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\DeclareMathOperator{\Spec}{Spec}
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\DeclareMathOperator{\Hom}{Hom}
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\begin{document}
\title[Part VI]{Morphisms}
\author{Kapil Hari Paranjape}
\maketitle
%\appendix
\section{Morphisms}
When we defined affine schemes as functors on the category of Finite
rings we were a bit vague on what exactly morphisms were. In the case of
projective schemes we didn't even try to define the notion. A morphism
of schemes $f:X\to Y$ is a natural transformation of functors such that
the graph is a closed subscheme. Recall that a natural transformation of
functors on the category of finite rings gives a map $f(A):X(A)\to Y(A)$ for
every finite ring $A$ and for every morphism $A\to B$ the following
diagram commutes.
\begin{eqnarray*}
X(A) & \to Y(A) \\
\downarrow && \downarrow \\
X(B) & \to & Y(B)
\end{eqnarray*}
The product $X\times Y$ has been given a natural structure of a scheme.
The graph $\Gamma_{f(A)}$ is a sub-functor of $X(A)\times
Y(A)=(X\times Y)(A)$ and so it makes sense to require that there is a
closed subscheme $Z$ of $X\times Y$ such that $Z(A)=\Gamma_{f(A)}$ for
all $A$.
We note that for any pair of rings $R$ and $S$, the product
$\Spec(R)\times\Spec(S)$ is just $\Spec(R\tensor S)$. We have so far
been using ring homomorphisms $R\to S$ to obtain morphisms
$\Spec(S)\to\Spec(R)$; it is clear that this gives a natural
transformation of functors.. Note that such a ring homomorphism gives a
surjective homomorphism $R\tensor S \to S$ by ``multiplication''. This
gives $\Spec(S)$ as a closed sub-scheme of $\Spec(R)\times\Spec(S)$ and
this closed sub-scheme is the graph of the above natural transformation.
We will now prove the converse. First of all, exactly as above, we note
that what is required in oder to obtain a morphism is a closed subscheme
$Z$ of $\Spec(R)\times\Spec(S)$ such that $Z(A)\to\Spec(S)(A)$ is a
bijection for all finite rings $A$ (the graph of a map of sets projects
bijectively onto the the domain of the map). Thus we will conclude our
proof if we can show that any such map is an isomorphism. Since $Z$ is
itself the closed sub-scheme of an affine scheme it too is $\Spec(S')$
for some ring $S$; moreover, we have a natural homomorphism $S\to
R\tensor S$ that induces the projection. Hence we have reduced our
problem to the following statement. Let $S\to S'$ be a homomorphism of
finitely generated rings such that for every finite ring $A$ the
resulting map $\Hom(S',A)\to\Hom(S,A)$ is a bijection, then $S\to S'$ is
an isomorphism.
The proof of this statement will be in many parts. First of all we will
prove a condition for finite-ness. We will use this to show that $S\to
S'$ is finite. An application of Nakayama's lemma and another
finite-ness theorem will then conclude the result.
\subsection{Finite-ness}
Let $f:R\to S$ be a homomorphism of rings such that the induced
homomorphism $f:R[X]\to S[X]$ is closed. Recall that this means that for
any ideal $I$ in $S[X]$, and any maximal ideal $M$ in $R[X]$ containing
$f^{-1}(I)$, there is a maximal ideal $N$ in $S[X]$ such that
$f(M)\subset N$. We claim that this condition implies that $R\to S$ is
finite.
It is enough to show that any element $a$ in $S$ satisfies a monic
polynomial with coefficients in $R$. Consider the ideal $(1-aX)$ in
$S[X]$; if $P(X)$ lies in $f^{-1}((1-aX))$, and
$Q(X)=X^{\deg(P)}P(X^{-1})$ then $P(a)=0$. So we need to show that
$f^{-1}((1-aX))$ contains a monic polynomial. Let $J$ be the ideal in
$R$ generated by the leading coefficient of elements of
$f^{-1}((1-aX))$. This ideal is the image in $R$ of the ideal
$f^{-1}((1-aX))$ under the map $R[X]\to R$ that sends $X$ to 0. Now, if
$J$ is not the unit ideal then its inverse image $f^{-1}((1-aX))+(X)$
must be contained in a maximal ideal $M$ of $R[X]$. By the closed-ness,
there is a maximal ideal $N$ of $S[X]$ that lies over it and contains
$((1-aX))$. But then $N$ contains $1-aX$ and $X$ as well; a
contradition. Thus $J$ must be the unit ideal as required.
\subsection{Finite rings}
Let $R\to S$ be a morphism of finitely generated rings with
$\Hom(S,A)\to\Hom(R,A)$ a bijection for all finite rings $A$.
We now prove the result in case one of the rings is finite.
Now if $R$ is finite then we can take $A=R$ and obtain a bijection
$\Hom(S,R)\to\Hom(R,R)$. In particular, we obtain a homomorphism $S\to
R$ such that the composite $R\to S\to R$ is identity on $R$. Consider
the homomorphism $S\to R$ and the induced map $\Hom(R,A)\to\Hom(S,A)$.
By composing this with the bijection $\Hom(S,A)\to\Hom(R,A)$ we obtain
the identity map $\Hom(R,A)\to\Hom(R,A)$. It follows that
$\Hom(R,A)\to\Hom(S,A)$ is a bijection as well. Thus if $S$ is finite as
well, we see that $R\to S$ is an isomorphism as required.
So we can assume that we are in the situation of the theorem where in
addition $S$ is finite but we are not sure if $R$ is. Let $M$ be any
maximal ideal of $R$ of $S$. As proved in the previous section $R/M$ is
finite and so taking $A=R/M$ we have a natural homomorphism $R\to
A=R/M$. By assumption this factors as $R\to S\to A$, so $f(M)$ is
contained in the kernel of $S\to A$ which is a maximal ideal $N$ of $S$.
Thus every maximal ideal in $R$ is of the form $f^{-1}(N)$ for some
maximal ideal $N$ of $S$. Since $S$ is finite it has only finitely many
maximal ideals; thus $R$ too has only finitely many maximal ideals. As
we have seen in the previous section this means that $R$ is finite.
\subsection{Co-finite ideals}
Let $R\to S$ be as before. Let $I$ be an ideal of $R$. We obtain a
homomorphism $R/I\to S/f(I)S$; consider the map
$\Hom(S/f(I)S,A)\to\Hom(R/I,A)$. The former term is contained in
$\Hom(S,A)$ and the latter in $\Hom(R,A)$ thus the map is injective. On
the other hand a homomorphism $R/I\to A$ is the same as a homomorphism
$R\to A$ which is zero on $I$. By assumption, this factors as a
homomorphism $R \to S\to A$ which means that the homomorphism $S\to A$
is 0 on $f(I)$. Thus $\Hom(S/f(I)S,A)\to\Hom(R/I,A)$ is a bijection.
In particular, if $I$ is such that $R/I$ is finite, we see that $R/I\to
S/f(I)S$ is an isomorphism.
Now suppose that $J$ is an ideal of $S$ such that $S/J$ is finite and put
$I=f^{-1}(J)$ so that $R/I\to S/J$ is injective. Then $R/I$ is finite
and so $R/I\to S/f(I)S$ is an isomorphism. But $J$ contains $f(I)S$ and
the intersection of $J/f(I)S$ with $R/I$ is 0. Thus $J=f(I)S$.
What we have proved is that the maps $J\mapsto f^{J}$ and $I\mapsto
f(I)S$ induce a bijection between co-finite ideals in $R$ and $S$.
Moreover, we have $R/I\to S/J$ is an isomorphism in this situation.
\subsection{Relative Noether Normalisation}
Let $R$ be a finitely generated domain and $f:R\to S$ be any injective
morphism with $S$ finitely generated as an $R$-algebra. Let $X_1$,
$X_2$. \dots, $X_n$ be a maximal subset (of some fixed set of
generators) of $S$ that is transcendental over the quotient field of
$R$. We have an injective morphism $R_1=R[X_1,\dots,X_n]\to S$ and for
any generator $g$ of $S$, there is a polynomial $P(T)\in R_1[T]$ and
$a\in R_1$ with $a\neq 0$ such that $P(ag)=0$. Collecting denominators,
there is a non-zero element $d$ in $R_1$ such that $(R_1)_d\to S_d$ is
finite and injective.
Let $d_0$ be any coefficient of $d$ that is non-zero. As $R$ is a
finitely generated domain there is a maximal ideal $M$ that does not
contain $d_0$. Now $M[X_1,\dots,X_n]$ is a prime ideal $P$ in $R_1$
that $d$ does not lie in. Hence it gives a prime ideal in $(R_1)_d$.
Since $(R_1)_d\to S_d$ is finite and injective we have $PS_d\cap
(R_1)_d=P_d$. It follows that, if $K$ denotes $R/M$, $K[X_1,\dots,
X_n]_d\to (S_d)/(PS_d)$ is finite and injective. If $n>0$ then this
would mean that there are infinitely many maximal ideals is $S_d$ (and
hence in $S$) that lie over $M$.
If particular, applying this in our situation $R\to S$ as above, we see
that $n=0$; equivalently $R_1=R$. In other words, if $R$ is a domain and
$f:R\to S$ is an injective homomorphism that induces a bijection between
co-finite ideals in $R$ and $S$ then there is a non-zero element $d$ in
$R$ so that $R_d\to S_d$ is finite.
If $Q'$ is any prime such that $Q'\cap R=(0)$, the element $d$ does not
lie in $Q'$ so that $Q'_d$ is a prime in $S_d$. Since $R_d\to S_d$ is
finite it follows that $Q'_d$ is minimal in $S_d$; hence $Q'$ is minimal
in $S$ as well.
Suppose that $Q$ and $Q'$ are two distinct minimal primes in $S_d$. Let
$a$ be an element of $Q'$ that is not in $Q$ and consider a maximal
ideal $N$ in $S_d$ whose image in $S_{ad}$ is a non-trivial maximal
ideal containing the $Q_{ad}$. Let $M=N\cap R$; by the going up theorem
we can find a maximal ideal $N'$ in $S_d$ that contains $Q'$. Since $a$
lies in $N'$ and not in $N$, these are distinct maximal ideals in $S_d$
over $M$. This contradicts the bijection between co-finite ideals in $R$
and $S$.
The minimal prime $Q$ in $S$ such that $Q\cap R=(0)$ is thus unique. We
now want to show that there are no minimal primes of the other type
either.
\subsection{Krull's Intersection Theorem}
Let $M$ be any maximal ideal in a finitely generated ring $R$. Consider
the graded ring $\tilde{R}=R\oplus M\oplus M^2 \oplus \cdots$. Let $I=\cap M^n$;
then $\tilde{I}=I\oplus I \oplus \cdots$ is a graded ideal in $\tilde{R}$. In
particular, it is finitely generated. Let $x_1$, \dots, $x_n$ be a
homogeneous system of generators and $d$ be larger than all their
degrees. Expressing elements of degree $d$ in $\tilde{I}$ in terms of
the generators, we conclude that $I=MI$. By Nakayama's lemma we get
$I=0$.
Let $f:R\to S$ be a homomorphism of finitely generated rings which
induces a bijection between co-finite ideals such that $R/I\to S/f(I)S$
is an isomorphism for every co-finite ideal $I$. Let $M$ be any maximal
ideal in $R$ and $N=f(M)S$ be the corresponding maximal ideal in $S$. We
have an isomorphism $R/M^k\to S/N^k$ for all $k$. Applying the
intersection theorem we conclude that $\ker(f)=(0)$.
Now let us assume in addition that $R$ is a domain. If possible, let $Q$
be a minimal prime in $S$ such that $Q\cap R\neq (0)$. Let $a$ be an
element of the product of the remaining minimal primes that is not in
$Q$. Then $Q_a$ is the unique minimal prime in $S_a$ so it consists of
nilpotent elements. Since $R$ is a domain, we see that $R\to S_a$ has a
non-trivial kernel. Let $N$ be a maximal ideal of $S$ not containing $a$
and $M=f^{-1}(N)$. Then $(S_a)/(N^kS_a)=S/N^k$ for all $k$ so that
$R/M^K\to (S_a)/(N^kS_a)$ is an isomorphism for all $k$. But this
contradicts the existence of a kernel for $R\to S_a$.
\subsection{Closed-ness}
Let $R\to S$ be a morphism of finitely generated rings that induces a
bijection between co-finite ideals as above. Let $Q$ be a prime ideal in
$S$ and $P=f^{-1}(Q)$. We can consider the map $R/P\to S/f(P)S$ which
has the same properties as $R\to S$; in addition $R/P$ is a domain
and the map is injective. Replacing $R$ by $R/P$ and $S$ by $S/f(P)S$ we
see that we have an injective homomorphism $R\to S$ that induces a
bijection between co-finite ideals and $R$ is a domain. Moreover, $Q$ is
a prime ideal in $S$ such that $Q\cap R=(0)$. Note that we have shown
above that there is a non-zero $d$ in $R$ such that $R_d\to S_d$ is
finite.
In this context we would like to show that given any maximal ideal $M$
in $R$, the maximal ideal $N$ in $S$ that lies over it also contains
$Q$. By what we have proved above, $Q$ is the unique minimal prime in
$S$. Then $N$ must contain $Q$.
Now, let $J$ be any ideal in $S$ and $I=f^{-1}(J)$; clearly
$f^{-1}(\sqrt{J})=\sqrt{I}$. Let $P$ be a minimal
prime containing $I$ then $P$ contains the intersection of the prime
ideals $f^{-1}(Q)$ where $Q$ runs over the minimal primes containing
$J$; it thus equals one of them. Let $M$ be a maximal ideal in $R$ that
contains $I$. Then $M$ contains a minimal prime ideal $P$ for $I$; let
$Q$ be the minimal prime containing $J$ such that $f^{-1}(Q)=P$. As
shown above $N=f(M)S$ contains $Q$ and this contains $J$ as well. Hence
we have shown that $R\to S$ is closed.
\subsection{Conclusion}
What we have shown above is that if $R\to S$ is a homomorphism of
finitely generated rings such that $\Hom(S,A)\to\Hom(R,A)$ is a
bijection for all finite rings $A$ then $R\to S$ is closed. Now,
consider the induced homomorphism $R[X]\to S[X]$. Since
$\Hom(S[X],A)=\Hom(S,A)\times A$ and $\Hom(R[X],A)=\Hom(R,A)\times A$
we see that $\Hom(S[X],A)\to\Hom(R[X],A)$ is also a bijection. Then
$R[X]\to S[X]$ is also closed; this means that $R\to S$ is finite as
demonstrated earlier. Now $S/R$ is a finitely generated $R$-module such
that $(S/R)/M(S/R) =(0)$ for all maximal ideals $M$ in $R$. By
Nakayama's lemma we see that $S/R=0$. We have already seen that $R\to S$
is injective. Thus it follows that $R\to S$ is an isomorphism as
required.
\end{document}