\classheader{2013-03-13}

On Friday, attendence will be optional. I'll give the definition of etale cohomology.

Let $R$ be a commutative ring, and let $f_1,\ldots,f_m\in R[T_0,\ldots,T_n]$ be homogeneous polynomials. The projective scheme over $R$ defined by $f_1=\cdots=f_m=0$ is $\bigcup_{r=0}^n\Spec(A_r)$, where
\[A_r=R[T_1,\ldots,T_{r-1},T_{r+1},\ldots,T_n]/(f_{1,r},\ldots,f_{m,r})\]
and
\[f_{i,r}=f_i(T_1,\ldots,T_{r-1},T_{r+1},\ldots,T_n).\]
For $r\neq s$, we have
\[\Spec(A_r)\cap\Spec(A_s)=\Spec(A_r[\tfrac{1}{x_s}]),\]
which is an open subset of $\Spec(A_r)$. Here $x_s$ denotes the image of $T_s$ in $A_r$.

\begin{example-N}
We can take $R=\C[t]$, and $f=T_2^2T_0-T_1^2-tT_0^3$. For any $a\in\C$, we have a map $\C[t]\to\C$ defined by sending $t\mapsto a$, and then we get the projective scheme over $\C$ defined by the image of $f$, i.e. the projective scheme over $\C$ defined by $T_2^2T_0-T_1^1-aT_0^3$. In general the closed points of this projective scheme over $\C$ are
\[X_a:=\{(x,y)\in\C^2\mid y^2=x^3+a\}\cup\{\infty\}.\]
If $a=0$, then this becomes $y^2=x^3$, which has a singularity. We can see this problem in the cohomology:
\[H_1(X_a,\Q)\cong H^1(X_a,\Q)\cong\begin{cases}
\Q^2 & \text{ if }a\neq 0,\\
0 & \text{ if }a=0.
\end{cases}\]
\end{example-N}
\begin{example-N}
An analogous example in the case of $R=\Z$ would be $f=T_2^2T_0-T_1^3-3T_0^3$.
\end{example-N}

If $R=k$ is an algebraically closed field, then we can identify
\[\{x=(x_0:\cdots:x_n)\in\P^n(k)\mid f_1(x)=\cdots=f_m(x)=0\}=\bigcup_{r=0}^n\max(A_r)\subset\bigcup_{r=0}^n\Spec(A_r),\]
where
\[\bigcup_{r=0}^n\max(A_r)=\left\{x\in\bigcup_{r=0}^n\Spec(A_r)\;\middle|\; \{x\}\text{ is closed}\right\}.\]

We can define a condition $(\ast\ast)$: for any $r$ and any $\frak{p}\in\Spec(A_r)$ (you can also use $\max(A_r)$ without changing the condition), the image of $\left(\frac{\partial f_{i,r}}{\partial T_j}\right)_{1\leq i\leq m, 0\leq j\leq n, j\neq r}$ in $\kappa(\frak{p})$ has rank $m$.

\begin{itemize}
\item If $R=\C[t,\frac{1}{t}]$, and $f$ is the same as in our earlier example, and $(\ast\ast)$ is satisfied.
\item If $R=\Z[\frac{1}{6}]$, and $f$ is the same as in our earlier example, then $(\ast\ast)$ is satisfied.
\item If $R=\C[t]$ and we consider the maps $R\to\C$ sending $t\mapsto a$ and $t\mapsto b$ respectively, where $a,b\neq 0$, then when $(\ast\ast)$ is satisfied, we have $H^1(X_a,\Q)\cong H^1(X_b,\Q)$. In general, if $(\ast\ast)$ is satisfied and if $R$ is an integral domain finitely generated over $\C$, then $H^i(X_a,\Q)\cong H^i(X_b,\Q)$ for any ring homomorphisms $a,b:R\to\C$ and any $i$.
\end{itemize}

\subsection*{Grothendieck's \'etale cohomology}

Let $k$ be a separably closed field (this means algebraically closed when $\characteristic(k)=0$). Let $\ell$ be a prime number, $\ell\neq\characteristic(k)$. Let $X$ be a scheme of finite type over $k$, which just means that there is an open covering $X=\bigcup_{s=1}^n\Spec(A_s)$ where each $A_s$ is finitely generated over $k$. Then we have finite-dimensional vector spaces over $\Q_\ell$, denoted $H_{et}^i(X,\Q_\ell)$, for all $i\geq 0$.

If $k=\C$, then $H_{et}^i(X,\Q_\ell)\cong H^i(X_{\text{cl}},\Q)\otimes_\Q \Q_\ell$, where $X_\text{cl}$ is the closed points of $X$ with the topology coming from $\C$.

If $R$ is an integral domain satisfying condition $(\ast\ast)$ and we have any ring homomorphisms $a:R\to k_1$, $b:R\to k_2$ where the $k_i$ are separably closed fields, then
\[H_{et}^m(\frak{X}_a,\Q_\ell)\cong H_{et}^m(\frak{X}_b,\Q_\ell),\]
where $\frak{X}_a$ is the scheme over $k_1$ defined by the images of the $f_i$ in $k_1[T_0,\ldots,T_n]$.


If $R$ is a finite field and $(\ast\ast)$ is satisfied, then we can express the zeta function as
\[\zeta_X(s)=\prod_{\substack{x\in X\\ x\text{ closed point}}}\left(1-\frac{1}{\#\kappa(x)^s}\right)^{-1}=\prod_{i=0}^{2\dim(X)}\det\left(1-\varphi^{-1}u:H_{et}^i(X_{\overline{\F}_q},\Q_\ell)\right)^{(-1)^{i-1}}\]
where $u=q^{-s}$. There is an action of
\[\Gal(\overline{\F}_q/\F_q)\cong\projlim_{n}\Gal(\F_{q^n}/\F_q)\cong\projlim_n\Z/n\Z.\]
Let $P_i(u)=\det(1-\varphi^{-1}u:H^i)$ as in above. Then
\[\zeta_X(s)=\frac{P_1(q^{-s})\cdots P_{2d-1}(q^{-s})}{P_0(q^{-s})\cdots P_{2d}(q^{-s})},\]
where $d=\dim(X)$.

If $K$ is a number field and $X$ is a scheme of finite type over $K$, then because we have $K\hookrightarrow \overline{K}=\overline{\Q}\hookrightarrow\C$, we have
\[H_{et}^m(X_{\overline{K}},\Q_\ell)\cong H_{et}^m(X_\C,\Q_\ell)\cong H^m(X_{\C\,cl},\Q)\otimes_\Q\Q_\ell.\]
and there is a Galois action of $\Gal(\overline{K}/K)$ on the left.

If we have a donut over $\C$, then we cannot hear the action of the Galois group. We can eat it and enjoy the taste though. 

The Langlands correspondence 







