\(\renewcommand{\Im}{\mathrm{im}}\)
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\(\newcommand{\prop}{\textbf{Proposition:}}\)
\(\newcommand{\lm}{\textbf{Lemma:}}\)
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Algebraic Varieties Seminar
This seminar hold in QZC is about Algebraic varieties, reference is An Invitation to Algebraic Geometry published by Springer.
\(\S1\) Affine Algebraic Varieties
\(\S1.1\) Basic Definition
\(\def\) Let \(S=\{F_i\}_{i\in I}\subseteq \mathbb C[X_1,\cdots,X_n]\) be a class of polynomials of \(n\) variables. $$ V=\mathbb V(S):=\bigcap_{i\in I}\lbrace\boldsymbol x\in\mathbb A^n|F_i(\boldsymbol x)=0,\forall i\in I\rbrace$$ is called an Affine Algebraic Variety.
Where \(\mathbb A^n\) is equal to \(\mathbb C^n\) , but it is no distinguished "origin". It's called an Affine Space.
\(\eg\)
- A hyperspace \(V=\mathbb V(f)\) is an affine algebraic variety.
- \(\mathrm{SL}_n(\mathbb C)\) is a hyperspace, so it's an affine algebraic variety.
\(\prop\) \(\mathbb V(S)=\mathbb V\left((S)\right)\), where \((S)\) is the ideal generated by \(S\).
\(\prop\) A nontrivial affine algebraic variety is closed under standard topology in \(\mathbb C^n\) , and has no interior points.
\(\S1.2\) Zariski Topology
\(\prop\) Let \(\{I_i\}_{i\in J}\) be a class of ideal in \(\mathbb C[X_1,\cdots,X_n]\). Then $$\bigcap\limits_i\mathbb V(I_i)=\mathbb V\left((I_i)_i\right)$$ Where $(I_i)_i $ is the ideal generated by \(\{I_i\}_{i\in J}\).
\(\blacktriangleleft\) Indeed, for \(\boldsymbol x\in\bigcap\limits_i\mathbb V(I_i)\), we have \(f(\boldsymbol x)=0\) for all \(f\in\bigcup\limits_{i\in I}\mathbb V(I_i)\). Then for all \(f\in(I_i)_i\), there exists \(n\in\mathbb N_+,f_1,\cdots,f_n\in\bigcup\limits_{i\in I} I_i,g_1,\cdots,g_n\in\mathbb C[X_1,\cdots,X_n]\) such that$$f=\sum\limits_{i=1}^ng_if_i$$ So \(f(\boldsymbol x)=\sum\limits_{i=1}^ng_i(\boldsymbol x)f_i(\boldsymbol x)=0\). Thus $$\bigcap\limits_i\mathbb V(I_i)\subseteq \mathbb V\left((I_i)_i\right)$$
For the another direction, since \(I_j\subseteq (I_i)_i\) for each \(j\in I\), we have \(\mathbb V(I_j)\supseteq \mathbb V((I_i)_i)\). Then $$\bigcap\limits_i\mathbb V(I_i)= \mathbb V\left((I_i)_i\right)$$
\(\blacktriangleright\)
\(\prop\) Let \(I,J\) be two ideals in \(\mathbb C[X_1,\cdots,X_n]\). Then $$\mathbb V(I)\cup \mathbb V(J)=\mathbb V(IJ)=\mathbb V(I\cap J)$$
\(\blacktriangleleft\) Indeed, since \(I\cap J\subseteq I,J\), we have $ \mathbb V(I)\cup \mathbb V(J)\subseteq \mathbb V(I\cap J)$.
For all \(\boldsymbol x\in \mathbb V(I\cap J)\), we have \(f(\boldsymbol x)=g(\boldsymbol x)=0\) for all \(f\in I,g\in J\). Then for all \(f\in IJ,g\in I,h\in J\), there exists \(n\in\mathbb N_+,f_1,\cdots,f_n\in I, g_1,\cdots,g_n\in J\) such that \(f=\sum\limits_{i=1}^nf_ig_i\). Then $$f(\boldsymbol x)=\sum\limits_{i=1}^nf_i(\boldsymbol x)g_i(\boldsymbol x)=0=g(\boldsymbol x)=h(\boldsymbol x)$$ Thus \(\mathbb V(I\cap J)\subseteq \mathbb V(IJ),\mathbb V(I)\cup V(J)\). And for \(\boldsymbol x\in\mathbb V(IJ)\), for all \(f\in I\cap J\), we have \(f^2\in IJ\), so \(f^2(\boldsymbol x)=0\), thus \(f(\boldsymbol x)=0\), then $$\mathbb V(I)\cup \mathbb V(J)=\mathbb V(IJ)=\mathbb V(I\cap J)$$
\(\blacktriangleright\)
\(\prop\) \(\mathbb V(\{\boldsymbol0\})=\mathbb A^n, \mathbb V(\mathbb C[X_1,\cdots,X_n])=\varnothing\).
\(\def\) The topology on \(\mathbb A^n\) with closed set of all affine algebraic varieties is called Zariski Topology.
\(\eg\) When \(n=1\), we have \(|\mathbb V(f)|\le \deg(f)\) for \(f\neq0\). Then \(X\subseteq \mathbb A^1\) is closed if and only if \(X\) is finite or \(X=\mathbb A^1\).
\(\rmk\) For \(m,n\in\mathbb N_+\), the Zariski topology on \(\mathbb A^{m+n}\) is not the product topology of \(\mathbb A^m\) and \(\mathbb A^n\). For example, when \(m=n=1\), \(\mathbb V(xy)\) is closed in \(\mathbb A^2\). But \(\pi_1^{-1}\left(\mathbb V (xy)^c\right)=\{0\}\) is not open in \(\mathbb A^1\).
\(\prop\) \(U,V\) are open in \(\mathbb A^n\) and nonempty. Then \(U\cap V\neq\varnothing\).
\(\blacktriangleleft\) Indeed, let \(U^c=\mathbb V(I_1),V^c=\mathbb V(I_2)\). Then \(I_1,I_2\neq\{0\}\) , so \(I_1I_2\ne\{0\}\). Thus $$U\cap V=\left(\mathbb V(I_1)\cup \mathbb V(I_2)\right)^c=\mathbb V(I_1I_2)^c\neq\varnothing$$
$\blacktriangleright$
\(\cor\) Zariski topology is not Hausdorff.
\(\S 1.3\) Morphisms
\(\def\) Let \(V\subseteq \mathbb A^m\) and \(W\subseteq \mathbb A^n\) be two affine algebraic varieties. Then a map \(f:V\to W\) is called a morphism if there exists \(g_1,\cdots,g_n\in\mathbb C[X_1,\cdots,X_m]\) such that $$f(x_1,\cdots,x_m)=\left(g_1(x_1,\cdots,x_m),\cdots,g_n(x_1,\cdots,x_m)\right),\forall (x_1,\cdots,x_m)\in V$$
If \(f\) is bijective and its inverse is also a morphism, \(f\) is called an isomorphism, and we say \(V\) is isomorphic to \(W\).
\(\S 1.4\) Dimension
\(\def\) Let \(V\) be an affine algebraic variety. If \(V\) cannot be written as \(V=W\sqcup U(W,U\neq V)\), where \(W,U\) are also affine algebraic varieties, then \(V\) is called irreducible.
\(\def\) For an affine algebraic variety \(V\) and \(x\in V\), define the dimension of \(V\) is $$\dim(V)=\sup\lbrace L\in\mathbb N_+|\exists\text{a chain of irreducible subvarieties }V_L\supset V_{L-1}\supset\cdots\supset V_0 \rbrace$$
and the dimension of \(V\) near \(x\) is $$\dim(V)=\sup\left\lbrace L\in\mathbb N_+|\exists\text{a chain of irreducible subvarieties }V_L\supset V_{L-1}\supset\cdots\supset V_0=\lbrace x\rbrace \right\rbrace$$
We'll prove dimension is always finite later.
\(\S2\) Algebraic Foundations
\(\S2.1\) Noetherian Ring & Modules
\(\prop\) Let \((\Sigma,\le)\) be a partial order set, then the followings are equivalent:
- Every increasing sequence \(x_1\le x_2\le\cdots\) in \(\Sigma\) is stabling.
- EverWy non-empty subset of \(\Sigma\) has a maximum.
If \(\Sigma\) satisfies them, then we say \(\Sigma\) satisfies ascending chain condition( a.c.c.).
\(\def\) A module \(M\) is called Noetherian if \((\Sigma,\subseteq)\) satisfies a.c.c., where \(\Sigma\) is the set of submodules of \(M\).
\(\prop\) \(M\) is Noetherian if and only if every submodule of \(M\) is finite generated.
\(\blacktriangleleft\) Indeed, for necessity, if \(M\) is Noetherian, let \(N\) be a submodule of \(M\), let \(\Sigma\) be the set of all finite generated submodules of \(M\). Then since \(M\) is Noetherian, \(\Sigma\) has a maximum \(N_0\). Then if \(N\neq N_0\), let \(x\in N\backslash N_0\), the module generated by \(N_0\) and \(x\) is also finite generated, lead to contradiction! So \(N\) is finite genrated.
For sufficiency, if all submodules of \(M\) is finite generated, let \(M_0\subseteq M_1\subseteq\cdots\) is an increasing sequence. Let \(N=\bigcup\limits_{n=0}^\infty M_n=\left\langle f_1,\cdots,f_r\right\rangle\). Then there exists \(n\in\mathbb N\) such that \(f_1,\cdots,f_r\in M_n\). Thus \(N=M_n\).
$\blacktriangleright$
\(\prop\) If there's a \(A\)-Module short exact sequence
(it means, \(\alpha\) is injective and \(\beta\) is surjective, \(\Im(\alpha)=\ker(\beta)\). So we consider \(M^\prime\) as submodule of \(M\), then \(M/M^\prime\cong M^{\prime\prime}\). )
Then \(M\) is Noetherian if and only if \(M^\prime\) and \(M^{\prime\prime}\) are Noetherian.
\(\blacktriangleleft\) Indeed, if \(M\) is Noetherian, for \(M_0\subseteq M_1\subseteq \cdots\), a sequence of submodules of \(M^\prime\), we have \(\alpha(M_0)\subseteq \alpha(M_1)\subseteq\cdots\) is a sequence of submodules of \(M\). Then there exists \(N\in\mathbb N_+,\forall n>N\), \(\alpha(M_n)\) are the same. Since \(\alpha^{-1}\left(\alpha(M_n)\right)=M_n\), \(M_n(n>N)\) are the same, so \(M^\prime\) is Noetherian. Similarly, since \(\beta\left(\beta^{-1}(N)\right)=N\) for \(N\le M^{\prime\prime}\), \(M^{\prime\prime}\) is also Noetherian.
If \(M^\prime\) and \(M^{\prime\prime}\) is Noetherian, then assume \(M_0\subseteq M_1\subseteq \cdots\) is a increasing sequence of submodules of \(M\). Then there exists \(N\in\mathbb N_+,\forall n>N\), \(\alpha^{-1}(M_n)=M^\prime_{\infty},\beta(M_n)=M^{\prime\prime}_\infty\). We only need a lemma:
\(\lm\) Let \(X\le Y\le M\), \(\beta(X)=\beta(Y)\) and \(X\cap \Im\alpha=Y\cap\Im\alpha\), then \(X=Y\).
\(\blacktriangleleft\) Indeed, let \(m\in Y\), then there exists \(n\in X\) such that \(\beta(m)=\beta(n)\). Then \(m-n\in\ker\beta=\Im \alpha\). But \(m-n\in X\), so \(m-n\in X\cap\Im\alpha=Y\cap\Im\alpha\subseteq Y\), thus \(m-n\in Y\Longrightarrow m\in X\). So \(X=Y\).
\(\blacktriangleright\)
So, \(M_n\) are the same for \(n> N\). Thus \(M\) is Noetherian.
$\blacktriangleright$
\(\cor\) \(M_1,\cdots, M_n\) are Noetherian Modules, then so is \(\mathop{\oplus}\limits_{i=1}^n M_i\).
\(\blacktriangleleft\) Indeed, we have short exact sequence $$0\rightarrow M_n\rightarrow \mathop{\oplus}\limits_{i=1}^{n} M_i\rightarrow \mathop{\oplus}\limits_{i=1}^{n-1} M_i\rightarrow0$$
So \(\mathop{\oplus}\limits_{i=1}^n M_i\) is Noetherian by induction and the proposition.
\(\blacktriangleright\)
\(\def\) If \(A\) is a ring, and \(A\) is Noetherian module as \(A\)-Module, then we say \(A\) is a Noetherian ring.
\(\prop\) \(A\) is a Noetherian ring, \(M\) is a finite generated \(A\)-module, then \(M\) is Noetherian.
\(\blacktriangleleft\) Indeed, assume \(M\) has \(n\) generators \(f_1,\cdots,f_n\), \(N=\lbrace(r_1,\cdots,r_n)\in A^n|r_1f_1+\cdots+r_nf_n=0\rbrace\). Then \(N,A^n\) are Noetherian, and \(M\sim A^n/ N\) , so we have short exact sequence $$0\to N\to A^n\to M\to0$$ Thus \(M\) is Noetherian.
$\blacktriangleright$
\(\S2.2\) Hilbert Basis Theorem
\(\thm\)(Hilbert basis Theorem) If \(A\) is a Noetherian ring, then so do \(A[X]\).
\(\blacktriangleleft\) Indeed, consider \(\mathfrak a\) be a ideal of \(A[X]\). Let \(I\) be the set of leading coefficient of elements in \(\mathfrak a\). Then \(I\) is an ideal of \(A\). Thus since \(A\) is Noetherian, let \(I=(a_1,\cdots, a_m)\). Then for exists \(f_i\in \mathfrak a,f_i=a_ix^{r_i}+o(x^{r_i})\).
Let \(r=\max\limits_{1\le i\le m}r_i,M=<1,\cdots,x^r>\), then \(\mathfrak a=(\mathfrak a\cap M)+(f_1,\cdots,f_m)\). We only need to prove \(N:=\mathfrak a\cap M\) is finite generated.
Prove by induction. When \(r=0\), \(N\) is an ideal of \(A\), then \(N\) is finite generated. Assume \(<r\) is proved, for \(r\), if \(N\) does not contain any polynomials of degree \(r\), then proved; if not, let \(J\) be the set of the coefficient of \(x^r\) of elements in \(N\). Then \(J\) is an ideal of \(N\), let \(N=(b_1,\cdots,b_k)\). Thus similarly, \(N=(N\cap<1,\cdots,x^{r-1}>)+(g_1,\cdots,g_k)\), where \(g_i=b_ix^r+o(x^r)\in N\). So by induction, \(N\) is finite generated.
\(\blacktriangleright\)
\(\rmk\) By induction, we can prove \(A[X_1,\cdots,X_N]\) is also Noetherian. And if we consider the coefficient of the smallest \(x^r\), we can also prove \(A[[X]]\) is Noetherian.
\(\S2.3\) Integral extension and Hilbert's Nullstellensatz
\(\def\) For an affine algebraic variety \(V\), define $$\mathbb I(V)=\lbrace f\in\mathbb C[X_1,\cdots,X_N]|f(\boldsymbol x)=0,\forall f\in V\rbrace$$ Then \(\mathbb I(V)\) is an ideal of \(\mathbb C[X_1,\cdots,X_N]\) contained \(I\), where \(V=\mathbb V(I)\).
\(\prop\) \(\mathbb V(\mathbb I(V))=V\) for any affine algebraic variety.
Then, since \(\mathbb C[X_1,\cdots,X_n]\) is Noetherian, we have:
\(\prop\) For affine algebraic variety \(V\), there exists \(F_1,\cdots,F_r\in\mathbb C[X_1,\cdots,X_r]\) such that \(V=\mathbb V(F_1,\cdots,F_r)\).
\(\def\) Let \(\varphi:A\to B\) is a ring morphism. We say \(b\in B\) is integral over \(A\) if there exists \(a_0,\cdots, a_{n-1}\in A\), such that \(b^n+a_{n-1}\cdot b^{n-1}+\cdots+a_0=0\), where \(a\cdot b\) is defined as \(\varphi(a)\cdot b\).
\(\def\) \(\varphi:A\to B\) is called an integral extension, if for all \(b\in B\), \(b\) is integral over \(B\).
\(\lm\) If \(\varphi:A\hookrightarrow B\) be an injective morphism from domain \(A\) to domain \(B\), and it's an integral extension, then \(A\) is a field if and only if \(B\) is a field.
\(\blacktriangleleft\) Indeed, if \(A\) is a field, then for all \(b\in B\backslash\{0\}\), there exists \(a_0,\cdots,a_{n-1}\in A\) such that \(b^n+a_{n-1}b^{n-1}+\cdots+a_0=0\), and \(n\) is minimal. Then \(a_0\neq0\), thus
If \(B\) is a field, then for all \(a\in A\backslash \{0\}\), there exists \(a_0,\cdots, a_{n-1}\in A\) such taht \(a^{-n}+a_{n-1}a^{1-n}+\cdots+a_0=0\). So
$\blacktriangleright$
\(\rmk\) Assume \(\varphi:A\hookrightarrow B\) be an injective morphism from domain \(A\) to domain \(B\), and it's an integral extension. Then for all \(Q\in\text{spec } B\) (that means, \(B\) is a prime adeal), let \(P=\varphi^{-1}(Q)\), then \(P\) is the maximal ideal of \(A\) if and only if \(Q\) is the maximal ideal of \(B\).
\(\thm\)( The weak form of Hilbert's Nullstellensatz) \(\mathbb K\) is an algebraic closed field, \(\mathfrak m\) is the maximal ideal of \(\mathbb K[X_1,\cdots,X_n]\). Then there exists \(a_1,\cdots,a_n\in\mathbb K\) such that $$\mathfrak m=(X-a_1,\cdots,X_n-a_n)$$
\(\blacktriangleleft\) Indeed, we can prove it by induction. When \(n=1\), since \(\mathbb K[X]\) is PID and \(\mathbb K\) is algebraic closed, it's true.
Assume the work for \(<n\), for \(n\), let \(0\ne f\in\mathfrak m\). We can change the variables by
\(\left\lbrace\begin{array} \ X_1&=&Y_1+Y_n^{N_1}\\ X_2&=&Y_2+Y_n^{N_2}\\ \vdots &=&\vdots\\X_{n-1}&=&Y_{n-1}+Y_n^{N_{n-1}}\\X_n&=&Y_n\end{array}\right.\)
We can choose \(N_1,\cdots,N_{n-1}\in\mathbb N\) such that \(f\) is a monic polynomial of \(Y_n\) in \(\mathbb K[Y_1,\cdots,Y_{n-1}][Y_n]\).
Consider the nature map
Then it's an integral extension.(WHY?) \(\overline{\mathfrak m}=\mathfrak m+(f)\) is a maximal ideal, consider \(\mathfrak n\), the preimage of \(\overline{\mathfrak m}\). Then \(\mathfrak n\) is also maximal.
\(\S2.4\) Coordinate Ring
$\S2.5 $ Equivalence of Algebra and Geometry
\(\S2.6\) Spectrum of Ring
\(\S3\) Projective Variety
$\S3.1 $ Projective Space
For the \(n\)-dimensional projective space \(\mathbb P^n\), we have the following definition:
\(\def\) \(\mathbb P^n=\mathbb CP^n\) is defined as all \(1-\)dimensional subspaces of \(\mathbb C^{n+1}\). That is, \(\mathbb P^n=\frac{\mathbb C^{n+1}\backslash\lbrace0\rbrace}{\sim}\), where \(\boldsymbol x\sim \boldsymbol y\) if and only if \(\exists \lambda\ne0, \boldsymbol x=\lambda \boldsymbol y\).
Let \([z_0:z_1:\cdots:z_n]\) be the equivalence class of \((z_0,\cdots,z_n)\), it's called the homogenous coordinate of \((z_0,\cdots,z_n)\).
Let \(U_i=\{[z_0:z_1:\cdots:z_n]|z_i\neq0\}\), then we have a bijective from \(U_i\) to \(\mathbb A^n\):
Thus we get a injection \(\mathbb A^n\hookrightarrow \mathbb P^n\).
\(\mathbb P^n=U_0\cup(\mathbb P^n\backslash U_0)\) can be consider as \(\mathbb P^n=\mathbb A^n\cup\mathbb P^{n-1}\), thus we have
In fact, \(\mathbb P^n\) is the compactification of \(\mathbb A^n\) under quotient space topology. Indeed, \(\mathbb A^n\cong U_0\) is dense in \(\mathbb P^n\), and \(\mathbb P^n\) is compact.(WHY?)
\(\newcommand{\aff}{\text{aff}}\)
\(\newcommand{\proj}{\text{proj}}\)
\(\S3.2\) Projective Variety
\(\def\) A projective variety in \(\mathbb P^n\) is the common root set of an arbitrary collection of homogenous polynomials in \(\mathbb C[X_0,\cdots,X_n]\).
Since \(\mathbb C[X_0,\cdots,X_n]\) is Noetherian, it has the form
The affine cone of \(V\) is defined as \(\mathbb V_\aff\left(\{F_i\}_{1\le i\le m}\right)\), and for each \(0\le i\le n\), \(U_i\cap V\) can be considered as an affine variety.
And we define \(\mathbb I(V):=\{F\in\mathbb C[X_0,\cdots,X_n]|F(V)=0\}\), then \(\mathbb I(\mathbb V_\proj(I))=\mathbb I(\mathbb V_\aff(I))=\sqrt I\) by Hilbert's nullstellensatz.
\(\S3.3\) Projective Closure
\(\def\) \(f\in\mathbb C[X_1,\cdots,X_n]\) with degree \(d\), define its homogenization
For an ideal \(I\) in \(\mathbb C[X_1,\cdots,X_n]\), define its homogenization
Let \(V=\mathbb V_\proj(I)\) is a projective variety, define its projective closure
It's also the closure of \(V\) under both Zariski topology and Euclidean topology(WHY?).
\(\S 3.4\) Morphisms
$\S3.5 $ Automorphisms
\(\S 4\) Quasi-Projective Varieties
\(\S4.1\) Quasi-Projective Varieties
\(\def\) A quasi-projective variety in \(\mathbb P^n\) is a locally closed set in \(\mathbb P^n\) under Zariski topology.( Equivalent to an intersection of an open subset and a closed subset.)
\(\eg\) All projective/affine varieties, all Zariski-open subsets of them are quasi-projective variety.
Now we want to define the morphisms between quasi-projective varieties.
\(\def\) A quasi-projective variety is called affine if it's isomorphic to an affine algebraic variety as projective varieties. The origin affine algebraic varieties is called Zariski closed.
\(\eg\) \(\mathbb A^{1}\backslash\{0\}\) is affine since it's isomorphic to \(\mathbb V(xy-1)\).
\(\def\) Let \(V\subseteq \mathbb P^n\) and \(W\subseteq \mathbb P^m\) be quasi-projective varieties. \(F:V\to W\) is called a morphism if for all \(p\in V\), there exists homogenous \(F_0,\cdots,F_m\) such that \(V\to\mathbb P^m\), \(q\mapsto[F_0(q):\cdots:F_m(q)]\) is well-defined on a neighborhood of \(p\) and equal to \(F(p)\) at \(p\).
\(\def\) Let \(W\) be an affine variety. Assume \(W\xrightarrow{F}V\) is an isomorphism to a Zariski-closed set \(V\). Then the coordinate ring of \(W\) is defined as
That is, the pull backs of \(\mathbb C[V]\).
\(\rmk\) This definition is not depend on the choice of \(V\). We need to show if two Zariski-closed set is isomorphic as quasi-projective varieties, then so are they as Zariski-closed sets. See at 4.3.
\(\eg\) \(\mathbb C[\mathbb A^1\backslash\{0\}]\cong\mathbb C\left[x,\frac1x\right]\).
\(\S4.2\) A basis for Zariski Topology
\(\lm\) If \(V\) is a Zariski-closed subset, \(f\in\mathbb C[V]\), then \(U=V\backslash\mathbb V(f)\) is an affine variety.
\(\blacktriangleleft\) Indeed, consider \(U\) as subset of \(\mathbb A^n\), consider the map \(F:U\to\mathbb A^{n+1}\),
This map is well-defined. Assume \(V=\mathbb V(F_1,\cdots,F_r)\), then \(F\) is an injective morphism, and \(F(U)=\mathbb V(F_1,\cdots,F_r,x_{n+1}f(x_1,\cdots,x_n)-1)\), so \(U\) is affine.
$\blacktriangleright$
\(\rmk\) Open sets are not always affine. For example, \(\mathbb A^2\backslash\{0\}\) is not affine( we cannot prove it now).
\(\prop\) Affine sets form a basis of quasi-projective varieties set.
\(\blacktriangleleft\) Indeed, for a quasi-projective variety \(V\), we have \(V=(V\cap U_0)\cup\cdots\cup(V\cap U_n)\).
Since \(V\cap U_i\) is locally compact, it's the open subset of closed subset. That is,
So \(V\) can be covered by affine open sets.
$\blacktriangleright$
\(\S4.3\) Regular Functions
\(\def\) \(U\) is an open subset of affine variety \(V\). \(f:U\to\mathbb C\) is called regular at \(p\in U\) if \(\exists g,h\in\mathbb C[V]\), such that \(h(p)\neq0\), and \(f=\frac gh\) in a neighborhood of \(p\). If \(f\) is regular at every point in \(U\), then \(f\) is called regular on \(U\). The set of regular function on \(U\) is denoted by \(\mathcal O_V(U)\).
\(\eg\) \(\mathbb A^2\backslash\mathbb V(x)\to\mathbb C,(x,y)\to\frac yx\) is regular.
\(\thm\) Let \(V\) be an irreducible Zariski closed set of \(\mathbb A^n\), then every regular function \(f:V\to\mathbb C\) can be expressed as \(g|V\), where \(g:\mathbb A^n\to\mathbb C\) is a polynomial. That is, \(\mathcal O_V(V)=\mathbb C[V]\).
\(\blacktriangleleft\) Indeed, \(\mathbb C[V]\subseteq \mathcal O_V(V)\), so we only need to show \(\mathcal O_V(V)\subseteq \mathbb C[V]\).
Let \(f:V\to\mathbb C\) regular, then for each \(p\in V\), there exists open \(U_p\supseteq p\) such that \(f|_{U_p}\) is a rational function. Since \(V\) is compact, there exists \(p_1,\cdots,p_n\in V\), and \(U_i=U_{p_i}\), such that \(\bigcup\limits_{i=1}^nU_i=V\). Assume \(f|_{U_i}=\frac{h_i}{k_i}\). Since affine sets are a basis, we can assume \(U_i=V\backslash\mathbb V(F_i)\), where \(F_i\) are polynomials.
Since \(V=\bigcup\limits_{i=1}^n U_i\), \(k_i\) will not all simultaneously vanish on \(V\), thus \(\mathbb V(k_1,\cdots,k_n)=\varnothing\), so if \(I=(k_1,\cdots,k_n)\), then \(\sqrt I=\mathbb I(\varnothing)=(1)\), so \(I=(1)\), thus there exists \(l_1,\cdots,l_i\in\mathbb C[V]\) such taht \(1=\sum\limits_{i=1}^nl_ik_i\), so
on \(U_j\). Now we'll prove \(\frac{h_i}{k_i}\) are the same for \(1\le i\le n\). For \(1\le i<j\le n\), \(U_i\cap U_j=V\backslash(\mathbb V(F_i)\cup\mathbb V(F_j))=V\backslash\mathbb V(F_iF_j)\) is dense in \(V\), ( since \(V\) is irreducible) so \(\frac{h_i}{k_i}\) and \(\frac{h_j}{k_j}\) must be the same.
So \(f=\sum\limits_{i=1}^nl_ih_i\in\mathbb C[V]\).
\(\blacktriangleright\)
\(\rmk\) If \(V\) is not irreducible, this theorem is also right. See GTM 52.
Now we'll define regular function on quasi-projective varieties.
\(\def\) Let \(U\) be an open subset of a quasi-projective variety \(V\). \(f:U\to\mathbb C\) is called regular at p if there exists an affine open set containing \(p\) such that \(f\) is regular at \(p\). If \(f\) is regular at all \(p\in U\), then we say \(f\) is regular on \(U\), all of these functions is denoted by \(\mathcal O_V(U)\).
\(\S5\) Classical Constructions
\(\S5.1\) Veronese Maps
We want to construct a non-trivial embed from \(\mathbb P^n\) to \(\mathbb P^m\).
\(\def\) The \(d\)-th Veronese map of \(\mathbb P^n\) is the morphism \(\nu_d:\mathbb P^n\to\mathbb P^m\),
Where it runs over all monomials of degree \(d\) in \(\mathbb C[X_0,\cdots,X_n]\), \(m=\binom{n+d}{d}-1\). It's easy to check \(\nu_d\) is well-defined.
\(\prop\) Image of \(\nu_d\) can be embedding to an algebraic variety of \(\mathbb P^m\). That is, \(\nu_d(\mathbb P^n)\) is an algebraic variety and \(\nu_d:\mathbb P^n\to\nu_d(\mathbb P^n)\) is isomorphism.
\(\blacktriangleleft\) Indeed, let \(W=\nu_d(\mathbb P^n)\). For \(I=(i_0,\cdots,i_n)\in\mathbb N^{n+1}\) such that \(i_0+\cdots+i_n=d\), let \(z_I=x_0^{i_0}\cdots x_n^{i_n}\).
For \(0\le i\le n\), we define morphism \(\varphi_i:W\supseteq U_i\to\mathbb P^n\), \([x_0^d:\cdots:x_n^d]\mapsto [x_0x_i^{d-1}:\cdots:x_0x_i^{d-1}]\), where \(U_i\) is the set such that the \(x_i^d\ne0\).
And for \(i,j\), \(\varphi_i,\varphi_j\) are the same on \(U_i\cap U_j\). Thus we have \(\varphi:W\to\mathbb P^n\) such that \(\varphi|_{U_i}=\varphi\). So since \(U_0\cup\cdots\cup U_n=W\), \(\varphi\) is the inverse of \(\nu _d\).
Now we show \(W\) is an algebraic variety. For \(I\in\mathbb N^{n+1}\) such that \(i_0+\cdots+i_n=d\), let \(z_I\in\mathbb C[X_0,\cdots,X_n]\). Then \(W=\mathbb V(X)\), where
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\(\S5.2\) Five Points Determine a Conic
This is an example of enumerative algebraic geometry.
\(\thm\) Each five points in \(\mathbb P^2\) can determine a conic, and it's unique unless four of them are collinear, conic is not degenerate unless three of points are collinear.
\(\blacktriangleleft\) Indeed, a conic in \(\mathbb P^2\) with vertex of origin is \(\mathbb V(ax^2+by^2+cz^2+dxy+eyz+fxz)\). It has an 1-1 map to \([a:b:c:d:e:f]\in\mathbb P^5\).
Let \(P_1,\cdots,P_5\in\mathbb P^5\), \(H_i\) is set of conics contain \(P_i\). Then \(H_i\) can be considered as varieties in \(\mathbb P^5\).
Thus since \(\dim H_i=5\), \(\dim\cap H_i=1\) so it contains exactly one point, or linear depend so the intersection do not change.
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