(I\) is a subring of \(R\) and \(I\) is closed under left multiplication by elements from \(R\), i.e., \(rI\subseteq I\) for all \(r\in R\). Similarly we can define right ideal.
(3) A subset \(I\) that is both a left ideal and a right ideal is called an ideal (or, for added emphasis, a two-sided ideal) of \(R\).

Prop. Let \(R\) be a ring and let \(I\) be an ideal of \(R\). Then the (additive) quotient group \(R/I=\{r+I\mid r\in R\}\) is a ring under the binary operations \((r+I)+(s+I)=(r+s)+I\) and \((r+I)\times (s+I)=(rs)+I\) for all \(r,s\in R\). Conversely, if \(I\) is any subgroup such that the above operations are well defined, then \(I\) is an ideal of \(R\).

这个概念类似于群论中的 normal subgroup。

Def. The ring \(R/I\) with the operations above is called the quotient ring of \(R\) by \(I\).

Thm.
(1) (The First Isomorphism for Rings) If \(\varphi:R\to S\) is a homomorphism of rings, then the kernel of \(\varphi\) is an ideal of \(R\), the image of \(\varphi\) is a subring of \(S\) and \(R/\ker \varphi\) is isomorphic as a ring to \(\varphi(R)\).
(2) If \(I\) is any ideal of \(R\), then the map \(R\to R/I\) defined by \(r\mapsto r+I\) is a surjective ring homomorphism with kernel \(I\) (called the natural projection of \(R\) onto \(R/I\)). Thus every ideal is the kernel of a ring homomorphism and vice versa.

Thm. Let \(R\) be a ring.
(1) (The Second Isomorphism Theorem for Rings) Let \(A\) be a subring and let \(B\) an ideal of \(R\). Then \(A+B=\{a+b\mid a\in A,b\in B\}\) is a subring of \(R\), \(A\cap B\) is an ideal of \(A\) and \((A+B)/B\cong A/(A\cap B)\).
(2) (The Third Isomorphism Theorem for Rings) Let \(I\) and \(J\) be ideals of \(R\) with \(I\subseteq J\). Then \(J/I\) is an ideal of \(R/I\) and \((R/I)/(J/I)\cong R/J\).
(3) (The Fourth or Lattice Isomorphism Theorem for Rings) Let \(I\) be an ideal of \(R\). The correspondence \(A\leftrightarrow A/I\) is an inclusion preserving bijection between the set of subrings \(A\) of \(R\) that contain \(I\) and the set of subrings of \(R/I\). Furthermore, \(A\) (a subring containing \(I\)) is an ideal of \(R\) if and only if \(A/I\) is an ideal of \(R/I\).

Def. Let \(I\) and \(J\) be ideals of \(R\).
(1) Define the sum of \(I\) and \(J\) by \(I+J=\{a+b\mid a\in I,b\in J\}\).
(2) Define the product of \(I\) and \(J\), denoted by \(IJ\), to be the set of all finite sums of elements of the form \(ab\) with \(a\in I\) and \(b\in J\).

It's easy to see that the sum \(I+J\) is the smallest ideal of \(R\) containing both \(I\) and \(J\), and that the product \(IJ\) is an ideal contained in \(I\cap J\).

Properties of ideals

Def. Suppose \(R\) is a ring with \(1\neq 0\). Let \(A\) be any subset of the ring \(R\).
(1) Let \((A)\) denote the smallest ideal of \(R\) containing \(A\), called the ideal generated by \(A\).
(2) Let \(RA=\{r_1a_1+r_2a_2+\cdots+r_na_n\mid r_i\in R,a_i\in A\}\) and similarly \(AR,RAR\).
(3) An ideal generated by a single element is called a principal ideal.
(4) An ideal generated by a finite set is called a finitely generated ideal.

Equivalently, \((A)\) is the intersection of all ideals of \(R\) that contain the set \(A\), i.e. \((A)=\bigcap\limits_{\substack{I\text{ an ideal}\\A\subseteq I}}I\). If \(R\) is commutative, then \(RA=AR=RAR=(A)\).

Prop. Let \(I\) be an ideal of \(R\).
(1) \(I=R\) if and only if \(I\) contains a unit.
(2) Assume \(R\) is commutative, then \(R\) is a field if and only if its only ideals are \(0\) and \(R\).

If \(u\) is a unit in \(I\) with inverse \(v\), then for any \(r\in R\), \(r=r1=r(vu)=(rv)u\in I\).

Corollary. If \(R\) is a field then any nonzero ring homomorphism from \(R\) into another ring is an injection.

Def. An ideal \(M\) in an arbitrary ring \(S\) is called a maximal ideal if \(M\neq S\) and the only ideals containing \(M\) are \(M\) and \(S\).

A general ring need not have maximal ideals, for example, take any Abelian group which has no maximal subgroups (for example, \(\mathbb Q\)) and make it into a trivial ring by define \(ab=0\) for all \(a,b\).

Prop. In a ring with identity, every proper ideal is contained in a maximal ideal.

Prop. Assume \(R\) is commutative. The ideal \(M\) is a maximal ideal iff the quotient ring \(R/M\) is a field.

By Lattice Isomorphism Theorem we know \(M\) is a maximal ideal iff the only ideals of \(R/M\) are \(0\) and \(R/M\). By the prop. above we see that \(M\) is maximal iff \(R/M\) is a field.

Exp. Let \(n\) be a non-negative integer, then \(n\Z\) is an ideal of \(\Z\), and \(n\Z\) is maximal iff \(\Z/n\Z\) is a field, iff \(n\) is a prime number.

Def. Assume \(R\) is commutative. An ideal \(P\) is called a prime ideal if \(P\neq R\) and whenever \(ab\in P\) for \(\forall a,b\in R\), then at least one of \(a\) and \(b\in P\).

Similar to prime in integers: if \(p\mid ab\), then \(p\mid a\) or \(p\mid b\).

Prop. Assume \(R\) is commutative. Then the ideal \(P\) is a prime ideal in \(R\) iff the quotient ring \(R/P\) is an integral domain.

Corollary. Let \(R\) be commutative. Every maximal ideal of \(R\) is a prime ideal.

Rings of Fractions

Throughout this section, \(R\) is a commutative ring.

Thm. Let \(R\) be a commutative ring. Let \(D\) be any nonempty subset of \(R\) that does not contain \(0\) or any zero divisors and is closed under multiplication. Then \(\exists\) a commutative ring \(Q\) with \(1\) such that \(Q\) contains \(R\) as a subring and every element of \(D\) is a unit in \(Q\). The ring \(Q\) has the following additional properties:
(1) every element of \(Q\) is of the form \(rd^{-1}\) for some \(r\in R\) and \(d\in D\). In particular, if \(D=R-\{0\}\) then \(Q\) is a field.
(2) (uniqueness) The ring \(Q\) is the smallest ring containing \(R\) in which all element of \(D\) become units, in the sense that: any ring containing an isomorphic copy of \(R\) in which all the elements of \(D\) become units must also contain an isomorphic copy of \(R\).

Let \(F=\{(r,d)\mid r\in R,d\in D\}\) and define the releation \(\sim\) on \(F\) by \((r,d)\sim (s,e)\) if and only if \(re=sd\). Then \(\sim\) is an equivalence relation. Denote the equivalence class of \((r,d)\) by \(\frac{r}{d}\), i.e., \(\frac{r}{d}=\{(a,b)\mid a\in R,b\in D,rb=ad\}\).

Let \(Q\) be the set of equivalence classes under \(\sim\). Define an additive and multiplicative structure on \(Q\): \(\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\) and \(\frac{a}{b}\times\frac{c}{d}=\frac{ac}{bd}\). It's easy to check that \(Q\) is a commutative ring.

Next we embed \(R\) into \(Q\) by defining \(\iota:R\to Q\) by \(\iota:r\mapsto\frac{rd}{d}\) where \(d\) is any element of \(D\), then \(\iota\) is a ring homomorphism and is injective. Note that each \(d\in D\) has a multiplicative inverse in \(Q\), so if \(D=R-\{0\}\), every element of \(Q\) has a multiplicative inverse and \(Q\) is a field.

To establish the uniqueness property of \(Q\), let \(\varphi:R\to S\) be an injective ring homomorphism such that \(\varphi(d)\) is a unit in \(S\) for all \(D\). Extend \(\varphi\) to a map \(\Phi:Q\to S\) by defining \(\Phi(rd^{-1})=\varphi(r)\varphi(d)^{-1}\) for all \(r\in R,d\in D\), then \(\Phi\) is a ring homomorphism and is injective, and this completes the proof.

Def. Let \(R,D,Q\) be as in last Theorem.
(1) The ring \(Q\) is called the ring of fractions of \(D\) with respect to \(R\) and is denoted \(D^{-1}R\).
(2) If \(R\) is an integral domain and \(D=R-\{0\}\), \(Q\) is called the field of fractions or quotient field of \(R\).

The Chinese Remainder Theorem

Throughout this section, all rings are commutative with \(1\neq 0\).

Def. The ideals \(A\) and \(B\) of the ring \(R\) are said to be comaximal if \(A+B=R\).

Thm. (Chinese Remainder Theorem) Let \(A_1,A_2,\cdots,A_k\) be ideals in \(R\). The map

\[R\to R/A_1\times R/A_2\times \cdots\times R/A_k~~~\text{defined by }~~~r\mapsto (r+A_1,r+A_2,\cdots,r+A_k) \]

is a ring homomorphism with kernel \(A_1\cap A_2\cdots\cap A_k\). If for each \(i,j\in\{1,2,\cdots,k\}\) with \(i\neq j\) the ideals \(A_i\) and \(A_j\) are comaximal, then this map is surjective and \(A_1\cap A_2\cdots\cap A_k=A_1A_2\cdots A_k\), so

\[R/(A_1A_2\cdots A_k)=R/(A_1\cap A_2\cap \cdots\cap A_k)\cong R/A_1\times R/A_2\times \cdots\times R/A_k \]

Corollary. Let \(n\) be a positive integer and let \(p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}\) be its factorization into powers of distinct primes. Then

\[\Z/n\Z \cong (\Z/p_1^{\alpha_1}\Z)\times (\Z/p_2^{\alpha_2}\Z)\times\cdots\times(\Z/p_k^{\alpha_k}\Z) \]

so in particular

\[(\Z/n\Z)^{\times} \cong (\Z/p_1^{\alpha_1}\Z)^{\times}\times (\Z/p_2^{\alpha_2}\Z)^{\times}\times\cdots\times(\Z/p_k^{\alpha_k}\Z)^{\times} \]

This Corollary gives the formula for the Euler \(\varphi\)-function.

Several types of special rings

All rings in this chapter are commutative.

Euclidean Domains

Def. Any function \(N:R\to \Z^{+}\cup \{0\}\) with \(N(0)=0\) is called a norm on the integral domain \(R\). If \(N(a)>0\) for \(a\neq 0\) define \(N\) to be a positive norm.

Def. The integral domain \(R\) is said to be a Euclidean Domain (or possess a Division Algorithm) if there is a norm \(N\) on \(R\) such that for any two elements \(a\) and \(b\) of \(R\) with \(b\neq 0\), there exist elements \(q\) and \(r\) in \(R\) with

\[a=qb+r~~~~\text{and}~~~~r=0~~\text{or}~~N(r)<N(b) \]

The element \(q\) is called the quotient and the element \(r\) the remainder of the division.

The Division Algorithm on an integral domain \(R\) allows a Eucliean Algorithm for two elements \(a\) and \(b\) of \(R\): by successive divisions we can write

\[\begin{aligned} a&=q_0b+r_0\\ b&=q_1r_0+r_1\\ r_0&=q_2r_1+r_2\\ \vdots\\ r_{n-2}&=q_nr_{n-1}+r_n\\ r_{n-1}&=q_{n+1}r_n \end{aligned} \]

where \(r_n\) is the last nonzero remainder.

Exp.
(1) The integer \(\Z\) are a Euclidean Domain with norm given by \(N(a)=|a|\), the usual absolute value.
(2) If \(F\) is a field, then \(F[x]\) is a Euclidean Domain.

Prop. Every ideal in a Euclidean Domain is principal. More precisely, if \(I\) is any nonzero ideal in the Euclidean Domain \(R\) then \(I=(d)\), where \(d\) is any nonzero element of \(I\) of minimum norm.

Recall that an ideal is principal if it can be generated by one single element.

Exp. Let \(R=\Z[x]\). Since the ideal \((2,x)\) is not principal, we know \(\mathbb Z[x]\) is not a Euclidean Domain (for any choice of norm), even though \(\mathbb Q[x]\) is.

Def. Let \(R\) be a commutative ring and let \(a,b\in R\) with \(b\neq 0\).
(1) \(a\) is said to be a multiple of \(b\) if \(\exists x\in R\) s.t. \(a=xb\). In this case \(b\) is said to divide \(a\) or be a divisor of \(a\), written \(b\mid a\).
(2) A greatest common divisor of \(a\) and \(b\) is a nonzero element \(d\) such that (i) \(d\mid a\) and \(d\mid b\), and (ii) if \(d'\mid a\) and \(d'\mid b\) then \(d'\mid d\). A greatest common divisor of \(a\) and \(b\) will be denoted by \(\operatorname{g.c.d.}(a,b)\) or (abusing the notation) simply \((a,b)\).

The defining properties (i) and (ii) translated into language of ideals becomes:

If \(I\) is the ideal of \(R\) generated by \(a\) and \(b\), then \(d\) is a greatest common divisor of \(a\) and \(b\) if (i) \(I\) is contained in \((d)\), and (ii) if \((d')\) is any principal ideal containing \(I\) then \((d)\subseteq (d')\).

Prop. If \(a\) and \(b\) are nonzero elements in the commutative ring \(R\) such that the ideal generated by \(a\) and \(b\) is a principal ideal \((d)\), then \(d=\operatorname{g.c.d.}(a,b)\).

An integral domain which every ideal \((a,b)\) is a principal is called a Bezout Domain.

Prop. Let \(R\) be an integral domain. If \(d\) and \(d'\) generate the same princile ideal, i.e., \((d)=(d')\), then \(d'=ud\) for some unit \(u\in R\). In particular, if \(d\) and \(d'\) are both \(\text{g.c.d.}\) of \(a\) and \(b\), then \(d'=ud\) for some unit \(u\).

Thm. Let \(R\) be a Euclidean Domain and let \(a\) and \(b\) be nonzero elements of \(R\). Let \(d=r_n\) be the last nonzero remainder in the Euclidean Algorithm for \(a\) and \(b\). Then
(1) \(d=\text{g.c.d.}(a,b)\)
(2) \((d)=(a,b)\). In particular, \(d\) can be written as \(R\)-linear combination of \(a\) and \(b\), i.e., \(\exists x,y\in R\) s.t. \(d=ax+by\).

Princial Ideal Domain (P.I.D.)

Def. A Principal Ideal Domain (P.I.D.) is an integral domain in which every ideal is principal.

Exp. \(\mathbb Z\) is P.I.D. but \(\mathbb Z[x]\) is not; \(\mathbb Z[\sqrt{-5}]\) is not P.I.D., in fact the ideal \((3,1+\sqrt{-5})\) is non-principal; a E.D. is a P.I.D..

It's is not true that every P.I.D. is a Euclidean Domain, for example, \(\mathbb Z[(1+\sqrt{-19})/2]\).

Prop. Let \(R\) be a P.I.D., and \(a,b\) be nonzero elements of \(R\). Let \(d\) be a generator for the principal ideal generated by \(a\) and \(b\). Then
(1) \(d\) is \(\text{g.c.d.}\) of \(a\) and \(b\);
(2) \(\exists x,y\in R\) s.t. \(d=ax+by\);
(3) \(d\) is unique up to multiplication by a unit of \(R\).

Prop. Every nonzero prime ideal in a P.I.D. is a maximal ideal.

Corollary. If \(R\) is any commutative ring such that the polynomial ring \(R[x]\) is a P.I.D., then \(R\) is necessarily a field.

Since \(R\) is a subring of \(R[x]\) then \(R\) must be integral domain. \((x)\) is a nonzero prime ideal in \(R[x]\) because \(R[x]/(x)\) is isomorphic to \(R\). So \((x)\) is maximal and hence \(R\) is a field.

Unique Factorization Domain (U.F.D.)

Def. Let \(R\) be an integral domain.
(1) Suppose \(r\in R\) is nonzero and is not a unit. Then \(r\) is called irreducible in \(R\) if whenever \(r=ab\) with \(a,b\in R\), at least one of \(a\) or \(b\) must be a unit in \(R\). Otherwise \(r\) is said to be reducible.
(2) The nonzero element \(p\in R\) is called prime in \(R\) if the ideal \((p)\) generated by \(p\) is a prime ideal. In other words, a nonzero element \(p\) is a prime if it is not a unit whenever \(p\mid ab\) for any \(a,b\in R\), then either \(p\mid a\) or \(p\mid b\).
(3) Two elements \(a\) and \(b\) of \(R\) differing by a unit are said to be associate in \(R\) (i.e., \(a=ub\) for some unit \(u\) in \(R\)).

Prop. In an integral domain a prime element is always irreducible.

It is NOT true in general that an irreducible element is necessarily prime. For example, let \(R=\mathbb Z[\sqrt{-5}]\), then \(3\in R\) is irreducible but not prime since \((2+\sqrt{-5})(2-\sqrt{-5})=3^2\).

Prop. In a P.I.D. a nonzero element is a prime iff it is irreducible.

Def. A Unique Factorization Domain (U.F.D.) is an integral domain \(R\) in which every nonzero element \(r\in R\) which is not a unit obeys
(1) \(r\) can be written as a finite product of irreducibles \(p_i\) of \(R\): \(r=p_1p_2\cdots p_n\) and
(2) the decomposition in (1) is unique up oo associates.

Exp. A field \(F\) is trivially a U.F.D.; \(\mathbb Z[\sqrt{-5}]\) is not a U.F.D. since \(6=2\cdot 3=(1+\sqrt{-5})(1-\sqrt{-5})\).

Prop. In a U.F.D. a nonzero element is a prime iff it is irreducible.

Prop. Let \(a\) and \(b\) be two nonzero elements of the U.F.D \(R\) and suppose

\[a=up_1^{e_1}p_2^{e_2}\cdots p_n^{e_n},~~~b=vp_1^{f_1}p_2^{f_2}\cdots p_n^{f_n} \]

are prime factorizations for \(a\) and \(b\), where \(u\) and \(v\) are units, the primes \(p_1,p_2,\cdots,p_n\) are distinct and the exponents \(e_i\) and \(f_i\) are \(\ge 0\). Then the element

\[d=p_1^{\min(e_1,f_1)}p_2^{\min(e_2,f_2)}\cdots p_n^{\min(e_n,f_n)} \]

(where \(d=1\) if all the exponents are \(0\)) is a \(\text{g.c.d.}\) of \(a\) and \(b\).

Thm. Every P.I.D. is a U.F.D..

Corollary. (Fundamental Theorem of Arithmetic): \(\mathbb Z\) is a U.F.D..

To sum up, we have fields \(\subset\) Euclidean Domain \(\subset\) P.I.D. \(\subset\) U.F.D. \(\subset\) integral domain. All containments are proper.

Polynomial Rings

Recall the definition of polynomial rings.

Prop. Let \(I\) be an ideal of the ring \(R\) and let \((I)=I[x]\) denote the ideal of \(R[x]\) generated by \(I\). Then

\[R[x]/(I)\cong (R/I)[x] \]

In particular, if \(I\) is a prime ideal of \(R\) then \((I)\) is a prime ideal of \(R[x]\).

Def. The polynomial ring in variables \(x_1,x_2,\cdots,x_n\) with coefficients in \(R\), denoted \(R[x_1,x_2,\cdots,x_n]\), is defined inductively by

\[R[x_1,x_2,\cdots,x_n]=R[x_1,x_2,\cdots,x_{n-1}][x_n] \]

i.e., a finite sum of nonzero monomial terms（单项式） \(ax_1^{d_1}x_2^{d_2}\cdots x_n^{d_n}\).

Polynomial Rings over fields

Thm. Let \(F\) be a field. The polynomial ring \(F[x]\) is a Euclidean Domain. Specifically, if \(a(x)\) and \(b(x)\) are two polynomials in \(F[x]\) with \(b(x)\) nonzero, then there are unique \(q(x)\) and \(r(x)\) in \(F[x]\) such that

\[a(x)=q(x)b(x)+r(x)~~~~\text{with}~~ r(x)=0 ~~\text{or}~~ \deg r(x) < \deg b(x) \]

Corollary. If \(F\) is a field, then \(F[x]\) is a E.D..

Polynomials that are U.F.D.

Prop. (Gauss's Lemma) Let \(R\) be a U.F.D. with field of fractions \(F\) and let \(p(x)\in R[x]\). If \(p(x)\) is reducible in \(F[x]\) then \(p(x)\) is reducible in \(R[x]\). More precisely, if \(p(x)=A(x)B(x)\) for some nonconstant polynomials \(A(x),B(x)\in F[x]\), then there are nonzero elements \(r,s\in F\) such that \(rA(x)=a(x)\) and \(sB(x)=b(x)\) both lie in \(R[x]\) and \(p(x)=a(x)b(x)\) is a factorization in \(R[x]\).

Corollary. Let \(R\) be a U.F.D., let \(F\) be its field of fractions and let \(p(x)\in R[x]\). Suppose the \(\text{g.c.d.}\) of the coefficients of \(p(x)\) is \(1\). Then \(p(x)\) is irreducible in \(R[x]\) iff it is irreducible in \(F[x]\). In particular, if \(p(x)\) is a monic（首一的）polynomial that is irreducible in \(R[x]\), then \(p(x)\) is irreducible in \(F[x]\).

Thm. \(R\) is a U.F.D. iff \(R[x]\) is a U.F.D..

Corollary. If \(R\) is a U.F.D. then a polynomial ring in an arbitrary number of variables with coefficients in \(R\) is also a U.F.D..

Irreducibility criteria

Prop. Let \(F\) be a field and let \(p(x)\in F[x]\). Then \(p(x)\) has a factor of degree one if and only if \(p(x)\) has a root in \(F\), i.e., there is an \(\alpha\in F\) with \(p(\alpha)=0\).

Prop. A polynomial of degree two or three over a field \(F\) is reducible if and only if it has a root in \(F\).

Prop. Let \(p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0\) be a polynomial of degree \(n\) with integer coefficients. If \(r/s\in\mathbb Q\) is in lowest terms (i.e., \(r\) and \(s\) are relatively prime integers) and \(r/s\) is a root of \(p(x)\), then \(r\) divides the constant term and \(s\) divides the leading coefficient of \(p(x)\): \(r\mid a_0\) and \(s\mid a_n\). In particular, if \(p(x)\) is a monic polynomial with integer coefficients and \(p(d)\neq 0\) for all integers \(d\) dividing the constant term of \(p(x)\), then \(p(x)\) has no roots in \(\mathbb Q\).

This technique is limited to polynomials of low degree, because it relies on the presence of a factor of degree one.

Prop. Let \(I\) be a proper ideal in the integral domain \(R\) and let \(p(x)\) be a nonconstant monic polynomial in \(R[x]\). If the image of \(p(x)\) in \((R/I)[x]\) cannot be factored in \((R/I)[x]\) into two polynomials of smaller degree, then \(p(x)\) is irreducible in \(R[x]\).

以 \(R=\mathbb Z\)，\(I=(n)\) 为例，定理相当于说，如果 \(p(x)\) 的各项系数 \(\bmod n\) 之后不可约，那么 \(p(x)\) 不可约。

Prop. (Eisenstein's Criterion) Let \(P\) be a prime ideal of the integral domain \(R\) and let \(f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\) be a polynomial in \(R[x]\) (here \(n\ge 1\)). Suppose \(a_{n-1},\cdots,a_1,a_0\) are all elements of \(P\) and suppose \(a_0\) is not an element of \(P^2\), then \(f(x)\) is irreducible in \(R[x]\).

Corollary. (Eisenstein's Criterion for \(\mathbb Z[x]\)) Let \(p\) be a prime in \(\mathbb Z\) and let \(f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in\mathbb Z[x],n\ge 1\). Suppose \(p\) divides \(a_i\) for all \(i\in\{0,1,\cdots,n-1\}\) but \(p^2\) does not divide \(a_0\). Then \(f(x)\) is irreducible in both \(\mathbb Z[x]\) and \(\mathbb Q[x]\).

Exp.
(1) \(x^4+10x+5\) in \(\mathbb Z[x]\) is irreducible.
(2) For prime \(p\), consider \(\Phi_p(x)=\frac{x^p-1}{x-1}=x^{p-1}+\cdots+x+1\). Eisenstein's Criterion does not imeediately apply, but it does apply for the prime \(p\) to the polynomial \(\Phi_p(x+1)\), so \(\Phi_p\) is irreducible.

Polynomial Rings over Fields II

Let \(F\) be a field.

Prop. The maximal ideals in \(F[x]\) are the ideals \((f(x))\) generated by irreducible polynomials \(f(x)\). In particular, \(F[x]/(f(x))\) is a field if and only if \(f(x)\) is irreducible.

Recall that every nonzero prime ideal in a P.I.D. is a maximal ideal.

Prop. Let \(g(x)\) be a nonconstant monic element of \(F[x]\) and let

\[g(x)=f_1(x)^{n_1}f_2(x)^{n_2}\cdots f_k(x)^{n_k} \]

be its factorization into irreducibles, where the \(f_i(x)\) are distinct. Then we have the following isomorphism of rings:

\[F[x]/(g(x))\cong F[x]/(f_1(x)^{n_1})\times F[x]/(f_2(x)^{n_2})\times\cdots\times F[x]/(f_k(x)^{n_k}) \]

Prop. If the polynomial \(f(x)\) has roots \(\alpha_1,\alpha_2,\cdots,\alpha_k\) in \(F\) (not necessarily distinct), then \(f(x)\) has \((x-\alpha_1)\cdots(x-\alpha_k)\) as a factor. In particular, a polynomial of degree \(n\) in one variable over a field \(F\) has at most \(n\) roots in \(F\), even counted with multiplicity.

Prop. A finite subgroup of the multiplicative group of a field is cyclic. In particular, if \(F\) is a finite field, then the multiplicative group \(F^{\times}\) of nonzero elements of \(F\) is a cyclic group.

By the Fundamental Theorem of Finitely Generated Abelian Groups, the finite subgroup can be written as the direct product of cyclic groups \(\mathbb Z/n_1\mathbb Z\times\mathbb Z/n_2\mathbb Z\times \cdots\times\mathbb Z/n_k\mathbb Z\) where \(n_k\mid n_{k-1}\mid\cdots\mid n_2\mid n_1\). In general, if \(G\) is a cyclic group and \(d\mid |G|\), then \(G\) contains precisely \(d\) elements of order dividing \(d\). Since \(n_k\) divides the order of each of the cyclic groups in the direct product, it follows that each direct factor contains \(n_k\) element of order dividing \(n_k\). If \(k\) were greater than \(1\), there would therefore be a total of more than \(n_k\) such elements. But then there would be more than \(n_k\) roots of the polynomial \(x^{n_k}-1\) in the field \(F\), which is a contradiction. Hence \(k=1\) and the group is cyclic.

Corollary. Let \(p\) be a prime, then the multiplicative group \((\mathbb Z/p\mathbb Z)^{\times}\) is cyclic.

Corollary. Let \(n\ge 2\) be an integer with factorization \(n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_r^{\alpha_r}\) in \(\mathbb Z\), where \(p_1,\cdots,p_r\) are distinct primes. We have the following isomorphisms of (multiplicative) groups:
(1) \((\mathbb Z/n\mathbb Z)^{\times}\cong (\mathbb Z/p_1^{\alpha_1}\mathbb Z)^{\times}\times (\mathbb Z/p_2^{\alpha_2}\mathbb Z)^{\times}\times\cdots\times(\mathbb Z/p_r^{\alpha_r}\mathbb Z)^{\times}\)
(2) \((\mathbb Z/2^{\alpha}\mathbb Z)^{\times}\) is the direct product of a cyclic group of order \(2\) and a cyclic group of order \(2^{\alpha-2}\), for all \(\alpha\ge 2\).
(3) \((\mathbb Z/p^{\alpha}\mathbb Z)^{\times}\) is a cyclic group of order \(p^{\alpha-1}(p-1)\), for all odd primes \(p\).

Field Theory

Recall that a field \(F\) is a commutative ring with identity in which every nonzero element has an inverse. \(\def\ch{\text{ch}}\)

Introduction to Field Theory

Basic Theory of field extensions

Def. The characteristic of a field \(F\), denoted \(\ch(F)\), is the smallest positive integer \(p\) such that \(p\cdot 1_F=0\) if such a \(p\) exists and is defined to be \(0\) otherwise.

\(p\cdot 1_F\) 是指 \(p\) 个 \(F\) 中的单位元求和的结果，这里的乘法是数乘。

Prop. \(\ch(F)\) is either \(0\) or a prime number. If \(\ch(F)=p\), then \(p\cdot \alpha=0\) for \(\forall \alpha\in F\).

Remark. This notion of a characteristic makes sense also for any integral domain and its characteristic will be the same as for its field of fractions.

Exp. \(\ch(\mathbb Q)=\ch(\mathbb R)=\ch(\mathbb Z)=0,\ch(F_p=\mathbb Z/p\mathbb Z)=p,\ch(F_p[x])=p\).

Def. The prime subfield of a field \(F\) is the subfield of \(F\) generated by the multiplicative identity \(1_F\) of \(F\). It is isomorphic to either \(\mathbb Q\) (if \(\ch(F)=0\)) or \(\mathbb F_p\) (if \(\ch(F)=p\)).

Exp. The prime subfield of both \(\mathbb Q\) and \(\mathbb R\) is \(\mathbb Q\), while the prime subfield of \(\mathbb F_p[x]\) is isomorphic to \(\mathbb F_p\), given by the constant polynomials.

Def. If \(K\) is a field containing the subfield \(F\), then \(K\) is said to be an extension field (or simply an extension) of \(F\), denoted \(K/F\) or by the diagram \(\small{\begin{matrix}K\\|\\F\end{matrix}}\). In particular, every field is an extension of its prime subfield. The field \(F\) is sometimes called the base field of the extension.

If \(K/F\) is any extension of fields, then the multiplication defined in \(K\) makes \(K\) into a vector space over \(F\). In particular, every field \(F\) can be considered as a vector space over its prime field.

Recall the definition of vector space.

Def. The degree (or relative degree or index) of a field extension \(K/F\), denoted \([K:F]\), is the dimension of \(K\) as a vector space over \(F\) (i.e., \([K:F]=\dim_FK\)). The extension is said to be finite if \([K:F]\) is finite and is said to be infinite otherwise.

Prop. Let \(\varphi:F\to F'\) be a homomorphism of fields. Then \(\varphi\) is either identically \(0\) or is injective, so that the image of \(\varphi\) is either \(0\) or isomorphic to \(F\).

Thm. Let \(F\) be a field and \(p(x)\in F[x]\) be an irreducible polynomial. Then there exists a field \(K\) containing an isomorphic copy of \(F\) in which \(p(x)\) has a root. Identifying \(F\) with this isomorphic copy shows that there exists an extension of \(F\) in which \(p(x)\) has a root.

Thm. Let \(p(x)\in F[x]\) be an irreducible polynomial of degree \(n\) over the field \(F\) and let \(K\) be the field \(F[x]/(p(x))\). Let \(\theta=x\bmod (p(x))\in K\), then the elements \(1,\theta,\theta^2,\cdots,\theta^{n-1}\) are a basis for \(K\) as a vector space over \(F\). So the degree of the extension is \(n\), i.e., \([K:F]=n\). Hence

\[K=\{a_0+a_1\theta+a_2\theta^2+\cdots+a_{n-1}\theta^{n-1}\mid a_0,a_1,\cdots,a_{n-1}\in F\} \]

consists of all polynomials of degree \(<n\) in \(\theta\).

Corollary. Let \(K\) be as in the Theorem, and let \(a(\theta),b(\theta)\in K\) be two polynomials of degree \(<n\) in \(\theta\). Then addition in \(K\) is defined simply by usual polynomial addition and multiplication in \(K\) is defined by \(a(\theta)b(\theta)=r(\theta)\) where \(r(x)\) is the remainder (of degree \(<n\)) obtained after dividing the polynomial \(a(x)b(x)\) by \(p(x)\) in \(F[x]\).

Def. Let \(K\) be an extension of the field \(F\) and let \(\alpha,\beta,\cdots\in K\) be a collection of elements of \(K\). Then the smallest subfield containing both \(F\) and the elements \(\alpha,\beta,\cdots\), denoted \(F(\alpha,\beta,\cdots)\) is called the field generated by \(\alpha,\beta,\cdots\) over \(F\).

Def. If \(K\) is generated by a single element \(\alpha\) over \(F\), \(K=F(\alpha)\), then \(K\) is said to be a simple extension of \(F\) and the element \(\alpha\) is called a primitive element for the extension.

Thm. Let \(F\) be a field and let \(p(x)\in F[x]\) be an irreducible polylnomial. Suppose \(K\) is an extension field of \(F\) containing a root \(\alpha\) of \(p(x)\): \(p(\alpha)=0\). Let \(F(\alpha)\) denote the subfield of \(K\) generated over \(F\) by \(\alpha\), then \(F(\alpha)\cong F[x]/(p(x))\).

Consider the natural homomorphism \(\varphi:F[x]\to F(\alpha)\) defined by \(\varphi:a(x)\mapsto a(\alpha)\). The element \(p(x)\) is in the kernel of \(\varphi\).

Corollary. Suppose in the Thm that \(p(x)\) is of degree \(n\). Then

\[F(\alpha)=\{a_0+a_1\alpha+a_2\alpha^2+\cdots+a_{n-1}\alpha^{n-1}\mid a_0,a_1,\cdots,a_{n-1}\in F\}\subseteq K \]

Thm. Let \(\varphi:F\to F'\) be an isomorphism of fields. Let \(p(x)\in F[x]\) be an irreducible polynomial and let \(p'(x)\in F'[x]\) be the irreducible polynomial obtained by applying the map \(\varphi\) to the coefficients of \(p(x)\). Let \(\alpha\) be a root of \(p(x)\) (in some extension of \(F\)) and let \(\beta\) be a root of \(p'(x)\) (in some extension of \(F'\)). Then there is an isomorphism \(\sigma:F(\alpha)\to F'(\beta)\) mapping \(\alpha\) to \(\beta\) and extending \(\varphi\), i.e., such that \(\sigma\) restrict to \(F\) is the isomorphism \(\varphi\).

Algebraic Extensions

Let \(F\) be a field and let \(K\) be an extension of \(F\).

Def. \(\alpha\in K\) is said to be algebraic over \(F\) if \(\alpha\) is a root of some nonzero polynomial \(f(x)\in F[x]\). Otherwise, \(\alpha\) is transcendental over \(F\). The extension \(K/F\) is said to be algebraic if every element of \(K\) is algebraic over \(F\).

Prop. Let \(\alpha\) be algebraic over \(F\). Then \(\exists\) a unique monic irreducible polynomial \(m_{\alpha,F}(x)\in F[x]\) which has \(\alpha\) as a root. A polynomial \(f(x)\in F[x]\) has \(\alpha\) as a root if and only if \(m_{\alpha,F}(x)\) divides \(f(x)\) in \(F[x]\).

Corollary. If \(L/F\) is an extension of fields and \(\alpha\) is algebraic over both \(F\) and \(L\), then \(m_{\alpha,L}(x)\) divides \(m_{\alpha,F}(x)\) in \(L[x]\).

Def. The polynomial \(m_{\alpha,F}(x)\) (or just \(m_{\alpha}(x)\) if \(F\) is clear) in the Prop is called the minimal polynomial for \(\alpha\) over \(F\). The degree of \(m_{\alpha}(x)\) is called the degree of \(\alpha\).

Prop. Let \(\alpha\) be algebraic over \(F\) and let \(F(\alpha)\) be the field generated by \(\alpha\) over \(F\). Then \(F(\alpha)\cong F[x]/(m_{\alpha}(x))\) so that in particular \([F(\alpha):F]=\deg m_{\alpha}(x)=\deg \alpha\).

Prop. The element \(\alpha\) is algebraic over \(F\) iff the simple extension \(F(\alpha)/F\) is finite.

More precisely, the dimension \([F(\alpha):F]\) is finite.

Corollary. If the extension \(K/F\) is finite, then it is algebraic.

Thm. Let \(F\subseteq K\subseteq L\) be fields. Then \([L:F]=[L:K][K:F]\), i.e. extension degrees are multiplicative, where if one side of the equation is infinite, the other side is also infinite.

Corollary. Suppose \(L/F\) is a finite extension and let \(K\) be any subfield containing \(F\), \(F\subseteq K\subseteq L\). then \([K:F]\) divides \([L:F]\).

Def. An extension \(K/F\) is finitely generated if \(\exists \alpha_1,\cdots,\alpha_k\) such that \(K=F(\alpha_1,\cdots,\alpha_k)\).

Lemma. \(F(\alpha,\beta)=(F(\alpha))(\beta)\).

Thm. The extension \(K/F\) is finite iff \(K\) is generated by a finite number of algebraic elements over \(F\). More precisely, a field generated over \(F\) by a finite number of algebraic elements of degrees \(n_1,n_2,\cdots,n_k\) is algebraic elements of degree \(\le n_1n_2\cdots n_k\).

Corollary. Suppose \(\alpha\) and \(\beta\) are algebraic over \(F\), then \(\alpha\pm \beta,\alpha\beta,\alpha/\beta\) (for \(\beta\neq 0\)) are all algebraic.

Corollary. Let \(L/F\) be an arbitrary extension. Then the collection of elements of \(L\) that are algebraic over \(F\) form a subfield \(K\) of \(L\).

Thm. If \(K\) is algebraic over \(F\) and \(L\) is algebraic over \(K\), then \(L\) is algebraic over \(F\).

Def. Let \(K_1\) and \(K_2\) be two subfields of a field \(K\). Then the composite field of \(K_1\) and \(K_2\), denoted \(K_1K_2\), is the smallest subfield of \(K\) containing both \(K_1\) and \(K_2\). Similarly, the composite of any collection of subfields of \(K\) is the smallest subfield containing all the subfields.

Prop. Let \(K_1\) and \(K_2\) be two finite extension of \(F\) contained in \(K\), then \([K_1K_2:F]\le [K_1:F][K_2:F]\) with equality iff an \(F\)-basis for one the fields remains linearly independent over the other field.

Corollary. Suppose \([K_1:F]=n,[K_2:F]=m\) where \((n,m)=1\), then \([K_1K_2:F]=[K_1:F][K_2:F]=nm\).

Splitting fields and algebraic closures

Def. The extension field \(K\) of \(F\) is called a splitting field for the polynomial \(f(x)\in F[x]\) if \(f(x)\) factors completely into linear factors (or splits completely) in \(K[x]\) and \(f(x)\) does not factor completely into linear factors over any proper subfield of \(K\) containing \(F\).

Thm. For any field \(F\), if \(f(x)\in F[x]\), then there exists an extension \(K\) of \(F\) which is a splitting field for \(f(x)\).

Prove by induction.

Exp. The splitting field for the polynomial \(x^4+4\) over \(\mathbb Q\) is smaller than one might suspect at first. In fact, it has four roots \(\pm1\pm i\), so the splitting field is \(\mathbb Q(i)\), an extension of degree \(2\) of \(\mathbb Q\).

Prop. A splitting field of a polynomial of degree \(n\) over \(F\) is of degree at most \(n!\) over \(F\).

Prove along the induction.

Thm. Let \(\varphi:F\to F'\) be an isomorphism of fields. Let \(f(x)\in F[x]\) be a polynomial and let \(f'(x)\in F'[x]\) be the polynomial obtained by applying \(\varphi\) to the coefficients of \(f(x)\). Let \(E\) be a splitting field for \(f(x)\) over \(F\) and let \(E'\) be a splitting field for \(f'(x)\) over \(F\). Then the isomorphism \(\varphi\) extends to an isomorphism \(\sigma:E\to E'\), i.e., \(\sigma\) restricted to \(F\) is the isomorphism \(\varphi\).

Corollary. Any two splitting fields for a polynomial \(f(x)\in F[x]\) over a field \(F\) are isomophic.

Def. The field \(\bar F\) is called an algebraic closure of \(F\) if \(\bar F\) is algebraic over \(F\) and if every polynomial \(f(x)\in F[x]\) splits completely over \(\bar F\).

Def. A field \(K\) is said to be algebraically closed if every polynomial with coefficients in \(K\) has a root in \(K\).

Prop. Let \(\bar F\) be an algebraic closure of \(F\). Then \(\bar F\) is algebraically closed.

Prop. For any field \(F\) there exists an algebraically closed field \(K\) containing \(F\).

Prop. The field \(\mathbb C\) is algebraically closed.

The fundamental theorem of algebra.

It suffices to prove that every polynomial \(f(x)\in\mathbb R[x]\) has root in \(\mathbb C\). Prove by induction on \(k\), where \(n=2^km\) is the degree. The case \(k=0\) is trivial.

Let \(\alpha_1,\alpha_2,\cdots,\alpha_n\) be the roots of \(f(x)\). For all \(t\in \mathbb R\), consider the polynomial

\[L_t=\prod_{1\le i<j\le n}[x-(\alpha_i+\alpha_j+t\alpha_i\alpha_j)] \]
Note that \(L_t\in \mathbb R[x]\) and has a degree of \(\frac{n(n-1)}{2}=2^{k-1}m(2^{k}m-1)=2^{k-1}m'\), so by induction hypothesis \(L_t\) has a root in \(\mathbb C\).

Since there exists infinitely many choices for \(t\) and only finitely many values of \(i\) and \(j\), we see that for some \(i\) and \(j\) (say \(i=1,j=2\)), there exists distinct real number \(s\) and \(t\) with \(\alpha_1+\alpha_2+s\alpha_1\alpha_2, \alpha_1+\alpha_2+t\alpha_1\alpha_2\in\mathbb C\). It follows that \((s-t)\alpha_1\alpha_2\in\mathbb C\Rightarrow \alpha_1\alpha_2,\alpha_1+\alpha_2\in \mathbb C \Rightarrow \alpha_1,\alpha_2\in\mathbb C\). This completes the induction.

Separable and inseparable extensions

Let \(F\) be a field and let \(f(x)\in F[x]\) be a polynomial. Over a splitting field for \(f(x)\) we have the factorization

\[f(x)=a_n(x-\alpha_1)^{n_1}(x-\alpha_2)^{n_2}\cdots(x-\alpha_k)^{n_k} \]

where \(\alpha_1,\alpha_2,\cdots,\alpha_k\) are distinct elements of the splitting field and \(n_i\ge 1\) for all \(i\). Recall that \(\alpha_i\) is called a multiple root if \(n_i>1\) and is called a simple root if \(n_i=1\). The integer \(n_i\) is called the multiplicity of the root \(\alpha_i\).

Def. A polynomial over \(F\) is called separable is it has no multiple roots. A polynomial which is not separable is called inseparable.

Def. The derivative of the polynomial

\[f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in F[x] \]

is defined to be the polynomial

\[D_xf(x)=na_nx^{n-1}+(n-1)a_{n-1}x^{n-2}+\cdots+2a_2x+a_1\in F[x] \]

Prop. A polynomial \(f(x)\) has a multiple root \(\alpha\) if and only if \(\alpha\) is also a root of \(D_xf(x)\), i.e., \(f(x)\) and \(D_xf(x)\) are both divisible by the minimal polynomial for \(\alpha\). In particular, \(f(x)\) is separable if and only if it is relatively prime to its derivative.

Corollary. Every irreducible polynomial over a field of characteristic \(0\) is separable. A polynomial over such a field is separable if and only if it is the product of distinct irreducible polynomials.

Prop. Let \(F\) be a field of characteristic \(p\). Then for any \(a,b\in F\),

\[(a+b)^p=a^p+b^p~~~\text{and}~~~(ab)^p=a^pb^p \]

Put another way, the \(p^{\text{th}}\)-power map defined by \(\varphi(a)=a^p\) is an injective field homomorphism from \(F\) to \(F\).

Def. The map in this Prop. is called the Frobenius endomorphism of \(F\).

Corollary. Suppose that \(\mathbb F\) is a finite field of characteristic \(p\), then every element of \(\mathbb F\) is a \(p^{\text{th}}\) of \(\mathbb F\) (notationally, \(\mathbb F=\mathbb F^p\)).

Prop. Every irreducible polynomial over a finite field \(\mathbb F\) is separable. A polynomial over such a field is separable if and only if it is the product of distinct irreducible polynomials.

Galois Theory

Basic Definitions

Def. An isomorphism \(\sigma\) of \(K\) with itself is called an automorphism of \(K\). The collection of automorphisms of \(K\) is denoted by \(Aut(K)\).
An automorphism \(\sigma\in Aut(K)\) is said to fix an element \(\alpha\in K\) if \(\sigma(\alpha)=\sigma\alpha=\alpha\). If \(F\) is a subset of \(K\), then an automorphism \(\sigma\) is said to fix \(F\) if it fixes all the elements of \(F\), i.e., \(\sigma(a)=\sigma a=a\) for all \(a\in F\).

The prime field of \(K\) is generated by \(1\in K\), and since any automorphism \(\sigma\) takes \(1\) to \(1\) (and \(0\) to \(0\)), it follows that \(\sigma a=a\) for all \(a\) in the prime field. Hence any automorphism of a field \(K\) fixes its prime field. In particular, \(Aut(\mathbb Q)=Aut(\mathbb F_p)=\{1\}\), where \(1\) is the identity map or the trivial automorphism.

Def. Let \(K/F\) be an extension of fields. Let \(Aut(K/F)\) be the collection of automorphisms of \(K\) which fix \(F\).

If \(F\) is the prime subfield of \(K\), then \(Aut(K)=Aut(K/F)\).

Prop. \(Aut(K)\) is a group under composition and \(Aut(K/F)\) is a subgroup.

Prop. Let \(K/F\) be a field extension and let \(\alpha\in K\) be algebraic over \(F\). Then for any \(\sigma\in Aut(K/F)\), \(\sigma\alpha\) is a root of the minimal polynomial for \(\alpha\) over \(F\), i.e., \(Aut(K/F)\) permutes the roots of irreducible polynomials. Equivalently, any polynomial with coefficients in \(F\) having \(\alpha\) as a root also has \(\sigma\alpha\) as a root.

Exp. Let \(K=\mathbb Q(\sqrt 2)\). If \(\tau\in Aut(\mathbb Q(\sqrt 2))=Aut(\mathbb Q(\sqrt 2)/\mathbb Q)\), then \(\tau(\sqrt2)=\pm \sqrt 2\) since these are the two roots of \(x^2-2\), the minimal polynomial for \(\sqrt 2\). Since \(\tau\) fixes \(\mathbb Q\), this determines \(\tau\) completely: \(\tau(a+b\sqrt 2)=a\pm b\sqrt{2}\).

Prop. Let \(H\le Aut(K)\) be a subgroup of the group of automorphisms of \(K\). Then the collection \(F\) of elements of \(K\) fixed by all the elements of \(H\) is a subfield of \(K\).

Note that it is not important in this proposition that \(H\) actually be a subgroup of \(Aut(K)\).

Def. If \(H\) is a subgroup of \(Aut(K)\), the subfield of \(K\) fixed by all the elements of \(H\) is called the fixed field of \(H\).

Prop. The association of groups to fields and fields to groups defined above is inclusion reversing, namely
(1) If \(F_1\subseteq F_2\subseteq K\) are two subfields of \(K\) then \(Aut(K/F_2)\le Aut(K/F_1)\), and
(2) If \(H_1\le H_2\le Aut(K)\) are two subgroups of automorphisms with associated fixed fields \(F_1\) and \(F_2\), respectively, then \(F_2\subseteq F_1\).

Prop. Let \(E\) be the splitting field over \(F\) of the polynomial \(f(x)\in F[x]\), then \(|Aut(E/F)|\le [E:F]\) with equality if \(f(x)\) is separable over \(F\).

Def. Let \(K/F\) be a finite extension. Then \(K\) is said to be Galois over \(F\) and \(K/F\) is a Galois extension if \(|Aut(K/F)|=[K:F]\). If \(K/F\) is Galois, the group of automorphisms \(Aut(K/F)\) is called the Galois group of \(K/F\), denoted \(Gal(K/F)\).

Corollary. If \(K\) is a splitting field over \(F\) of a separable polynomial \(f(x)\), then \(K/F\) is Galois.

Def. If \(f(x)\) is separable polynomial over \(F\), then the Galois group of \(f(x)\) over \(F\) is the Galois group of the splitting field of \(f(x)\) over \(F\).

Exp. (1) The extension \(\mathbb Q(\sqrt 2)/\mathbb Q\) is Galois with Galois group \(Gal(\mathbb Q(\sqrt 2)/\mathbb Q)=\{1,\sigma\}\cong \mathbb Z/2\mathbb Z\) where \(\sigma\) is the automorphism \(\sigma:a+b\sqrt{2}\mapsto a-b\sqrt{2}\).
(2) The extension \(K=\mathbb Q(\sqrt 2,\sqrt 3)\) is Galois over \(\mathbb Q\) since it is splitting field of the polynomial \((x^2-2)(x^2-3)\). Define the automorphisms \(\sigma\) and \(\tau\) by

\[\sigma:a+b\sqrt2+c\sqrt3+d\sqrt6\to a-b\sqrt2+c\sqrt3-d\sqrt6 \\\tau:a+b\sqrt2+c\sqrt3+d\sqrt6\to a+b\sqrt2-c\sqrt3-d\sqrt6 \]

Then \(Aut(K)=Aut(K/\mathbb Q)=\{1,\sigma,\tau,\sigma\tau\}\).

The fundamental theorem of Galois theory

Def. A character \(\chi\) of group \(G\) with values in a field \(L\) is a homomorphism from \(G\) to the multiplicative group of \(L\):

\[\chi:G\to L^{\times} \]

Def. The characters \(\chi_1,\chi_2,\cdots,\chi_n\) of \(G\) are said to be linearly independent over \(L\) if they are linearly independent as functions of \(G\), i.e., if there is no nontrivial relation

\[a_1\chi_1+a_2\chi_2+\cdots+a_n\chi_n=0~~~~~~(a_1,\cdots,a_n\in L\text{ not all }0) \]

as a function fo \(G\) (that is, \(a_1\chi_1(g)+a_2\chi_2(g)+\cdots+a_n\chi_n(g)=0\) for all \(g\in G\)).

Thm. (Linearly Independence of Characters) If \(\chi_1,\chi_2,\cdots,\chi_n\) are distinct characters of \(G\) with values in \(L\) then they are linearly independent over \(L\).

Suppose \(a_1\chi_1+a_2\chi_2+\cdots+a_m\chi_m=0\) and \(a_1,a_2,\cdots,a_m\neq 0\), where \(m\) is the minimal possible number.

Let \(g_0\) be an element such that \(\chi_1(g_0)\neq \chi_m(g_0)\), then for all \(g\in G\),

\[\begin{aligned} &a_1\chi_1(g)+a_2\chi_2(g)+\cdots+a_m\chi_m(g)=0\\ \Rightarrow &a_1\chi_1(g_0)\chi_1(g)+a_2\chi_2(g_0)\chi_2(g)+\cdots+a_m\chi_m(g_0)\chi_m(g)=0\\ \Rightarrow &(\chi_m(g_0)-\chi_1(g_0))a_1\chi_1(g)+(\chi_m(g_0)-\chi_2(g_0))a_2\chi_2(g)+\cdots =0 \end{aligned} \]
This is a relation with fewer nonzero coefficients, a contradiction.

Thm. Let \(G=\{\sigma_1=1,\sigma_2,\cdots,\sigma_n\}\) be a subgroup of automorphisms of a field \(K\) and let \(F\) be its fixed field. Then \([K:F]=n=|G|\).

Corollary. Let \(K/F\) be any finite extension, then \(|Aut(K/F)|\le [K:F]\) with equaltiy if and only if \(F\) is the fixed field of \(Aut(K/F)\). Put another way, \(K/F\) is Galois if and only if \(F\) is the fixed field of \(Aut(K/F)\).

Corollary. Let \(G\) be a finite subgroup of automorphisms of a field \(K\) and let \(F\) be the fixed field. Then every automorphism of \(K\) fixing \(F\) is contained in \(G\), i.e., \(Aut(K/F)=G\), so that \(K/F\) is Galois, with Galois group \(G\).

Corollary. If \(G_1\neq G_2\) are distinct finite subgroups of automorphisms of a field \(K\) then their fixed fields are also distinct.

Thm. The extension \(K/F\) is Galois if and only if \(K\) is the splitting field of some separable polynomial over \(F\). Furthermore, if this is the case then every irreducible polynomial with coefficients in \(F\) which has a root in \(K\) is separable and has all its roots in \(K\) (so in particular \(K/F\) is a separable extension).

Def. Let \(K/F\) be a Galois extension. If \(\alpha\in K\) the element \(\sigma\alpha\) for \(\sigma\) in \(Gal(K/F)\) are called the conjugates (or Galois conjugates) of \(\alpha\) over \(F\). If \(E\) is a subfield of \(K\) containing \(F\), the field \(\sigma(E)\) is called the conjugate field of \(E\) over \(F\).

Thm. (Fundamental Theorem of Galois Theory) Let \(K/F\) be a Galois extension and set \(G=Gal(K/F)\). Then there is a bijection

\[\left\{\begin{matrix}&K\\\text{subfields }E&|\\\text{of }K&E\\\text{containing }F&|\\&F\end{matrix}\right\}\leftrightarrow\left\{\begin{matrix}&1\\\text{subgroups }H&|\\\text{of }G&H\\&|\\&G\end{matrix}\right\} \]

given by the correspondences

\[E\rightarrow \left\{\begin{matrix}\text{the elements of }G\\\text{fixing }E\end{matrix}\right\}\\ \left\{\begin{matrix}\text{the fixed field}\\\text{of }H\end{matrix}\right\}\leftarrow H \]

which are inverse to each other. Under this correspondence,
(1) (inclusion reversing) If \(E_1,E_2\) correspond to \(H_1,H_2\), respectively, then \(E_1\subseteq E_2\) if and only if \(H_2\le H_1\).
(2) \([K:E]=|H|\) and \([E:F]=|G:H|\), the index of \(H\) in \(G\).
(3) \(K/E\) is always Galois, with Galois group \(Gal(K/E)=H\).
(4) \(E\) is Galois over \(F\) if and only if \(H\) is a normal subgroup in \(G\). If this is the case, then the Galois group is isomorphic to the quotient group \(Gal(E/F)\cong G/H\).
(5) If \(E_1,E_2\) correspond to \(H_1,H_2\), respectively, then \(E_1\cap E_2,E_1E_2\) corresponds to \(H_1\cap H_2,H_1H_2\), respectively. Hence the lattice of subfields of \(K\) containing \(F\) and the lattice of subgroups of \(G\) are "dual".

Prop. Suppose \(K/F\) is a Galois extension and \(F'/F\) is any extension. Then \(KF'/F\) is a Galois extension, with Galois group \(Gal(KF'/F')\cong Gal(K/K\cap F')\) isomorphic to a subgroup of \(Gal(K/F)\).

Galois groups of polynomials

Def. Let \(x_1,x_2,\cdots,x_n\) be indeterminates, the elementary symmetric functions \(s_1,s_2,\cdots,s_n\) are defined by

\[\begin{aligned} s_1=&x_1+x_2+\cdots+x_n\\ s_2=&x_1x_2+x_1x_3+\cdots+x_{n-1}x_n\\ \vdots\\ s_n=&x_1x_2\cdots x_n \end{aligned} \]

i.e. the \(i\)-th symmetric function \(s_i\) of \(x_1,x_2,\cdots,x_n\) is the sum of all products of the \(x_j\)'s taken \(i\) at a time.

It's easy to see by induction that

\[(x-x_1)(x-x_2)\cdots(x-x_n)=x^n-s_1x^{n-1}+s_2x^{n-2}+\cdots+(-1)^{n}s_n \]

Prop. The fixed field of the symmetric group \(S_n\) acting on the field of rational functions in \(n\) variables \(F(x_1,x_2,\cdots,x_n)\) is the field of rational functions in the elementary symmetric functions \(F(s_1,s_2,\cdots,s_n)\).

Corollary. (Fundamental Theorem on Symmetric Functions) Any symmetric function in the variables \(x_1,x_2,\cdots,x_n\) is a rational function in the elementary symmetric functions \(s_1,s_2,\cdots,s_n\).

Thm. The general polynomial

\[x^n-s_1x^{n-1}+s_2x^{n-2}+\cdots+(-1)^{n}s_n \]

over the field \(F(s_1,s_2,\cdots,s_n)\) is separable with Galois group \(S_n\).

Def. Define the discriminant \(D\) of \(x_1,x_2,\cdots,x_n\) by formula \(D=\prod\limits_{i<j}(x_i-x_j)^2\). Define the discriminant of a polynomial to be the discriminant of the roots of the polynomial.

Solvable and Radical Extensions: Insolvability of the Quintic

Def. The extension \(K/F\) is said to be cyclic if it is Galois with a cyclic Galois group.

Prop. Let \(F\) be a field of characteristic not dividing \(n\) which contains the \(n\)-th roots of unity. Then the extension \(F(\sqrt[n]{a})\) for \(a\in F\) is cyclic over \(F\) of degree dividing \(n\).

Def. An element \(\alpha\) which is algebraic over \(F\) can be expressed by radicals or solved for in terms of radicals if \(\alpha\) is an element of a field \(K\) which can be obtained by a succesion of simple radical extensions

\[F=K_0\subseteq K_1\subseteq\cdots\subseteq K_i\subseteq K_{i+1}\subseteq\cdots\subseteq K_s=K \]

where \(K_{i+1}=K_i(\sqrt[n_i]{a_i})\) for some \(a_i\in K_i,i=0,1,\cdots,s-1\). Here \(\sqrt[n_i]{a_i}\) denotes some root of the polynomial \(x^{n_i}-a_i\). Such a field \(K\) will be called a root extension of \(F\). A polynomial \(f(x)\in F[x]\) can be solved by radicals if all its roots can be solved in terms of radicals.

Lemma. If \(\alpha\) is contained in a root extension \(K\) as above, then \(\alpha\) is contained in a root extension which is Galois over \(F\) and where each extension \(K_{i+1}/K_i\) is cyclic.

Thm. The polynomial \(f(x)\) can be solved by radicals if and only if its Galois group is a solvable group.

Corollary. The general equation of degree \(n\) cannot be solved by radicals for \(n\ge 5\).

Recall that for \(n\ge 5\), \(S_n\) is not solvable!

posted @ 2025-06-13 14:44 by_chance 阅读(84) 评论(0) 收藏举报

刷新页面返回顶部

by-chance

Abstract Algebra

Group Theory

Introduction

Dihedral Group

Symmetric Group

The quaternion group

Homomorphisms & Isomorhpisms

Group actions

Subgroups

Cyclic Group

Subgroup generated by subsets

The Lattice of subgroups of a group

Quotient Groups

Multiplication of fibers

Lagrange Theorem

The Isomorphism Theorems

Composition series and the Holder program

Transposition and the alternating group

Group Action

Group acting on themselves by left multiplication and conjugation

Automorphism

Sylow's Theorem

Direct and Semidirect Products and Abelian Groups

The fundamental theorem of finitely generated Abelian groups

Recognizing Direct Products

Semidirect products

Ring Theory

Introduction to Rings

Polynomial Rings

Group Rings

Ring homomorphisms and quotient rings

Properties of ideals

Rings of Fractions

The Chinese Remainder Theorem

Several types of special rings

Euclidean Domains

Princial Ideal Domain (P.I.D.)

Unique Factorization Domain (U.F.D.)

Polynomial Rings

Polynomial Rings over fields

Polynomials that are U.F.D.

Irreducibility criteria

Polynomial Rings over Fields II

Field Theory

Introduction to Field Theory

Basic Theory of field extensions

Algebraic Extensions

Splitting fields and algebraic closures

Separable and inseparable extensions

Galois Theory

Basic Definitions

The fundamental theorem of Galois theory

Galois groups of polynomials

Solvable and Radical Extensions: Insolvability of the Quintic

公告