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\author{Bryden R. Cais}
\title{Algebraic Number Theory, Lecture 2}
\begin{document}
\maketitle

\section{Traces, Norms, and Discriminants}

Let $A\subset B$ be rings, and suppose that $B$ is a finitely generated free $A$ module, that is $B\simeq A^n$.
For an element $b\in B$, we can define
$$\tr_{B/A}(b):=\tr(m_b),$$
where $m_b:B\rightarrow B$ is the matrix of multiplication by $b$.  One can check that this is independent of a choice of basis.
We also define
$$\n_{B/A} (b):=\det(m_b).$$
If $A\subset B\subset C$ are modules (each free and finitely generated over $A$) then a straightforward matrix argument
gives
\begin{align*}
    \tr_{C/A}&=\tr_{C/B}\circ \tr_{B/A}\\
    \n_{C/A}&=\n_{C/B}\circ \n_{B/A}.
\end{align*}

\begin{lem}
    Suppose that $L/K$ is a finite separable extension of fields of degree $d$ and that $\sigma_1,\ldots,\sigma_d$
    are the $d$ distinct embeddings of $L$ into $\bar{K}$ fixing $K$.  Then
    \begin{align}
        \tr_{L/K}=\sum_{i=1}^d \s_i\\
        \n_{L/K}=\prod_{i=1}^d \s_i.
    \end{align}
\end{lem}
\begin{proof}
    Let $\alpha\in L$ have minimal polynomial
    $$\m{\alpha}(x)=x^n+a_{n-1}x^{n-1}+\ldots+a_0\in K[x]$$ and set $F=K(\alpha)$.
    It follows at once from the definition of the characteristic polynomial that
    $\m{\alpha}(x)$ is the characteristic polynomial of multiplication by $\alpha$, and consequently that
    \begin{align}
        \tr_{L/K}(\alpha)&=-a_{n-1}\label{tr}\\
        \n_{L/K}(\alpha)&=(-1)^n a_{0}\label{nor}.
    \end{align}
    Now we know that
    $$F\simeq K[x]/(\m{\alpha}(x)).$$  Therefore, if $\alpha=\alpha_1,\alpha_2,\ldots,\alpha_n$ are the
    $n$ distinct roots of $\m{\alpha}(x)$ in $\bar{K}$, then the $n$ distinct embeddings of $F$ in $\bar{K}$
    are given by
    $$x\mapsto \alpha_i.$$
    It now follows from (\ref{tr}) and (\ref{nor}) that
    \begin{align*}
        \tr_{L/K}(\alpha)&=\sum_{i=1}^n \alpha_i \\
        \n_{L/K}(\alpha)&=\prod_{i=1}^n \alpha_i.
    \end{align*}
    We are therefore done if $F=L$.  If not, it is not hard to see that
    $$\sum_{i=1}^d \s_i(\alpha)=\sum_{j=1}^n \sum_{\s_i(\alpha)=\alpha_j}\alpha_j=\sum_{j=1}^{n}\deg(L/F)\alpha_j=\deg(L/K)\tr_{F/K}(\alpha).$$
    Since $\alpha\in F$, multiplication by $\alpha$ as a linear map of $L$ as an $F$ vector space is just diagonal; therefore,
    the corresponding matrix for $L$ as a $K$ vector space is block diagonal.  The claim for $\tr{L/K}(\alpha)$ follows from this,
    and the proof for $\n_{L/K}(\alpha)$ is similar so we omit it.
\end{proof}

The following properties of $\tr_{L/K}$ and $\n_{L/K}$ follow easily from the definitions:
\begin{enumerate}
    \item $\tr_{L/K}(\alpha+\beta)=\tr_{L/K}(\alpha)+\tr_{L/K}(\beta)$.
    \item If $a\in K$ then $\tr_{L/K}(a\alpha)=a\tr_{L/K}(\alpha)$.
    \item If $\alpha\in K$ then $\tr_{L/K}(\alpha)=\deg(L/K)$.
    \item $\n_{L/K}(\alpha\beta)=\n_{L/K}(\alpha)\n_{L/K}(\beta)$.
    \item If $a\in K$ then $\n_{L/K}(a\alpha)=a^{\deg(L/K)}\n_{L/K}(\alpha)$.
    \item We have that $\langle x,y \rangle\mapsto \tr_{L/K}(xy)$ is a $K$ bilinear pairing.
\end{enumerate}

Now suppose that $A\subset B$ are rings with $B\simeq A^n$ and let $\b_1,\ldots,\b_n\in B$.
\begin{defn}
    We define the \textbf{discriminant} of $\b_1,\ldots,b_n$ to be
    $$\disc(\b_1,\ldots,\b_n)=\Delta(\b_1,\ldots,\b_n)=\det(\tr_{B/A}(\b_i\b_j)).$$
\end{defn}

Observe that if $M\in \mat_{n\times n}(A)$ and if $(\g_1,\ldots,\g_n)=(\b_1,\ldots,\b_n)M$, then since
$$\tr_{B/A}(\g_1\g_j)=M^T(\tr_{B/A}(\b_i\b_j))M,$$ we have
$$\disc(\g_1,\ldots,\g_n)=(\det(M))^2\disc(\b_1,\ldots,\b_n).$$

\begin{lem}
    Suppose that $B\simeq A^n$.
    \begin{enumerate}
        \item   If $\{\b_1,\ldots,\b_n\}$ and $\{\g_1,\ldots,\g_n\}$ are two $A$ bases for $B$ then
        $$\disc(\g_1,\ldots,\g_n)=u^2 \disc(\b_1,\ldots,\b_n)$$ for some unit $u\in A^*$.

        \item If $\{\b_1,\ldots,\b_n\}$ is an $A$ basis for $B$ and
        $$(\disc(\g_1,\ldots,\g_n))=(\disc(\b_1,\ldots,\b_n))\subset A$$
        then $\{\g_1,\ldots,\g_n\}$ is also an $A$ basis for $B$.

    \end{enumerate}
\end{lem}

\begin{proof}
    This is clear.
\end{proof}

\begin{exam}
    Let $D\neq 0,1$ be a square free integer and take $A=\Z$, $\Z[\sqrt{D}]$.  Then clearly $\{1,\sqrt{D}\}$ is a $\Z$ basis
    of $B$ and we have
    \begin{align*}
    \disc(1,\sqrt{D})&=\det\begin{pmatrix}\tr_{B/A}(1) & \tr_{B/A} (\sqrt{D})\\  \tr_{B/A} (\sqrt{D}) & \tr_{B/A}(D)\end{pmatrix}\\
    &=\det\begin{pmatrix}2 & 0\\  0 & 2D\end{pmatrix}\\
    &=4D.
    \end{align*}
    As another example, if $D\equiv 1\mod 4$ then it is not difficult to see that $\a=\frac{1+\sqrt{D}}{2}$ is an algebraic integer.
    Letting $B=\Z\oplus \a\Z$, we find that
    \begin{align*}
    \disc(1,\a)&=\det\begin{pmatrix}\tr_{B/A}(1) & \tr_{B/A} (\a)\\  \tr_{B/A} (\a) & \a^2\end{pmatrix}\\
    &=\det\begin{pmatrix}2 & 1\\  1 & \frac{1+D}{2}\end{pmatrix}\\
    &=D.
    \end{align*}
\end{exam}

\section{Integrality}

\begin{defn}
    An element $\a\in B$ is \textbf{integral} over $A$ if it satisfies $f(\a)=0$ for some monic $f\in A[x]$.  If every
    element of $B$ is integral over $A$, we say that $B$ is integral over $A$.
\end{defn}

\begin{prop}
    An element $\a\in B$ is integral over $A$ if and only if $A[\alpha]\subset B$ is a finitely generated $A$ module.
\end{prop}

\begin{proof}
    We have in any case a surjective map $$A[x]/(f(x))\twoheadrightarrow A[\alpha].$$  If $\a$ is integral, so that
    $f$ is monic, then $A[x]/(f)$ is a finitely generated $A$ module, with basis $1,x,\ldots,x^{n-1},$ where $n=\deg(f)$.
    Conversely, if $A[\a]$ is a finitely generated $A$ module, let $\omega_i$ for $i=1,\ldots, m$ be a basis for $A[\alpha]$
    over $A$.  Then
    $$\alpha\omega_i=\sum_{j=1}^m b_{ij}\omega_j$$ for each $i$, where the $b_{ij}\in A$.  We then see that
    $\alpha$ is a root if the characteristic polynomial of the matrix given by the $b_{ij}$, which is clearly
    monic.
\end{proof}

\begin{lem}
    \begin{enumerate}
        \item   If $B$ is a finitely generated $A$ module then every element of $B$ is integral over $A$.

        \item The $A$-integral elements of $B$ form a subring of $B$.
    \end{enumerate}
\end{lem}

\begin{proof}
    If $\a,\b$ are integral over $A$, then $A[\a,\b]$ is finitely generated since $A[\a,\b]=A[\a][\b]$
    and $\b$ is integral over $A[\a]$ since it is integral over $A$.  Since $\a+\b,\a\b\in A[\a,\b]$, it follows
    that $A[\a+\b],A[\a\b]$ are finitely generated $A$ modules, from which the lemma follows.
\end{proof}

\begin{defn}
    The \textbf{integral closure} of $A\subset B$ is the ring of all elements of $B$ that are integral over $A$.
\end{defn}

\begin{defn}
    The ring $A$ is said to be \textbf{integrally closed} in $B$ if it is its own integral closure in $B$.
\end{defn}

\begin{lem}
    The integral closure of $A$ in $B$ is integrally closed.
\end{lem}

\begin{proof}
    Let $C$ be the integral closure of $A$ in $B$ and suppose that $\a\in B$ is integral over $C$.  Then
    $\a^n+c_{n-1}\a^{n-1}+\cdots+c_0=0$ for some $c_i\in C.$  Then
    $$A[c_0,\ldots,c_{n-1},\a]$$ is obviously finitely generated over $A$ and hence $\a$ is integral over $A$.
\end{proof}

\begin{defn}
    A domain $A$ is integrally closed if it is integrally closed in its field of fractions.
\end{defn}
\begin{exam}
    $\Z$ is integrally closed since it is integrally closed in $\Q$.
\end{exam}

\begin{lem}\label{integral}
    Suppose that $A$ is an integrally domain with field of fractions $K$ and that $L/K$ is a finite separable extension of fields.
    Then $\a\in L$ is integral over $A$ if and only if $\m{\a}(x)\in K[x]$ is in $A[x]$.
\end{lem}
\begin{proof}
    One direction is obvious.  So suppose that $\a$ is integral over $A$.  Let $p(x)\in A[x]$ be any monic polynomial
    having $\a$ as a root.  Then for any embedding $\s:L \hookrightarrow \bar{K}$ fixing $K$ we see that $\s(\a)$
    is a root of $p(x)$.  It follows that $\s(\a)$ is integral and hence, since the $\s(\a)$ are precisely the roots
    of $\m{\alpha}(x)$, that the coefficients of $\m{\a}(x)$ (which are elementary symmetric polynomials in the $\s(\a)$) are integral over $A$.
    On the other hand, these coefficients are fixed by each $\s$ and therefore lie in $K$.
    Using the fact that $A$ is integrally closed shows that $\m{\alpha}(x)\in A[x]$, as required.
\end{proof}

\begin{defn}
    Let $K$ be a number field.  The \textbf{ring of integers}, $\O_k$ is the integral closure of $\Z$ in $K$.
\end{defn}

\begin{lem}
    If $\a\in\O_K$ then
    $$\tr_{K/\Q}(\a),\n_{K/\Q}(\a)\in \Z.$$
\end{lem}
More generally, we have
\begin{proof}
    The trace and norm are coefficients of $\m{\a}(x)\in \Z[x]$ by Lemma \ref{integral}.
\end{proof}
\begin{lem}
    If $L/K$ is a finite extension of number fields and $\a\in\O_L$ then
        $$\tr_{L/K}(\a),\n_{L/K}(\a)\in \O_K.$$
\end{lem}
\end{document}
