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\begin{center}
\bf
Math 254A, UC Berkeley, Fall 2001 (Kedlaya) \\
Lecture 23: Quadratic Fields and Quadratic Reciprocity \\
October 26, 2001
\end{center}

\head{Reference}
Neukirch, Section 1.8.

\head{Jargon watch}
Recall that a rational integer $D \neq 1$ is called a \emph{fundamental
discriminant} if one of the following holds:
\begin{enumerate}
\item[(a)] $D$ is odd, squarefree, and congruent to 1 mod 4;
\item[(b)] $D$ is divisible by 4, and $D/4$ is odd, squarefree, and congruent
to 3 mod 4.
\item[(c)] $D$ is divisible by 8, and $D/8$ is odd and squarefree.
\end{enumerate}
The point: $D$ is a fundamental discriminant if and only if the discriminant
of the quadratic field $\QQ(\sqrt{D})$ is equal to $D$.

\head{Outline of lecture}

\begin{enumerate}
\item
Let $D$ be a fundamental discriminant.
We may apply
the criterion from last time to test whether primes split in $\QQ(\sqrt{D})$,
since the ring of integers is $\QQ[\alpha]$ for $\alpha = \sqrt{D/4}$,
if $D \equiv 0 \pmod{4}$, or $\alpha = (1+\sqrt{D})/2$, if
if $D \equiv 1 \pmod{4}$.
\item
We compute the splitting behavior of a prime $p$ as follows:
\begin{enumerate}
\item If $p$ divides $D$, then $p$ is ramified. That is, there is one prime
$\gothp$ above $(p)$ with $e(\gothp/(p)) = 2$ and $f(\gothp/(p)) = 1$.
\item If $p$ is odd and does not divide $D$, then $(p)$ is unramified,
and splits into two primes if and only if the polynomial $x^2-D$ factors
modulo $p$.
\item If $p = 2$ and $D$ is odd, then $(p)$ is unramified, and 
splits into two primes if and only if the polynomial $x^2-x+(D-1)/4$
factors modulo $2$, that is, if and only if $D \equiv 1 \pmod{8}$.
\end{enumerate}
\item
For quadratic fields, it turns out there is a beautiful condition that
reduces the question of whether $D$ has a square root modulo $p$ to a question
about $p$ modulo the prime factors of $D$: the \emph{law of quadratic
reciprocity}. To state it, define the Legendre symbol
$\left( \frac{a}{p} \right)$, for integers $a,p$ with $p>0$ an odd prime, as
\[
\left( \frac{a}{p} \right) = 
\begin{cases}
  0 & a \equiv 0 \pmod{p} \\
  1 & a \not\equiv 0 \pmod{p} \mbox{and} a \equiv x^2 \pmod{p} \mbox{for some $x \in \ZZ$} \\
  -1 & a \not\equiv 0 \pmod{p} \mbox{and} a \not\equiv x^2 \pmod{p} 
\mbox{for all $x \in \ZZ$}.
\end{cases}
\]
In other words, $\left( \frac{a}{p} \right) \equiv a^{(p-1)/2} \pmod{p}$
(because $\FF_p^*$ is cyclic).
\item
State Gauss's law of quadratic reciprocity: for $p, q$ odd primes,
\[
\left( \frac{p}{q} \right) \left( \frac{q}{p} \right) = (-1)^{(p-1)(q-1)/4},
\qquad
\left( \frac{-1}{p} \right) = (-1)^{(p-1)/2},
\qquad \left( \frac{2}{p} \right) = (-1)^{(p^2-1)/8}.
\]
Of these, the second is elementary, and the third follows from the splitting
result above.
\item
Sketch the proof from Neukirch Section 1.8, using the Gauss sum
\[
\tau = \sum_{a \in \FF_p^*} \left( \frac{a}{p} \right) \zeta_p^a,
\]
where $\zeta_p$ is a primitive $p$-th root of unity. Namely, calculate
explicitly that $\tau^2 = \left( \frac{-1}{p} \right) p$.
We will
revisit this proof when we study cyclotomic fields, several lectures from
now, and give it a more conceptual interpretation.
\item
Advertise \emph{class field theory}, which can be viewed (among other things)
as a generalization of quadratic reciprocity to a splitting criterion for
any Galois extension of any number field with abelian Galois group. (I will
be covering class field theory next semester.)
\end{enumerate}

\head{Addendum to previous lecture}
A more conceptual way to see the relationship, in case $\gotho_L =
\gotho_K[\alpha]$, between the factorization of $\gothp \gotho_L$ and
the factorization of $P(x)$ modulo $\gothp$ (for $\gothp$ a prime of
$\gotho_K$ and $P(x)$ the minimal polynomial of $\alpha$) is as follows.
Put $\gothp = \prod_i \gothb_i^{e_i}$ and suppose
$P(x) \equiv \prod_j P_j(x)^{c_j} \pmod{\gothp}$, where each $P_j$ is
irreducible modulo $\gothp$. Then on one hand, by the Chinese remainder theorem,
$\gotho_L/(\gothp \gotho_L) \cong \prod_i \gotho_L/\gothb_i^{e_i}$.
On the other hand,
\[
\gotho_L/(\gothp \gotho_L)
\cong \gotho_K[x]/[(P(x)) + \gothp]
\cong (\gotho_K/\gothp)[x]/(P(x))
\cong \prod_j (\gotho_K/\gothp)[x]/(P_j(x)^{c_j}).
\]
We have now written the ring $\gotho_L/(\gothp \gotho_L)$ as a finite
product of
local rings in two ways. It's a standard fact from algebra that such a 
decomposition is unique if it exists, so we can pair off the $\beta_i$
with the $P_j$.

In case you haven't seen that standard fact before, here's how it's proved:
call an element $e$ of a ring an \emph{idempotent} if $e^2=e$. In a local
ring, this implies $e(e-1) = 0$, and $e$ and $e-1$ cannot both be in the
maximal ideal. Thus either $e$ is a unit, so $e-1=0$, or $e-1$ is a unit,
so $e=0$. That is, the only idempotents in a local ring are 0 and 1.

Suppose the ring $R$ is isomorphic both to $\prod_i S_i$ and $\prod_j T_j$,
where the $S_i$ and $T_j$ are local. Let $e_i$ be the element of $R$ which
has 1 in $S_i$ and 0 in the other $S$'s, and let $f_j$ be the element of $R$
which has 1 in $T_j$ and 0 in the other $T$'s. Now $e_i x \in S_i$ for all
$x \in R$ (that is, $e_i x$ has 0 in the $S$'s other than $S_i$), and likewise
$f_j x \in T_j$ for all $x \in R$. Also, $e_i f_j$ is an idempotent for all
$i,j$; since it's in $S_i$, it's either equal to $e_i$ or 0. But the
sum of the $f_j$ is 1, so $\sum_j e_i f_j = e_i\cdot 1 = e_i$, so the
$e_if_j$ cannot all be 0. So there is some $j$ such that $e_i f_j = e_i$;
by a similar argument, $e_i f_j = f_j$. Thus the rings $S_i$ and $T_j$
are isomorphic (they're both the image of multiplication by $e_i=f_j$),
so we can factor them off and keep going.

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