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\begin{document}
\smallskip
\centerline{\sc Math 632, Lecture 9\\ January 28, 2004}

\section{Last time}

Recall that last time we wanted to show that for $X=\Spec A$ with $A$ a ring, the association 
$X_f\rightarrow A_f$ is a $\B$-sheaf.  We reduced to the case of a finite open cover $\{X_{a_i}\}$
of $X$ with $(a_i)A=A$ and we want to show that 
\begin{align}
	\begin{diagram}
		A & \rTo^{a\mapsto (a)}& \prod_i A_{a_i} &\rTo^{(s_i)\mapsto (s_j-s_k)} & \prod_{(j,k)} A_{a_j a_k}  
	\end{diagram}
\label{seq}
\end{align}
is left exact.  Set $A'=\prod_{i=1}^n A_{a_i}$.  Then since localization and finite products preserve faithful flatness,
$A'$ is faithfully flat over $A$.  (N.B. the geometric meaning of $A'$ faithfully flat over $A$ is that $\Spec A'\rightarrow\Spec A$
is surjective).
One checks that the diagram (\ref{seq}) is equivalent to 
\begin{align}
	\begin{diagram}
		A & \rTo & A' &\rTo^{a'\mapsto a'\otimes 1-1\otimes a'} & A'\otimes_A A'
	\end{diagram}
\label{seq1}
\end{align}
Therefore, it suffices to prove the more general result:

\begin{theorem}
	Let $A$ be a ring and $A'$ a faithfully flat $A$-algebra.  Then $a'\in A'$ comes from $A$ if and only if $a'\otimes 1=1\otimes a'$
	in $A'\otimes_A A'$.
\end{theorem}

\begin{proof}
	First suppose that we have a section $\sigma:A'\rightarrow A$, {\em i.e.} that we have $A'=A\oplus I$ with I some ideal
	of $A'$.  Then the map $A'\rightarrow A'\otimes_A A'$ is given by
	\begin{diagram}
		A\oplus I & \rTo^{a+i\mapsto (a,i,0,0)-(a,0,i,0) = (0,i,-i,0)} & A\oplus (I\otimes_A A)\oplus (A\otimes_A I) \oplus (I\otimes_A I),
	\end{diagram}
	so that the image of $a+i$ in $A'\otimes_A A'$ is zero if and only if $i=0$.
	
	Now we reduce to this case.  We wish to show that the sequence of $A$-modules
	\begin{diagram}
		A & \rTo^{a\mapsto a} & A' & \rTo^{a'\mapsto a'\otimes 1-1\otimes a'} A'\otimes_A A'
	\end{diagram}
	is left exact.  For any $B$ faithfully flat over $A$, such exactness occurs if and only if
	\begin{diagram}
		B\otimes_A A & \rTo^{b \otimes 1\mapsto b\otimes 1} & B\otimes_A A' & \rTo^{b'\mapsto b'\otimes 1-1\otimes b'} B\otimes_A (A'\otimes_A A')=(B\otimes_A A')\otimes_B (B \otimes_A A')
	\end{diagram}
	is exact.  Thus, with $B'=B\otimes_A A'$,  we need only check exactness of
		\begin{diagram}
		B& \rTo^{b \mapsto b\otimes 1} & B' & \rTo^{b'\mapsto b'\otimes 1-1\otimes b'} B'\otimes_B B'. 
	\end{diagram}
	Taking $B=A'$, we obtain a section $\sigma A'\otimes_A A'\rightarrow A'=A'\otimes_A A$
	given by $\sigma(a_1'\otimes a_2') = a_1'a_2'$ (which evidently composes with $a'\mapsto a'\otimes 1$
	to give the identity on $A'$).  Thus, we are in the situation above and we are done.
\end{proof}

\section{$\Spec$ of a ring}

We can now define $\Spec A$:

\begin{definition}
	$\Spec A$ is the ringed space whose underlying topological space is the prime spectrum of the ring $A$
	and whose structure sheaf is given on a base of opens $D(f)$ by $D(f)\mapsto A_f$.
\end{definition}

\begin{theorem} Let $(X,\O_X)=\Spec A$.
	\begin{enumerate}
		\item For $a\in A$ there exists a canonical isomorphism $A_a\simeq \Gamma(X_a,\O_X)$, where
		$X_a=\{x\in X : a(x)\neq 0\ \text{in}\ k(x)\}$.
		
		\item For $x=\p\in X$ we have $\O_{X,x}\simeq A_{\p}$, so that $X$ is a locally ringed space.
	\end{enumerate}
\end{theorem}

\begin{proof}
	\begin{enumerate}
		\item This was how $\O_X$ was constructed as a $\B$-sheaf.
		
		\item We have 
		$$\O_{X,x}=\varinjlim_{U\ni x} \Gamma(U,\O_X)=\varinjlim_{X_a \ni x} A_a=\varinjlim_{a\not\in\p}A_a=A_{\p}$$
		by the universal mapping property of the localization $A_{\p}$,
		since the transition maps defining the direct limit are the natural localization maps.
	\end{enumerate}
\end{proof}

\begin{definition}
	An {\em affine scheme} is a locally ringed space isomorphic (as a locally ringed space) to $\Spec A$
	for some ring $A$.
\end{definition}

\begin{definition}
	A {\em scheme} is a locally ringed space $(X,\O_X)$ such that there exists an open cover $\{U_i\}$
	of $X$ with $(U_i,\O_X\big|_{U_i})$ affine schemes.
\end{definition}

We have already seen that schemes and affine schemes are full subcategories of the category 
of locally ringed spaces.

\begin{lemma}
	Any scheme has a base of affine opens.
\end{lemma}

\begin{proof}
	It is enough to treat the affine case, say $X=\Spec A$.  But $X$ has a base of opens
	$(X_a,\O_X\big|_{X_a})\simeq \Spec A_a$, as is evident from the definition of
	the structure sheaf of $\Spec A_a$.
\end{proof}

\begin{corollary}
	If $X$ is a scheme and $U\subseteq X$ open, then $(U,\O_X\big|_{U})$ is a scheme.
\end{corollary}

Now let $\varphi:A\rightarrow B$ be a map of rings.  Then we get a map of topological spaces
$f:X=\Spec B\rightarrow Y=\Spec A$ defined by $f(\q)=\varphi^{-1}(\q)$.  We want to 
define a map $f^{\#}:\O_Y\rightarrow f_*\O_X$.  For this, it is enough to 
find a base $\B$ of opens in $Y$ and maps $\O_Y(U)\rightarrow \O_X(f^{-1}(U))$
for all $U\in \B$ compatible with restriction.
We therefore take $\B=\{Y_a\}$ and we have $f^{-1}(Y_a)= X_{\varphi(a)}$, so that we must define
a map $A_a\rightarrow B_{\varphi(a)}$ compatible with restriction.  To do this, we take as our map the
map induced by $\varphi$ via the commutative diagram
\begin{diagram}
	A & \rTo^{\varphi} & B\\
	\dTo & & \dTo\\
	A_a & \rTo^{\exists !} & B_{\varphi_a}.
\end{diagram}
It is straightforward to check that this map is compatible with restriction.  Moreover, we claim that 
the induced map 
$$A_{\varphi^{-1}(\q)}=\O_{Y,f(x)}\rightarrow (f_*\O_{Y})_x\rightarrow \O_{X,x}=B_{\q}$$
is {\em local}.  Indeed, this follows easily from the definition of $B_{\q}$ as $\varinjlim_{b\not\in \q} B_b$,
or from the fact that the {\em only} map $A_{\varphi^{-1}(\q)}\rightarrow B_{\q} $ making 
\begin{diagram}
	A & \rTo^{\varphi} & B\\
	\dTo &  &\dTo \\
	A_{\varphi^{-1}(\q)} &\rTo  & B_{\q} 
\end{diagram}
commute {\em is} the local map induced by $\varphi$ (via the universal property of $A_{\varphi^{-1}(\q)}$).

\begin{theorem}
	Let $A,B$ be rings.  Then there is a bijection
	$$\Hom_{\rm ring}(A,B)\longleftrightarrow \Hom_{\rm l.r. spaces}(\Spec B,\Spec A).$$

\end{theorem}

\begin{proof}
	We wish to show that the composite functor
	\begin{diagram}
		\Hom(A,B) & \rTo^{\Spec} & \Hom(\Spec B,\Spec A) & \rTo^{\Gamma(\cdot,\O_{\cdot})} & \Hom(A,B)
	\end{diagram}
	is the identity.  Suppose that under the map $ \Hom(\Spec B,\Spec A)\rightarrow \Hom(A,B)$
	the map $f$ is carried to $\varphi.$  We must show that $f=\Spec\varphi$.  It is enough
	to show that for two maps 
	\begin{diagram}
		Y=\Spec B & \pile{\rTo^{(f,f^{\#})}\\ \rTo_{(g,g^{\#})}} & \Spec A=X
	\end{diagram} 
	which agree on global sections we have $f=g$ and $f^{\#}=g^{\#}$.  First assume that $f=g$.  Then 
	we want $f^{\#}=g^{\#}$, and it is enough to check this holds on a base of opens.  
	But the uniqueness of the induced map $A_a=\O_X(X_a)\rightarrow \O_Y(f^{-1}X_a)=B_{f^{\#}(a)}$
	in the diagram below shows that if $f^{\#},g^{\#}$ agree on global sections then they agree on a base of opens.
	\begin{diagram}
		A & \rTo & B\\
		\dTo & & \dTo\\
		A_{a} & \rTo & B_{f^{\#}(a)}.
	\end{diagram}
	Thus, we need only check that $f=g$ (as topological maps).  Again, there is a unique {\em local} map
	$A_{\varphi^{-1}(\q)}\rightarrow B_{\q}$ making the diagram
	\begin{diagram}
					A & \rTo^{\varphi} & B\\
		\dTo & & \dTo\\
		A_{\varphi^{-1}(\q)} & \rTo & B_{\q}.		
	\end{diagram}
	commute.  Thus, since $f^{\#}$ and $\varphi$ agree as maps $A\rightarrow B$,
	we must have $f(\q)=\varphi^{-1}(\q)$, so that $f,\Spec\varphi$ agree as topological maps.
\end{proof}

We will see next time that this bijection holds when $\Spec B$ is replaced by an arbitrary scheme $X$
and $B$ by $\Gamma(X,\O_X)$.








\end{document}