\documentclass{article}

\usepackage{amsthm}
\newtheorem{thm}{Theorem}
\newtheorem{lem}{Lemma}
\newcommand{\Q}{\mathbf{Q}}

\title{MATH 129, HW 3}
\author{Bryden Cais}

\begin{document}
\maketitle

\section{Discriminant}
\begin{enumerate}

	\item We have
	\begin{eqnarray*}
	    D(X^2+bX+c)&=&\mathrm{Res}(2X+b,X^2+bX+c)=\left|
	    \begin{array}{*{3}{c}}
		b & 0 & c \\
		2 & b & b \\
		0 & 2 & 1 \\
	    \end{array}
	    \right|\\
	    &=&-b^2+4c.
	\end{eqnarray*}

	\item 
	\begin{enumerate}
	    \item One direction is obvious: namely, if $\psi$ is a bijection then since 
	    $1\,\in\,K_{p+q-1}[X]$, there exist $A\,\in\,K_{q-1}[X]$ and $b\,\in\,K_{p-1}[X]$
	    such that $\psi(A,B)=1,$ that is $AP+BQ=1.$  This shows that $\mathrm{gcd}(P,Q)=1.$
	    Now conversely, suppose that $\mathrm{gcd}(P,Q)=1$.  If $\psi$ is not injective, then
	    there exist distinct $A_1,\ A_2\,\in\,K_{q-1}[X]$ and $B_1,\ B_2\,\in\,K_{p-1}[X]$ 
	    such that $A_1P+B_1Q=A_2P+B_2Q$, that is, $(A_1-A_2)P=(B_2-B_1)Q.$  Clearly,
	    $Q|(A_1-A_2)P.$  Since $\mathrm{gcd}(P,Q)=1$ and $K[X]$ is Euclidean, we have
	    $Q|(A_1-A_2),$ which is impossible since $\mathrm{deg}(A_1-A_2)\leq q-1 <\mathrm{deg}(Q).$
	    Thus we see that $\psi$ must be injective.  Now since the map $\psi$ is a $K$-vector space
	    homomorphism and since $\mathrm{dim}_K(K_{q-1}[X]\times K_{p-1}[X])=\mathrm{dim}_K(K_{p+q-1}[X])$,
	    we see that $\psi$ is also surjective, and hence bijective.

	    \item By the above, we know that $\mathrm{gcd}(P,Q)=1$ iff $\psi$ is a bijection.
	    Since $\psi$ is a $K$-linear map, $\psi$ is a bijection iff the determinant of its
	    matrix is not $0$.  Let $A=c_0+c_1X+\cdots+c_{q-1}X^{q-1},\ B=d_0+d_1X+\cdots+d_{p-1}X^{p-1}.$
	    Let $v\,\in\,K_{q-1}[X]\times K_{p-1}[X]$ be the column vector given by 
	    $c_0,\ldots,c_{q-1},d_0,\ldots,d_{p-1}.$  Then it follows from straightforward matrix multiplication
	    that the $p+q-1$ column vector given by the coefficiens of $AP+BQ$ is equal to
	    \begin{eqnarray*}
		\left(
		\begin{array}{*{6}{c}}
		    a_0 & & & b_0 & & \\
		    a_1 & \ddots & & \vdots & \ddots & \\
		    \vdots & \ddots & a_0 & b_{q-1} & \ddots & b_0 \\
		    a_p & \ddots & a_1 & b_q & \ddots & \vdots \\
		     & \ddots & \vdots & & \ddots & b_{q-1} \\
		     & & a_p & & & b_q \\
		\end{array}
		\right)\,v.
	    \end{eqnarray*}
	    Thus, we see that the determinant of $\psi$ is just the resultant $\mathrm{Res}(P,Q),$
	    and hence that $\psi$ is a bijection iff $\mathrm{Res}(P,Q)\not=0.$

	    \item Let $P=a_p(X-\alpha_1)\cdots (X-\alpha_p),\ Q=b_q(X-\beta_1)\cdots (X-\beta_q).$
	    Let $V(X_1,X_2,\ldots,X_n)$ denote the matrix
	    \begin{eqnarray*}
		\left(
		\begin{array}{*{4}{c}}
		1 & X_1 & \cdots & X_1^n \\
		1 & X_2 & \cdots & X_2^n \\
		\vdots & \vdots & \cdots & \vdots \\
		1 & X_n & \cdots & X_n^n \\
		\end{array}
		\right),
	    \end{eqnarray*}
	    set $M=V(\beta_q,\cdots,\beta_1,\alpha_p,\cdots,\alpha_1)$ and
	    \begin{eqnarray*}
		R\ =\ \left(
		\begin{array}{*{6}{c}}
		    a_0 & & & b_0 & & \\
		    a_1 & \ddots & & \vdots & \ddots & \\
		    \vdots & \ddots & a_0 & b_{q-1} & \ddots & b_0 \\
		    a_p & \ddots & a_1 & b_q & \ddots & \vdots \\
		     & \ddots & \vdots & & \ddots & b_{q-1} \\
		     & & a_p & & & b_q \\
		\end{array}
		\right).
	    \end{eqnarray*}
	    We first show that $\mathrm{det}(V(X_1,\ldots,X_n))=\prod_{1\leq i<j\leq n}(X_i-X_j).$
	    To see this, for $1\leq j\leq n-1$, subtract $X_1$ times the $j$ th column
	    from the $j+1$ st column of the matrix $V(X_1,\ldots,X_n).$ 
	    These operations do not alter the determinant, and we find that
	    \begin{eqnarray*}
		\mathrm{det}(V(X_1,\ldots,X_n))=\prod_{2\leq i\leq n}(X_i-X_1)\,
		\mathrm{det}(V(X_2,\ldots,X_n)).
	    \end{eqnarray*}
	    Induction then gives the proposed formula.  We now compute the determinant of $MR$ in
	    two different ways.  It is not difficult to see that
	    \begin{eqnarray*}
		\lefteqn{	
		\left(
		\begin{array}{*{4}{c}}
		1 & \beta_q & \cdots & \beta_q^{p+q-1}\\
		\vdots & \vdots & \cdots & \vdots \\
		1 & \beta_1 & \cdots & \beta_1^{p+q-1}\\
		1 & \alpha_p & \cdots & \alpha_p^{p+q-1}\\
		\vdots & \vdots & \cdots & \vdots \\
		1 & \alpha_1 & \cdots & \alpha_1^{p+q-1}\\
		\end{array}
		\right)\,\left(
		\begin{array}{*{6}{c}}
		    a_0 & & & b_0 & & \\
		    a_1 & \ddots & & \vdots & \ddots & \\
		    \vdots & \ddots & a_0 & b_{q-1} & \ddots & b_0 \\
		    a_p & \ddots & a_1 & b_q & \ddots & \vdots \\
		     & \ddots & \vdots & & \ddots & b_{q-1} \\
		     & & a_p & & & b_q \\
		\end{array}
		\right)}\\\\
		&=&\left(
		\begin{array}{*{6}{c}}
		    P(\beta_q) & \cdots & P(\beta_q)\beta_q^{q-1}  & & &\\
		    \vdots & \cdots & \vdots & & &\\
		    P(\beta_1) & \cdots & P(\beta_1)\beta_1^{q-1}  & & &\\
		    & & & Q(\alpha_p) & \cdots & Q(\alpha_p)\alpha_p^{p-1} \\
		    & & & \vdots & \cdots & \vdots \\
		    & & & Q(\alpha_1) & \cdots & Q(\alpha_1)\alpha_1^{p-1} \\
		\end{array}	
		\right),
	    \end{eqnarray*}
	    and hence that
	    \begin{eqnarray*}
	        \lefteqn{\mathrm{det}(MR)=}\\
	        &&\mathrm{det}(V(\beta_q,\ldots,\beta_1)V(\alpha_p,\ldots,\alpha_1))
		\prod_{1\leq j\leq q}P(\beta_j)\,\prod_{1\leq i\leq p}Q(\alpha_i).
	    \end{eqnarray*}
	    On the other hand, 
	    \begin{eqnarray*}
		\mathrm{det}(MR)&=&\mathrm{det}(M)\,\mathrm{det}(R)\\
		&=&\mathrm{det}(V(\beta_q,\ldots,\beta_1,\alpha_p,\ldots,\alpha_1))\mathrm{Res}(P,Q).
	    \end{eqnarray*}
	    Now, using our formula for the determinant of $V(X_1,\ldots,X_n)$, we see that
	    \begin{eqnarray*}
		\lefteqn{\mathrm{det}(V(\beta_q,\ldots,\beta_1,\alpha_p,\ldots,\alpha_1))}\\
		&=&\mathrm{det}(V(\beta_q,\ldots,\beta_1))\mathrm{det}(V(\alpha_p,\ldots,\alpha_1))
		\prod_{1\leq i\leq p}\prod_{1\leq j\leq q}(\alpha_i-\beta_j)\\
		&=&\mathrm{det}(V(\beta_q,\ldots,\beta_1))\mathrm{det}(V(\alpha_p,\ldots,\alpha_1))
		\frac{1}{b_q^p}\,\prod_{1\leq i\leq p}Q(\alpha_i).
	    \end{eqnarray*}
	    It thus follows that
	    \begin{eqnarray*}
		\mathrm{Res}(P,Q)=b_q^p\,P(\beta_1)\cdots P(\beta_q).
	    \end{eqnarray*}
	    Notice that $\mathrm{Res}(P,Q)=(-1)^{pq}\mathrm{Res}(Q,P)$ since to swap the roles
	    of the $a_i$'s and the $b_j$'s in the defining determinant of the resultant requires
	    a bubble sort involving $pq$ flips of columns.
	\end{enumerate}

	\item Let $P\,\in\,K[X]$ with $\mathrm{deg}(P)>0.$
	Clearly, $P$ has only simple roots in $\bar{K}$ iff $P,\ P^{\prime}$ have
	no common roots.  By part c above, this is the case iff $R(P^{\prime},P)=D(P)$
	is not zero.  Thus a and c are equivalent.  By part b above, $\mathrm{gcd}(P,\ P^{\prime})=1$
	iff $D(P)\not= 0,$ so that b and c are equivalent.  Furthermore, if $P$ has some root of
	multiplicity greater than $1$ then we can write $P=Q^kR$ for $Q,\ R\,\in\,K[X]$ with
	$P\not| Q$, $\mathrm{deg}(Q)\geq 1$, and $k\geq 2$. Thus $QR\,\in\,K[X]/P$ is not
	zero and is nilpotent since $P|(QR)^k.$  Thus, $K[X]/P$ is not reduced.  Conversely, if
	we have some nilpotent element $Q\,\in\,K[X]/P$ then we have $P|Q^k$ for some $k\geq 2$
	but $P\not| Q.$  The only way that this is possible is if $P$ has a root of multiplicity 
	greater than $1$.

	\item From class, we have the formula
	\begin{eqnarray*}
	    \mathrm{disc}(1,\alpha,\ldots,\alpha^{n-1})=(-1)^{\frac{n(n-1)}{2}}N_{L/K}(P^{\prime}(\alpha)),
	\end{eqnarray*}
	where $\alpha$ is a root of the monic irreducible polynimial $P$ and $L=K(\alpha)$.
	Since $P$ is irreducible, $\mathrm{Hom}_K(L,\bar{K})$ acts transitively on the roots of $P$.
	We thus have
	\begin{eqnarray*}
	    \mathrm{disc}(1,\alpha,\ldots,\alpha^{n-1})&=&(-1)^{\frac{n(n-1)}{2}}N_{L/K}(P^{\prime}(\alpha))\\
	    &=&(-1)^{\frac{n(n-1)}{2}}\prod_{\sigma\,\in\,\mathrm{Hom}_K(L,\bar{K})}\sigma(P^{\prime}(\alpha))\\
	    &=&(-1)^{\frac{n(n-1)}{2}}\prod_{1\leq i\leq n}\sigma(P^{\prime}(\alpha_i)),
	\end{eqnarray*}
	where $\alpha_1,\ldots,\alpha_n\,\in\,\bar{K}$ are the roots of $P$.  Thus, we have	
	\begin{eqnarray*}
	    \mathrm{disc}(1,\alpha,\ldots,\alpha^{n-1})=(-1)^{\frac{n(n-1)}{2}}\mathrm{Res}(P^{\prime},P)
	\end{eqnarray*}
	by 2c and the fact that $P$ is monic.  

	\item Let $K$ be a number field and $O_K$ its ring of integers.  Suppose that $O_K=\mathbf{Z}[\alpha]$
	for some $\alpha\,\in\,O_K$, and that $Q$ is the minimal polynomial of $\alpha$ (of degree $n$).
	In this case, we have from a result in class that
	$\mathrm{disc}(K)=\mathrm{disc}(1,\alpha,\ldots,\alpha^{n})=\pm D(P)$ by the previous item.
	Let $p$ be any prime.  Then we know that $\mathbf{Z}[X]/(p,Q)=(\mathbf{Z}/p)[X]/Q=(\mathbf{Z}[X]/Q)/p
	=O_K/pO_K$.  Since $\mathbf{Z}/p=\mathbf{F}_p$ is perfect, we have that $(\mathbf{Z}/p)[X]/Q$ is
	reduced iff $D(P)\not=0$ in $\mathbf{F}_p$.  That is, iff $p\not|D(P).$  Hence, $O_K/p$ is 
	reduced iff $p\not|\mathrm{disc}(K).$

\end{enumerate}

\section{A Special Case of Kronecker-Weber}
	For any positive integer $m$, set $\zeta_m=e^{2\pi i/m}.$  Let $p$ be an odd prime. 
	From class, we know that $\mathrm{disc}(\zeta_p)=\pm p^{p-2}$ is a square in $\Q(\zeta_p)$.
	Since $p$ is odd, this implies that $p-2$ is odd and hence that 
	$\Q(\sqrt{\pm p})\subset \Q(\zeta_p)$.  It is obvious that $\Q(i)=\Q(\zeta_4)$, and that
	$Q(\zeta_8)\supset \Q(\sqrt{2})$, since $\zeta_8+\zeta_8^7=\sqrt{2}.$  Thus, for any prime
	$p$ (even or odd), $\Q(\sqrt{p})\subset \Q(\zeta_{4p})$.  This follows for $p=2$ by the above,
	and for odd $p$ since $\Q(\zeta_{4p})\supset \Q(\zeta_p,\zeta_4)=\Q(\sqrt{\pm p})(i)\supset \Q(\sqrt{p}).$
	We now show that for any $m$, $\Q(\sqrt{m})$ is contained in some cyclotomic field.  Without 
	loss of generality, assume that $m$ is square free.  Then we may write $m=p_1p_2\cdots p_k$ for
	distinct primes $p_i$.  We have $\Q(\sqrt{m})\subset \Q(\sqrt{p_1},\sqrt{p_2},\ldots,\sqrt{p_n}).$
	However, we have shown that $\Q(\sqrt{p_i})\subset \Q(\zeta_{4p_i})\subset \Q(\zeta_{4p_1\cdots p_n}),$
	since $\zeta_{4p_1\cdots p_n}^{p_1\cdots \hat{p_i}\cdots p_n}=\zeta_{4p_i},$ where $\hat{p_i}$
	denotes that $p_i$ is omitted from the product.  Thus, every quadratic extension of $\Q$
	is contained in some cyclotomic field.

\section{Stickelberger's Criterion}
	Let $K$ be a number field of degree $n$ over $\Q$ and fix algebraic integers $\alpha_1,\ldots,\alpha_n.$
	Let $\sigma_1,\ldots,\sigma_n$ denote the $n$ embeddings of $K$ in $\mathbf{C}$.  We can write 
	$d=\mathrm{disc}(\alpha_1,\ldots,\alpha_n)=|\sigma_i(\alpha_j)|^2=(P-N)^2,$ where $P$ denotes
	the product
	\begin{eqnarray*}
	    \sum_{\tau\in A_n}\prod_{i=1}^n \sigma_i(\alpha_{\tau(i)}),
	\end{eqnarray*}
	and $N$ is given by
	\begin{eqnarray*}
	    \sum_{\tau\in S_n\setminus A_n}\prod_{i=1}^n \sigma_i(\alpha_{\tau(i)}),
	\end{eqnarray*}
	that is, $P$ is the sum in the determinant corresponding to even permutations, while $N$
	corresponds to the odd permutations.  Notice that since the $\alpha_i$ are algebraic
	integers, $\sigma_j(\alpha_i)$ are algebraic integers also (since any $\sigma_j$ must
	sent any $\alpha_i$ to a root of its minimal polynomial), and since the set of all
	algebraic integers forms a ring, $P,\ N$ must also be algebraic integers.  Now, let
	$L$ be some normal closure of $K$ in $\mathbf{C}$ with $\mathrm{Gal}(L/\Q)=
	\{\sigma_1,\ldots,\sigma_n}.$  Now the $\sigma_j$ act on the set $\{1,\ldots,n\}$
	by defining $\tilde{\sigma}_j(k)$ to be the integer $l$ such that $\sigma_j\sigma_k=\sigma_l.$
	Thus, clearly, $\tilde{\sigma_j}\,\in\,S_n$ for each $j$.  Now, for any \textit{even} $\tilde{\sigma}$,
	we have
	\begin{eqnarray*}
	    \sigma(P)&=&\sum_{\tau\in A_n}\prod_{i=1}^n \sigma\sigma_i(\alpha_{\tau(i)})\\
	    &=&\sum_{\tau\in A_n}\prod_{i=1}^n \sigma_{i}(\alpha_{\tau\tilde{\sigma}^{-1}(i)})\\
	    &=& P
	\end{eqnarray*} 
	upon reindexing.  Similarly, we have for any even $\tilde{\sigma}$ that $\sigma(N)=N,$  
	while for odd $\tilde{\sigma}$ we have $\sigma(P)=N$ and $\sigma(N)=P.$  Since every
	element of $\mathrm{Gal}(L/\Q)$ is either even or odd, we find that
	\begin{eqnarray*}
	    N+P,\\
	    NP,
	\end{eqnarray*}
	are both fixed by every embedding $\sigma_j,$ (and hence every element of $\mathrm{Gal}(L/\Q)$
	and hence must be in $\Q.$  Since they are algebraic integers in $\Q$, 
	they must be in $\mathbf{Z}.$  Hence, $d=(P-N)^2=(P+N)^2-4PN$ is congruent 
	to $1$ or $0$ modulo $4$.

\section{Ring of Integers of Biquadratic Extensions}
	\begin{enumerate}

	    \item (a) One direction is easy: if $\alpha\,\in\,R$ then the relative norm and trace
	    over $\Q(\sqrt{m})$ are algebraic integers.  This follows from the fact that the norm 
	    and trace are, respectively, a product and sum of
	\end{enumerate}

\end{document}
