\classheader{2013-03-01}


The homework that was due today will be the last one.

Today we'll talk about algebraic curves.

Let $k$ be a field, and let $K$ be a finite extension of $k(T)$. This is similar to letting $K$ be a finite extension of $\Q$. In the 19th century, people started to compare these two scenarios. Recall that $\ringofintegersof{K}$ denotes the integral closure of $\Z$ in $K$; analogously, let $A$ be the integral closure of $k[T]$ in $K$.

Both $A$ and $\ringofintegersof{K}$ are always Dedekind domains. However, the following does not occur in the case of $\ringofintegersof{K}$: we can let $A'$ be the integral closure of $k[\frac{1}{T}]$ in $K$, and let $A''$ be the integral closure of $k[T,\frac{1}{T}]$ in $K$. Then $U''=\max(A'')$ can be regarded as a subset of both $U=\max(A)$ and $U'=\max(A')$, and we can form $\P^1(k)$ as
\begin{center}
\begin{tikzcd}[row sep=-0.1in,column sep=-0.05in]
\P^1(k)=\max(k[T])\cup_{\max(k[T,\frac{1}{T}])}\max(k[\frac{1}{T}])=& k & \cup & k\\
 & & {\text{\,\scriptsize$k^\times$}} \ar[start anchor=base,end anchor=base,shift right=0.1in,mapsto]{ru}[swap,pos=0]{x}[swap,pos=1]{\!\!x^{-1}}
 \ar[start anchor=base,end anchor=base,shift  left=0.1in,mapsto]{lu}[pos=0]{x}[near end]{x}
\end{tikzcd}
\end{center}
It is always the case that $\max(A'')$ is just $\max(A)$ minus a finite set, and also $\max(A')$ minus a finite set.

When $k=\C$, we have that $\P^1(\C)=\C\cup\{\infty\}$ is the Riemann sphere. The ring $A'$ corresponds to $\P^1(\C)\setminus\{0\}\cong \C$. Note how looking at $A$ or $A'$ corresponds to stereographic projection from the north or south pole.

Localization is compatible with taking integral closure; in other words, if $A$ is a domain, $K$ is the fraction field of $A$, $L$ is a finite extension of $K$, and $B$ is the integral closure of $A$ in $L$, then if $S\subset A\setminus\{0\}$ is a multiplicative subset of $A$, then the integral closure of $S^{-1}A$ in $L$ is $S^{-1}B$. If $A$ is finitely generated over a field $k$, then $B$ is finitely generated as an $A$-module, and therefore also finitely generated over $k$.

In the case of $\Z$, the best analog we know for $\P^1$ is taking $X=\max(\Z)\cup\{\infty\}$ where $\infty$ denotes the embedding of $\Q\hookrightarrow\R$. More generally, for $\ringofintegersof{K}$, we take $\max(\ringofintegersof{K})\cup\{\infty_1,\ldots,\infty_r\}$, which we can regard as the set of all embeddings of $K$ into locally compact topological fields with dense image, considered up to a certain equivalence.

Going back to the case of $k=\C$, we have that $U$ and $U'$ are Riemann surfaces (one-dimensional complex analytic manifolds), and $X=U\cup U'$ is a compact Riemann surface. The field of meromorphic functions on $\P^1(\C)$ is just $\C(T)$.

For a less trivial example, suppose we have $K=\C(T)(\sqrt{T^3+1})$, so that $A=\C[T,\sqrt{T^3+1}]$. It may be surprising that $A'=\C[\frac{1}{T},\sqrt{(\frac{1}{T})^4+\frac{1}{T}}]$, so that the corresponding sets are 
\begin{align*}
U&=\{(x,y)\in\C^2\mid y^2=x^3+1\}\\
U'&=\{(u,v)\in\C^2\mid v^2=u^4+u\}\\
U''&=U-\{(x,y)\in U\mid x=0\}\\
&=U-\{(0,\pm 1)\}\\
&=U'-\{(u,v)\in U'\mid u=0\}\\
&=U'-\{(0,0)\}
\end{align*}
and $X$ is $U$, together with the point $(0,0)$ of $U'$.

\begin{theorem}
There is a bijection between (isomorphism classes of) function fields in one variable over $\C$ and (isomorphism classes of) compact connected Riemann surfaces, where $K\leftrightarrow X$ where $K$ is the field of meromorphic functions on $X$, and $X$ is constructed from $K$ as above.
\end{theorem}

A function field in one variable over $\C$ is a field which is isomorphic over $\C$ to a finite extension of $\C(T)$; however, note that you are free to choose a different $T$.

The complex topology on the surface $X$ corresponding to $K=\C(T)(\sqrt{T^3+1})$ is a torus. In general, $f$ is a polynomial with no repeated roots, the surface associated with $\C(T)(\sqrt{f})$ is a torus with $\frac{n-1}{2}$ holes if $n$ is odd, and $\frac{n-2}{2}$ holes if $n$ is even.

We say that the number of hole of a donut is $g$, the genus. Is that how ``donut'' is spelled? It is too bad, I can eat donuts, but I cannot write it. It turns out that
\[g=\dim_\C(A''/(A+A')),\]
and we always know that $A''/(A+A')$ is a finite-dimensional vector space over $\C$. For example, if $K=\C(T)$, then because $\C[T,\frac{1}{T}]=\C[T]+\C[\frac{1}{T}]$, we have that $g=0$;  if $K(T)(\sqrt{T^3+1})$, then $T^{-1}\sqrt{T^3+1}$ is a $\C$-basis for $A''/(A+A')$, so the genus is 1.