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\begin{document}

\title{Lecture 20}

\maketitle 
Last time we proved the following.

\begin{thm} \label{T:Val} Let L/K be a finite separable extension and $n = [L:K]$.  Let v be any absolute value of K.  Then there are finitely many extensions w of v to an absolute value of L.  If $L_w$ is the completion of L in w, then
$$
L \otimes_K K_v \cong \prod_{j=1}^{g} {L_{w_j}}
$$
where $w_1, \dots, w_g$ are the distinct valuations of L over v.
\end{thm}

Let us derive some corollaries from this theorem.

\begin{cor} \label{C:TrN} For $x \in L$, $Tr_{L/K}(x) = \sum_{i=1}^{g} Tr_{L_{w_i}/K_v}(x)$ and $N_{L/K}(x) = \prod_{i=1}^{g} N_{L_{w_i}/K_v}(x)$.
\end{cor}

\begin{proof} We can think of $Tr(x)$ (respectively norm) as the trace (respectively determinant) of the endomorphism $\alpha \mapsto x\alpha$ of the K-vector space L.  By extension of scalars from K to $K_v$, this endomorphism is the trace (respectively) of the $K_v$-vector space, $L \otimes_K K_v$.  By theorem \ref{T:Val}, this extended endomorphism can be put in block form for a suitable basis.
\end{proof}

\begin{cor} \label{C:Norm} Let L/K be a finite extension of number fields. Then
$$
|N_{L/K}|_v = \prod_{w|v} |x|^{n(w)}_w
$$ 
where $n(w) = [L_w:K_v]$
\end{cor}

\begin{proof} By Corollary \ref{C:TrN}, $|N_{L/K}|_v = \prod_{w|v} |N_{{L_w}/K_v}|_v$.  From Theorem 4.3, we have $|N_{{L_w}/K_v}|_v = |x|^{n(w)}_w$ and so we are done.
\end{proof}

\begin{cor} \label{C:nef} Let L/K be a finite extension of number fields and ${n = [L:K]}$.  Then
$$
n = \sum_{w|v} e(w|v)f(w|v).
$$
\end{cor}

\begin{proof} By Theorem \ref{T:Val}, $L \otimes_K K_v = \prod_{w|v} L_w$.  Now 
$$
\begin{aligned}
n 
& = dim_K L \\
& = dim_{K_v} (L \otimes_K K_v) \\
& = \sum_{w|v} dim_{K_v} L_w \\
& = \sum_{w|v} e(w|v)f(w|v) \hspace{.05in} \mbox{by Proposition 7.1}.
\end{aligned} 
$$
\end{proof}

Before we look at examples, let us address a general question.  Let L/K be a finite extension of number fields and let $\frak p$ be a prime ideal in K.  Then $\frak p\mathcal{O}_L$ is an ideal in K and we have the following factorization which is unique up to order.
$$
\frak p\mathcal{O}_L = \frak P_1^{e_1} \dots \frak P_g^{e_g}.
$$
How can we explicitly find these primes $\frak P_i$ ? To do this, we prove a general fact.  Let A be a local ring with unique maximal ideal $\frak m$   Consider the ring $B = A[X] / <f>$ where $f \in A[X]$ is a monic polynomial of degree m.  Note that as a module, B is free and finitely generated.  So 
$$
B = A\cdot1 \oplus A\cdot X \oplus \dots \oplus A\cdot X^{m-1}
$$
Note that $k = A/\frak m$ is a field.  Now consider $\bar f \in (A/\frak m)[X] = k[X]$.  Since k is a field, $k[X]$ is a unique factorization domain.  So we have
$$
\bar f = \prod_{i=1}^{g} \phi_i^{e_i}
$$
where $\phi_i$ are irreducible factors in $k[X]$.  Let $g_i \in A[X]$ be such that ${g_i \mapsto \phi_i \in k[X]}$.  Before stating a proposition, we first need to state two important results.

\begin{thm} \label{T:corres} Let R be a ring, $I \subset R$ an ideal of R, and $\lambda : R \to R/I$ the natural map.  Then the function $S \mapsto S/I$ defines a one-to-one correspondence between the set of all subrings of R containing I and the set of all subrings of $R/I$.  Under this correspondence, ideals of R containing I correspond to ideals of $R/I$.
\end{thm}

From this theorem, we have $R/\lambda^{-1}(\bar I) \cong \bar R/\bar I$ where $\bar R = R/I$ and $\bar I$ is an ideal of $\bar R$.  Also, $\lambda$ preserves prime and maximal ideals.  

\begin{lem} \label{L:Nak} (Nakayama's Lemma) Let R be a ring, $I \subset R$ an ideal of R.  Suppose I is contained in every maximal ideal of R.  Let M be an R-module and $N \subset M$ a submodule such that $M/N$ is finitely generated.  Then $M = N + IM$ implies $N=M$.
\end{lem}

\begin{pro} \label{P:max} Let $\frak m_i = <\frak m, g_i> \subset B$.  These ideals are maximal, pairwise distinct, and every maximal ideal of B is one of the $\frak m_i$.
\end{pro}

\begin{proof} We have the natural map $\lambda: B \to \bar B = B/\frak mB$.  Let $\bar \frak m_i = <\phi_i> \subseteq \bar B$.  We claim this ideal is maximal. Note that $\bar B /\bar \frak m_i = {B/<\frak m, \phi_i>} = {A[X]/<f,\frak m, \phi_i>} = k[X]/<\bar f, \bar\phi_i> \cong k[X]/<\phi_i> = k_i$.  So $k_i$ is a field and $\bar \frak m_i$ is maximal.  By theorem \ref{T:corres}, we have a bijection $\lambda^{-1}$ between the set of maximal ideals of $\bar B$, denoted Max$(\bar B)$ and the set ${\{\eta \in Max(B) : \eta \supseteq \frak m B \}}$.  Note that $\lambda^{-1}(\bar \frak m_i) = \frak m_i = <\frak m, g_i>$.  We claim that 
{Max$(\bar B) = \{\frak m_1, \dots, \frak m_g \}$}.  We now have
$$
\begin{aligned}
\bar B 
&= B/\frak m B \\
&= A[X]/<\frak m, f> \\
&\cong k[X]/<\bar f> \\
&\cong \prod_{i=1}^{g} k[X]/<\phi_i^{e_i}> \hspace{.05in} \mbox{by the Chinese Remainder Theorem}.
\end{aligned}
$$
Therefore Max$(\bar B) = \coprod Max(k[X]/<\phi_i^{e_i}>)$.
Now the last step.  To see that Max$(B) = \{\frak m_i, \dots, \frak m_g \}$, we need to see that any maximal ideal $\eta \subset B$ contains $\frak m B$.  Suppose $\frak m B \not \subset \eta$ and consider $\eta + \frak m B$.  We claim $\eta + \frak m B = B$.  To see this, suppose $\eta + \frak m B \not \subseteq B$.  Now $\frak m B \not \subset \eta$ implies that $\eta \not \subseteq \eta + \frak m B$.  This contradicts the maximality of $\eta$.  Now B is an A-module of finite type.  By Nakayama's Lemma (taking $R = A, M = B, I = \frak m$, and $N = \eta$), we have $\eta = B$, a contradiction. 
\end{proof}

Next time we will consider some examples.


\end{document}