Jacoboson 4.1 Work Problem

4.1 Work Problem

1.1 Let \(E=\mathbb{Q}(u)\) where \(u^3-u^2+u+2=0\). Express \((u^2+u+1)(u^2-u)\) and \((u-1)^{-1}\) in the form \(au^2+bu+c\) where \(a,b,c\in\mathbb{Q}\).
Solution: By \(u^3-u^2+u+2=0\), we have
\(u^3=u^2-u-2\).
Multiply \(u\) on both sides, we have
\(u^4=u^3-u^2-2u\).
Substitute \(u^3\) in this equality, we have
\(u^4=(u^2-u-2)-u^2-2u\), which can be simplified to
\(u^4=-3u-2\).
Expand \((u^2+u+1)(u^2-u)\), we have
\(u^4+u^3+u^2-u^3-u^2-u\), which can be simplified to
\(u^4-u\).
Substitute \(u^4\) by the previous equality, \((u^2+u+1)(u^2-u)\) is equal to
\(-4u-2\).
Factorize \(u^3-u^2+u+2\), we have
\(u^3-u^2+u+2=(u^2+1)(u-1)+3\).
Since \(u^3-u^2+u+2=0\), we have
\((u^2+1)(u-1)+3=0\).
Hence we have
\(1=-\frac{1}{3}(u-1)(u^2+1)\)
and
\((u-1)^{-1}=\frac{1}{3}(u^2+1)\).
1.2 Determine \([\mathbb{Q}(\sqrt{2},\sqrt{3}),\mathbb{Q}]\).
Solution: We have
\([\mathbb{Q}(\sqrt{2},\sqrt{3}),\mathbb{Q}]=[\mathbb{Q}(\sqrt{2},\sqrt{3}),\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}),\mathbb{Q}]=4.\)
1.3 Let \(p\) be a prime and let \(v\in\mathbb{C}\) satisfy \(v\neq1,v^p=1\) (e.g., \(v=\cos {2\pi}/p+i\sin{2\pi}/p\)). Show that \([\mathbb{Q}(v):\mathbb{Q}]=p-1.\) (Hint: Use exercise 3, p.154.)
Solution: By Eisentein's criterion, the rational polynomial \(x^{p-1}+x^{p-2}+\dots+x+1\) is an irreducible polynomial, and we have \(v^{p-1}+v^{p-2}+\dots+v+1=0\), since \(v\neq 1,v^p=1\).
Thus the degree \([\mathbb{Q}(v),\mathbb{Q}]\) equals to the degree of minimum polynomial of v, i.e., p-1.
1.4 Let \(w=\cos{\pi/6}+i\sin{\pi/6}\) (in \mathbb{C} ). Note that \(w^{12}=1\) but \(w^r\neq1\) if \(1\leq r<12\) (so \(w\) is a generator of the cyclic group of 12th roots of 1). Show that \([\mathbb{Q}:\mathbb{Q}]=4\) and determine the minimum polynomial of \(w\) over \(\mathbb{Q}\).
Solution: We consider the polynomial
\((x-\cos{\frac{\pi}{6}}-i\sin{\frac{\pi}{6}})(x-\cos{\frac{\pi}{6}}+i\sin{\frac{\pi}{6}})\cdot(x-\cos{\frac{5\pi}{6}}-i\sin{\frac{5\pi}{6}})(x-\cos{\frac{5\pi}{6}}+i\sin{\frac{5\pi}{6}})\) which equals to
\((x^2-\sqrt{3}x+1)(x^2+\sqrt{3}+1)=x^4-x^2+1\).
The polynomial \(x^4-x^2+1\) is irreducible over \(\mathbb{Q}\)(It is irreducible over \(\mathbb{Z}_2\)) and thus is the minimum polynomial of \(w\). Thus the degree \([\mathbb{Q}(w):\mathbb{Q}]=4\).
1.5 Let \(E=F(u)\) where \(u\) is algebraic of odd degree. Show that \(E=F(u^2).\)
Solution: We have
\([F(u):F]=[F(u):F(u^2)][F(u^2):F].\)
The degree \([F(u):F(u^2)]\) can only be 1 or 2 because \(u\) is root of the polynomial \(x^2-u^2\). Since \([F(u):F]\) is odd,\([F(u):F(u^2)]=1\). Thus \(F(u)=F(u^2)\).
1.6 Let \(E_i,i=1,2\) be a subfiled of \(K/F\) such that \([E_i:F]\) is finite. Show that if \(E\) is the subfield of \(K\) generated by \(E_1\) and \(E_2\) then \([E:F]\leq[E_1:F][E_2:F]\).
Solution: Assume \(E_2=span_F\{v_1,v_2,\dots,v_s\}\) with \(s=[E_2:F]\). It holds that
\(E=E_1(v_1,v_2,\dots,v_s)\).
Since \(v_1,v_2,\dots v_s\) are algebraic over \(F\), all elements in \(E\) are \(E_1-\)linear combination of monomials \(v_1^{n_1}v_2^{n_2}\cdots v_s^{n_s}\). A monomial \(v_1^{n_1}v_2^{n_2}\cdots v_s^{n_s}\) is a \(F-\)linear combination of \(v_1,v_2,\dots,v_s\). Thus an element in \(E\) is a \(E_1-\)linear combination of \(v_1,v_2,\dots,v_s\). Hence
\([E:F]\leq[E_1:F]s=[E_1:F][E_2:F]\).
1.7 Let \(E\) be an extension field of F which is algebraic over \(F\) in the sense that every element of \(E\) is algebraic over \(F\). Show that any subring of \(E/F\) is a subfield. Hence prove that any subring of finite dimensional extension field \(E/F\) is a subfield.
Solution: If \(u\) is algebraic over \(F\), we show that the inverse of \(u\) exist. This is because there exist a polynomial \(g(x)\) over \(F\), such that \(g(u)=0\), we just need to move the constant of this polynomial to the right and times the inverse of that constant directly. Thus the subring of \(E/F\) is closed under inverse operation and \(E\) is a subfield.
1.8 Let \(E=F(u),u\) transcendental, and let \(K\neq F\) be a subfield of \(E/F\). Show that \(u\) is algebraic over \(K\).
Solution: Since \(K\neq F\), it contains an element \(v\) of the form
\(\frac{f(u)}{g(u)},\qquad deg g(x)>deg f(x), (f(x),g(x))=1\)
such that \(f(x)=a_mx^m+\dots+a_1x+a_0\) and \(g(x)=b_nx^n+\dots+b_1x+b_0\) with \(a_i,b_j\in F\). It holds that
\(v(b_nu^n+\dots+b_1u+b_0)-a_mu^m-\dots-a_1u-a_0=0.\)
Since \(a_i,b_j,v\in K\), \(u\) is algebraic over \(K\).
1.9 Let \(E\) be an extension field of the field \(F\) such that
(i)\([E:F]<\infty\)
(ii)for any two subfields \(E_1\) and \(E_2\) containing \(F\), either \(E_1\subset E_2\) or \(E_2\subset E_1\).
Show that \(E\) has a primitive element over F.
Solution: Since \([E:F]<\infty\), we have \(u_1,u_2,\dots,u_r\in E\) such that
\(E=F(u_1,u_2,\dots,u_r).\)
by (ii), we can find a subfield \(F(u_i)\) such that for any integer \(j\) from 1 to \(r\), \(F(u_j)\subset F(u_i)\). For convenience, we assume \(u_i\) as \(u_1\). We show that \(F(u_1)=E\). Since \(F(u_1)\) contains all \(u_j\), it is clear that \(E\subset F(u_1)\). Hence we have \(F(u_1)=E\).