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\begin{document}
\centerline{Math 361 lecture for Monday, Week 4}
\bigskip

\centerline{Norms and traces}
\bigskip

We have the usual set-up: Let $K$ be a number field.  By the primitive element
theorem, write $K=\Q(\theta)$ for some $\theta\in K$.  Let
$\theta_{1},\ldots,\theta_{n}$ be the roots of the minimal polynomial for
$\theta$, taking $\theta_{1}=\theta$, and define the embeddings
$\sigma_{i}\colon K\to \C$ such that $\sigma_{i}(\theta)=\theta_{i}$ for
each~$i=1,\ldots,n$.  

For each $\alpha\in K$, define the \emph{field polynomial}:
\[
  f_{\alpha}:=\prod_{i=1}^{n}(x-\sigma_{i}(\alpha)).  
\]

\begin{eg} 
 With notation as above, the field polynomial of~$\theta$ is $p$, the minimal
 polynomial for $\theta$. 
\end{eg}

\begin{eg}
  Let $K=\Q(\sqrt{2})$.  The minimal polynomial for $\sqrt{2}$ over~$\Q$ is $p =
  x^{2}-2=(x-\sqrt{2})(x+\sqrt{2})$.  The embeddings of $K$ into $\C$ are given
  by the identity mapping $\sigma_{1}$,
  sending $\sqrt{2}\mapsto \sqrt{2}$, and $\sigma_{2}$, determined by
  $\sigma_{2}(\sqrt{2})=-\sqrt{2}$.  The field polynomial of
  $\alpha=2+3\sqrt{2}$ is
  \begin{align*}
    f_{\alpha} &= (x-\sigma_{1}(\alpha))(x-\sigma_{2}(\alpha))\\
    &=(x-(2+3\sqrt{2}))(x-(2-3\sqrt{2}))\\
    &=x^2-4x-14,
  \end{align*}
  which is the minimal polynomial for $\alpha$ over~$\Q$.
  The fact that the coefficients of $f$ are integers is a consequence
  of~$\alpha$ being an algebraic integer in~$K$ (see below).

  For another example, let~$\alpha=5\in \Z$.  We get
  \[
    f_{5} = (x-\sigma_{1}(5))(x-\sigma_{2}(5))=(x-5)^{2}.
  \]
  So $f_{5}$ is the square of the minimal polynomial~$x-5$ for $5$ over~$\Q$.
\end{eg}

\begin{thm}~\label{thm:fld poly}  With notation as above, let~$p_{\alpha}$ be the minimal polynomial
  for~$\alpha$ over~$\Q$.
  \begin{enumerate}
    \item\label{thm:04mon1} The field polynomial has rational coefficients: $f_{\alpha}\in\Q[x]$.
    \item\label{thm:04mon2} There exists a positive integer~$k$ such that
      $f_{\alpha}=p_{\alpha}^{k}$.
    \item\label{thm:04mon3} If~$\alpha$ is an algebraic integer, then~$f_{\alpha}$ has integer coefficients:
      $f_{\alpha}\in\Z[x]$.
  \end{enumerate}
\end{thm}
\begin{proof} (This proof follows the one given in our text.)
 We first prove part~\ref{thm:04mon1}.
 Since $\alpha\in K=\Q(\theta)=\Q[\theta]$, there exists $r\in \Q[x]$ such that
 $\alpha=r(\theta)$.  Then for each~$i$, 
 \[
   \sigma_{i}(\alpha)=\sigma_{i}(r(\theta))=r(\sigma_{i}(\theta))=r(\theta_{i})
 \]
 since $\sigma_{i}$ preserves addition and scalar multiplication and acts as the
 identity when restricted to~$\Q$.  Therefore,
 \begin{align*}
   f_{\alpha}&=\prod_{i=1}^{n} (x-\sigma_{i}(\alpha)) \\
	     &=\prod_{i=1}^{n} (x-r(\theta_{i}))\\
	     &=x^{n}-(r(\theta_{1})+r(\theta_{2})+\cdots+r(\theta_{n}))x^{n-1}+\cdots+(-1)^{n}r(\theta_{1})r(\theta_{2})\cdots
	     r(\theta_{n}).
\end{align*}
In the bottom expression, we see that each coefficient of~$f_{\alpha}$ is symmetric in
the $\theta_{i}$.  Hence, each coefficient is a rational polynomial combination
of the elementary symmetric functions in the~$\theta_{i}$, which we have seen
are rational numbers.

For part~\ref{thm:04mon2}, first note that $p_{\alpha}|f_{\alpha}$ in $\Q[x]$
since~$f\in\Q[x]$ and $f_{\alpha}(\alpha)=0$.  So we can write
$f_{\alpha}=p_{\alpha}^{k}h$ for some~$h\in \Q[x]$ sharing no polynomial factors
with $p_{\alpha}$ (recall that $\Q[x]$ is unique factorization domain and that
primes and irreducibles are the same).  Since~$f_{\alpha}$ and $p_{\alpha}$ are
monic, so is~$h$.  We complete the proof by showing that~$h$ is constant and,
thus, that $h=1$. For sake of contradiction, suppose that $h$ is not constant.
Since
\[
  f_{\alpha}=\prod_{i=1}^{n} (x-\sigma_{i}(\alpha))= \prod_{i=1}^{n}
  (x-r(\theta_{i}))=(p_{\alpha}(x))^{k}h(x),
\]
by unique factorization in $\C[x]$, there exists~$j$ such that $r(\theta_{j})$ is a zero of
$h$.  Define $g(x)=h(r(x))$.  Then $g(\theta_{j})=0$.  Let $p$ be the minimal
polynomial of~$\theta$ over~$\Q$.  Recall that
\[
  p=\prod_{i=1}^{n}(x-\theta_{i}). 
\]
So~$p$ is also the minimal polynomial for~$\theta_{j}$ over~$\Q$.  Hence, $p|g$.
It follows that~$g(\theta_{i})=0$ for all~$\theta_{i}$.  In particular,
$g(\theta)=0$ (since~$\theta=\theta_{1}$).  Therefore,
\[
  h(\alpha)=h(r(\theta))=g(\theta)=0.
\]
It follows that~$p_{\alpha}|h$, a contradiction since~$p_{\alpha}$ and~$h$ are coprime.

Finally, for part~\ref{thm:04mon3}, if $\alpha$ is an algebraic integer, then we
saw that its minimal polynomial has integer coefficients.  So
part~\ref{thm:04mon3} follows from part~\ref{thm:04mon2}.
\end{proof}

\begin{defn}
  Let $\alpha\in K$.  The \emph{norm} of $\alpha$ is
  \[
    N(\alpha)=\prod_{i=1}^{n} \sigma_{i}(\alpha),
  \]
  and the \emph{trace} of $\alpha$ is
  \[
    T(\alpha)=\sum_{i=1}^{n} \sigma_{i}(\alpha).
  \]
\end{defn}

\begin{eg}
  Consider~$\alpha=2+3\sqrt{2}\in K=\Q(\sqrt{2})$.  We have
  \[
    N(\alpha) = \sigma_{1}(2+3\sqrt{2})\sigma_{2}(2+3\sqrt{2})
    = (2+3\sqrt{2})(2-3\sqrt{2}) 
    = -14,
  \]
  and
  \[
    T(\alpha) = \sigma_{1}(2+3\sqrt{2})+\sigma_{2}(2+3\sqrt{2})
    =(2+3\sqrt{2})+(2-3\sqrt{2}) 
    = 4.
  \]
\end{eg}

\textbf{Properties of the norm and trace.}

Let~$\alpha,\beta\in K=\Q(\theta)$.
\begin{enumerate}[leftmargin=*]
  \item $N(\alpha \beta)=N(\alpha)N(\beta)$ and
    $T(\alpha+\beta)=T(\alpha)+T(\beta)$.  (Clear from definitions.)
  \item~\label{properties:2} The norm and trace appear as coefficients of the corresponding field
    polynomial:
    \begin{align*}
      f_{\alpha}&=\prod_{i=1}^{n} (x-\sigma_{i}(x)) \\[5pt]
		&=x^{n}-(\underbrace{\sigma_{1}(\alpha)+\cdots+\sigma_{n}(\alpha)}_{T(\alpha)})x^{n-1}+\cdots+(-1)^{n}\underbrace{\sigma_{1}(\alpha)\cdots\sigma_{n}(\alpha)}_{N(\alpha)}.
    \end{align*}
  \item $N(\alpha),T(\alpha)\in\Q$.  If $\alpha\in\mathfrak{O}_{K}$, then
    $N(\alpha),T(\alpha)\in\Z$.  (This follows from part~\ref{properties:2} and
    Theorem~\ref{thm:fld poly}.  We have~$f_{\alpha}\in\Z[x]$.)
  \item~\label{properties:4} Suppose that  $\alpha\in\mathfrak{O}_{K}$. Then $\alpha$ is a unit if and only if
    $N(\alpha)=\pm1$.
    \begin{proof}
      ($\Rightarrow$) Say there exists~$\beta\in\mathfrak{O}_{K}$ such that
      $\alpha \beta=1$.  Then
      \[
	1=N(1)=N(\alpha \beta)= N(\alpha)N(\beta).
      \]
      Since~$\alpha, \beta\in \mathfrak{O}_{K}$, we have $N(\alpha),
      N(\beta)\in\Z$.  It follows that~$N(\alpha)=\pm1$.

      ($\Leftarrow$) Suppose that
      \[
	\pm1=N(\alpha)
	=\sigma_{1}(\alpha)\cdots \sigma_{n}(\alpha)
	=\alpha\cdot \sigma_{2}(\alpha)\cdots \sigma_{n}(\alpha).
      \]
      Defining $\beta := \sigma_{2}(\alpha)\cdots \sigma_{n}(\alpha)$, we have $\alpha
      \beta=1$.  We need to show that~$\beta\in\mathfrak{O}_{K}$.
      Each $\sigma_{i}(\alpha)$ satisfies the minimal polynomial for~$\alpha$,
      which is a monic polynomial with integer coefficients.  Therefore, each
      $\sigma_{i}(\alpha)$ is an algebraic integer (not necessarily in~$K$).
      Since the algebraic integers in~$\C$ form a ring, we see that~$\beta$ is
      an algebraic integer.  On the other hand, 
      \[
	\beta=\pm\frac{1}{\alpha}\in K.
      \]
      Hence, $\beta\in\mathfrak{O}_{K}$, as required.
    \end{proof}
  \item Suppose that $\alpha\in\mathfrak{O}_{K}$.  If $N(\alpha)=q\in\Z$
    where~$q$ is a rational prime, then~$\alpha$ is irreducible.
    \begin{proof}
      Suppose that~$\alpha=\beta \gamma$ for some
      $\beta,\gamma\in\mathfrak{O}_{K}$.  Then
      \[
	q=N(\alpha)=N(\beta \gamma)=N(\beta)N(\gamma).
      \]
      Since~$N(\beta),N(\gamma)\in\Z$ and~$q$ is prime, one of $N(\beta)$ or
      $N(\gamma)$ is~$\pm1$.  Without loss of generality, say $N(\beta)=\pm1$.
      It follows from part~\ref{properties:4} that~$\beta$ is a unit.  Thus,
      $\alpha$ is irreducible.
    \end{proof}
  \item\label{properties:item6} Let $p$ be the minimal polynomial for $\theta$.  Then
    $\Delta[1,\theta,\ldots,\theta^{n-1}]=(-1)^{n(n-1)/2}N(p'(\theta))$ where
    $p'$ is the derivative of~$p$.
    \begin{proof}
      In the Wednesday Week 3, we saw that
      $\Delta[1,\theta,\ldots,\theta^{n-1}]=\prod_{1\leq i<j\leq
      n}(\theta_{i}-\theta_{j})^{2}$, where $\theta_{i}=\sigma_{i}(\theta)$.
      Since~$p(x)=\prod_{i=1}^{n} (x-\theta_{i})$, we have for
      each~$j=1,\ldots,n$,
      \begin{align*}
        p'(x)=\sum_{k=1}^{n} \prod_{i:i\neq k}(x-\theta_{i})
	&\quad\Rightarrow\quad
        p'(\theta_{j})=\prod_{i:i\neq j}(\theta_{j}-\theta_{i}).
      \end{align*}
      Therefore, since~$\sigma_{i}(p'(\theta))=p'(\theta_{i})$,
      \[
	N(p'(\theta))
	= \prod_{k=1}^{n}\sigma_{k}(p'(\theta)) 
        = \prod_{k=1}^{n}  p'(\theta_{k})
	= \prod_{\substack{i,j=1\\i\neq j}}^{n}(\theta_{j}-\theta_{i}).
      \]
      Note that on the right-hand side, each $\theta_{j}-\theta_{i}$ appears
      twice: once when~$j<i$, and once when $j>i$.  Grouping these two
      occurrences together gives
      \[
	N(p'(\theta))=\prod_{1\leq i<j\leq
	n}^{n}\left(-(\theta_{j}-\theta_{i})^{2}\right)   
	=(-1)^{\binom{n}{2}}\prod_{1\leq i<j\leq n}(\theta_{j}-\theta_{i})^{2},
      \]
      where $\binom{n}{2}=n(n-1)/2$ is the number of pairs $i,j$ with $i<j$.
    \end{proof}
  \item\label{properties:item7} Let $\alpha_{1},\ldots,\alpha_{n}$ be a $\Q$-basis for $K$. 
    Let~$S$ be the $n\times n$ matrix with $ij$-th entry
    $T(\alpha_{i}\alpha_{j})$.
    Then $\Delta[\alpha_{1},\ldots,\alpha_{n}]=\det S=\det(T(\alpha_{i}\alpha_{j}))$.
    \begin{proof}
       The $ij$-th entry of $S$ is
       \[
         S_{ij}=T(\alpha_{i}\alpha_{j})=\sum_{k=1}^{n}\sigma_{k}(\alpha_{i}\alpha_{j})
	 =\sum_{k=1}^{n} \sigma_{k}(\alpha_{i})\sigma_{k}(\alpha_{j}).
       \]
       Let $A$ be the $n\times n$ matrix with~$ij$-th entry
       $\sigma_{i}(\alpha_{j})$.  
       Then
       \[
         \Delta[\alpha_{1},\ldots,\alpha_{n}]
	 =\left(\det A\right)^{2}
	 = \det(A)\det(A)
	 = \det(A^{t})\det(A)
	 = \det(A^{t}A).
       \]
       The $ij$-th entry of $A^{t}A$ is
       \[
	 (A^{t}A)_{ij}
	 =\sum_{k=1}^{n} \left(A^{t}\right)_{ik}A_{kj}
	 =\sum_{k=1}^{n} A_{ki}A_{kj}
	 =\sum_{k=1}^{n} \sigma_{k}(\alpha_{i})\sigma_{k}(\alpha_{j})
	 = S_{ij}.
       \]
       Therefore,
       \[
         \Delta[\alpha_{1},\ldots,\alpha_{n}]
	 = \det(A^{t}A)
	 = \det(S).
       \]
    \end{proof}
\end{enumerate}

\begin{eg}
  Let $K=\Q(\sqrt{6})$ and embeddings
  \begin{align*}
    \sigma_{1}(a+b\sqrt{6})&=a+b\sqrt{6}\\  
    \sigma_{2}(a+b\sqrt{6})&=a-b\sqrt{6}.
  \end{align*}
  Since~$6\neq 1\bmod 4$, we have~$\mathfrak{O}_{K}=\Span_{\Z}\{1,\sqrt{6}\}$.  So the discriminant of~$K$
  is
  \[
    \Delta(K)
    = 
    \Delta[1,\sqrt{6}]
    =
    \left(    \det 
      \left(\begin{array}{rr}
	  1 & \sqrt{6} \\
	  1 & -\sqrt{6}
    \end{array}\right)\right)^{2}
    = \left(-2\sqrt{6}\right)^{2}=24.
  \]
  We check property~\ref{properties:item6} from above.  The minimal polynomial
  for $\sqrt{6}$ is $p=x^2-6$.  Therefore, $p'=2x$, and
  \begin{align*}
    \Delta[1,\sqrt{6}]
    &=(-1)^{2(2-1)/2}N(p'(\sqrt{6}))\\
    &=-N(2\sqrt{6})\\
    &=-\sigma_{1}(2\sqrt{6})\sigma_{2}(2\sqrt{6})\\
    &=-(2\sqrt{6})(-2\sqrt{6})\\
    &=24.
  \end{align*}

  Next, we check property~\ref{properties:item7} for the $\Q$-basis
  $\alpha_{1}=1, \alpha_{2}=\sqrt{6}$.  We have
  \begin{align*}
    T(\alpha_{1}^{2})&=T(1)=\sigma_{1}(1)+\sigma_{2}(1)=1+1=2\\
    T(\alpha_{1}\alpha_{2})&=T(\alpha_{2}\alpha_{1})=T(\sqrt{6})=\sigma_{1}(\sqrt{6})+\sigma_{2}(\sqrt{6})=\sqrt{6}-\sqrt{6}=0\\
    T(\alpha_{2}^{2})&=T(6)=\sigma_{1}(6)+\sigma_{2}(6)=12,
  \end{align*}
  and
  \[
    \Delta[1,\sqrt{6}]=\det 
    \left(\begin{array}{cc}
	2 & 0 \\
	0 & 12
    \end{array}\right)
    =
    24.
  \]
\end{eg}
\end{document}
