\classheader{2013-03-06}

\subsection*{Projective varieties; schemes}

Let $k$ be a field. Then $n$-dimensional projective space over $k$ is defined as
\[\P^n(k)=(k^{n+1}\setminus\{0\})/\sim\]
where $(a_0,\ldots,a_n)\sim(b_0,\ldots,b_n)$ when there is some $c\in k^\times$ such that $b_i=ca_i$ for all $i$. The equivalence class of $(a_0,\ldots,a_n)$ is denoted as $(a_0:\cdots:a_n)$. 

Suppose that $k$ is algebraically closed. We can define a Zariski topology on $\P^n(k)$ as follows: $Y\subseteq\P^n(k)$ is closed if and only if $Y\cap W_r$ is closed in $W_r$ for all $r$.

\begin{definition}
A closed subset $X\subseteq\P^n(k)$, i.e. a projective algebraic set, is said to be irreducible if
\begin{itemize}
\item[(i)] $X\neq\emptyset$
\item[(ii)] For any closed $Y,Z\subseteq \P^n(k)$ such that $X=Y\cup Z$, we have either $X=Y$ or $X=Z$.
\end{itemize}
An irreducible projective algebraic set is called a projective algebraic variety.
\end{definition}

\begin{example}
The subset
\[X=\{(x_0:x_1:x_2)\in\P^2(k)\mid x_2^2x_0=x_1^3+x_0^3\}\]
is a projective algebraic variety. Note that $U_0$ of this $X$ is
\[U_0=\{(x_1,x_2)\in k^2\mid x_2^2=x_1^3+1\},\]
and that $X\setminus U_0=\{(0:0:1)\}$.
\end{example}

Given a projective algebraic variety $X\subseteq\P^n(k)$, the function field of $X$ is defined as follows: we have $X=\bigcup_{r=0}^nU_r$, and $X\neq\emptyset$ so some $U_r\neq\emptyset$, and because $U_r\neq\emptyset$ we have that $U_r$ is an irreducible algebraic set in $k^n$. If $U_r\neq\emptyset$ and $U_s\neq\emptyset$, then $U_r\cap U_s\neq\emptyset$. The ring $\mathcal{O}(U_r)$, the ``coordinate ring'' of $U_r\subseteq k^n$, consists of the polynomial functions on $U_r$. We can identify $U_r=\max(\mathcal{O}(U_r))$, so that
\[X=\bigcup_{r=0}^n\max(\mathcal{O}(U_r)).\]
The intersection $U_r\cap U_s$ can be identified with $\max(\mathcal{O}(U_r[\frac{1}{x_s}]))$, which is an open subset of $\max(\mathcal{O}(U_r))=U_r$.

If $U_r\neq\emptyset$, then $\mathcal{O}(U_r)$ is an integral domain, and if $U_r\cap U_s\neq \emptyset$, then \[\mathcal{O}(U_r[\frac{1}{x_s}])=\mathcal{O}(U_s[\frac{1}{x_r}]).\]
Therefore their fraction fields are the same as well:
\[\mathrm{Frac}(\mathcal{O}(U_r))=\mathrm{Frac}(\mathcal{O}(U_r[\tfrac{1}{x_s}]))=\mathrm{Frac}(\mathcal{O}(U_s[\tfrac{1}{x_r}]))=\mathrm{Frac}(\mathcal{O}(U_s)).\]
We define the function field $K$ of $X$ t be $\mathrm{Frac}(\mathcal{O}(U_r))$ for any $U_r\neq\emptyset$. We then have that
\[\dim(X)=\dim(\mathcal{O}(U_r))=\mathrm{tr. deg.}_k(K).\]

\begin{conjecture}[Resolution of singularities]
Let $K$ be a finitely generated field over $k$. Then there is a non-singular projective algebraic variety $X\subseteq\P^n(k)$, such that $K$ is the function field of $X$.
\end{conjecture}

This was proved by Hironaka in the case that $\characteristic(k)=0$.

The conjecture is true if $\mathrm{tr. deg.}_k(K)=1$. In this case, we have that $X=\max(A)\cup \max(A')$ and that $X$ can be embedded in $\P^3(k)$; sometimes even $\P^2(k)$.

For example,
\[X=\{(x_0:x_1:x_2)\in\P^2(\C)\mid x_0^d=x_1^d+x_2^d\}\]
is a non-singular variety of genus $\frac{(d-1)(d-2)}{2}$. 

\begin{theorem}[Fermat's Last Theorem for $\C(T)$] For $n\geq 3$, there are no non-constant $f,g\in\C(T)$ such that $f^n+g^n=1$.
\end{theorem}
\begin{proof}
The function field $K$ is $\C(T_1,\sqrt[n]{1-T_1^n})$. This is the fraction field of $\C[T_1,T_2]/(T_1^n+T_2^n-1)$. If such an $f,g$ exist, then there is an embedding $K\hookrightarrow\C(T)$, defined by $T_1\mapsto f$ and $T_2\mapsto g$, induced from $\C[T_1,T_2]/(T_1^n+T_2^n-1)\to\C(T)$. But such an embedding would correspond to a covering $\P^1(\C)\twoheadrightarrow X$ of compact Riemann surfaces. This would then induce surjections in homology,
\[H_1(\P^1(\C),\Q)\twoheadrightarrow H_1(X,\Q).\]
But $H_1(\P^1(\C),\Q)=0$ and $H_1(X,\Q)=\Q^{2g}$ where $g=\frac{(n-1)(n-2)}{2}>0$ when $n\geq 3$, so this is impossible. It's important to use homology over a field, and in particular a field of characteristic 0; if we did it over $\F_p$ and the degree of the covering was a multiple of $p$, then the map would be the zero map and we would not have a contradiction. Integral homology can also mess up surjectivity, because for example we might have $\Z\to\Z$, $1\mapsto n$.
\end{proof}