\section{Comparison with ``classical'' definition}
In order to compare the given definition of schemes with the
``classical'' one, we will prove the following theorem:
\begin{thm}\label{main}
Let $f:R\to S$ be a homomorphism between finitely generated rings so
that for every finite ring $A$, the induced map
$\Hom(S,A)\to\Hom(R,A)$ is a bijection. Then $f$ is an isomorphism.
\end{thm}
In the paragraphs below $R$ and $S$ will always denote rings
satisfying the conditions of the theorem. We first prove a special
case:
\begin{lemma}
Let $f:R\to S$ be a homomorphism of finite rings so that for every
finite ring $A$ the induced map $\Hom(S,A)\to\Hom(R,A)$ is a
bijection. Then $f$ is an isomorphism.
\end{lemma}
\begin{proof}
Taking $A=R$ we see that there is a homomorphism $g:S\to R$ such
that the composite $g\circ f:R\to S\to R$ is identity. For any
finite ring $A$, consider the chain of maps
\[ \Hom(R,A) \to \Hom(S,A) \to \Hom(R,A) \]
The second map is a bijection by assumption. The composite is the
identity and in particular, a bijection. It follows that
$\Hom(R,A)\to\Hom(S,A)$ is a bijection as well. Now, taking $A=S$ we
see that we also have a homomorphism $h:R\to S$ so that the
composite homomorphism $S\by{g}R\by{h}S$ is the identity. We then
have
\[ f = \id_S\circ f = h\circ g\circ f = h \circ \id_R = h\]
Thus $f\circ g =\id_S$ and $g\circ f =\id_R$, hence $f$ and $g$ are
isomorphisms.
\end{proof}
Next, we show that the above condition is ``inherited'' by quotients.
\begin{lemma}
Let $f:R\to S$ be as above. Let $I$ be an ideal in $R$, then we
obtain a homomorphism $R/I\to S/f(I)S$. For any finite ring $A$, the
induced map $\Hom(S/f(I)S,A)\to\Hom(R/I,A)$ is a bijection.
\end{lemma}
\begin{proof}
Consider the diagram
\[
\begin{array}{lcr}
\Hom(S,A) & \to & \Hom(R,A) \\
\uparrow & & \uparrow \\
\Hom(S/f(I)S,A) & & \Hom(R/I,A)
\end{array}
\]
The top row is a bijection. Let $g:S/f(I)S\to A$ be any element in
the bottom left corner then the corresponding element $h:S\to A$ in
the top left corner satisfies $h(f(I)S)=0$. Thus $h\circ f:R\to A$
satisfies $h\circ f(I)=0$. Thus it factors through a homomorphism
$e:R/I\to A$. Thus we see that the elements in the bottom left
corner are mapped to elements in the bottom right corner.
Conversely, let $g:R/I\to A$ be an element in the bottom right
corner and $h:R\to A$ be its image in the top right corner; then
$h(I)=0$. By assumption there is a homomorphism $e:S\to A$ such that
$h=e\circ f$. It follows $e(f(I))=0$ and thus $e(f(I)S)=0$. Thus $e$
factors through an element $d:S/f(I)S\to A$ in the bottom left
corner. In other words we have a bijection
$\Hom(S/f(I)S,A)\to\Hom(R/I,A)$.
\end{proof}
Combining the above two lemmas we see that if $I$ is any ideal in
$R$ such that $R/I$ and $S/f(I)S$ are finite, then the map $R/I\to
S/f(I)S$ is an isomorphism. We will now show that if $R/I$ is finite
then $S/f(I)S$ is ``automatically'' finite as well.
\begin{lemma}
Let $f:R\to S$ be as in the theorem. For any maximal ideal $m$ in
$R$, the ideal $f(m)S$ in $S$ also a maximal ideal.
\end{lemma}
\begin{proof}
Since $R$ is finitely generated $R/m$ is a finite field by Hilbert's
Nullstellensatz. Thus $\Hom(S,R/m)\to\Hom(R,R/m)$ is a bijection and
so the homomorphism $R\to R/m$ must factor through $S$; moreover,
this factorisation is unique. Let $n$ be the kernel of this
factorisation. Then $n$ is a maximal ideal containing $f(m)S$ such
that $R/m\to S/n$ is an isomorphism. Now, let $n'$ be any maximal
ideal in $S$ containing $f(m)S$. Then, the composite $R\to S\to
S/n'$ factors through $R/m$. Thus, $S/n'$ is a finite field
extension of $R/m$. If this extension has degree $>1$ then if $q$ is
the cardinality of $R/m$, the map $x\mapsto x^q$ is a non-trivial
automorphism of $S/n'$ which is identity on $R/m$. Thus we obtain
two maps $S\to S/n'$ which restrict to the same map $R\to S/n'$
contradicting the hypothesis. Thus $R/m\to S/n'$ is an isomorphism.
But then this isomorphism gives a map $S\to R/m$ which restricts to
the natural map $R\to R/m$; there is a unique such map by
hypothesis. Since that map has kernel $n$, we see that $n'=n$.
In other words, we see that $f(m)S$ is contained in a unique maximal
ideal $n$ in $S$. Thus $S/f(m)S$ is an Artinian ring. By the earlier
discussion we see that $R/m\to S/f(m)S$ is an isomorphism. In other
words $f(m)S=n$ is a maximal ideal for every maximal ideal $m$ in
$R$. Conversely, if $n$ is any maximal ideal in $S$, then
$f^{-1}(n)=m$ is the kernel of the composite $R\to S\to S/n$ which
is a map to a finite field; hence $m$ is a maximal ideal. It follows
that {\em every} maximal ideal in $S$ is of the form $f(m)S$ for a
maximal ideal $m$ in $R$.
\end{proof}
Now, if $I$ is any ideal such that $R/I$ is finite then there are
finitely many maximal ideals $m_1$, \dots, $m_k$ and positive integers
$r_1$, \dots, $r_k$ such that $I\contains m_1^{r_1}\cdot
m_2^{r_2}\cdots m_k^{r_k}$. As seen above $n_i=f(m_i)S$ is a
maximal ideal. The relations
\[
f(I)S \contains f(m_1^{r_1}\cdots m_k^{r_k})S =
n_1^{r_1}\cdots n_k^{r_k}
\]
shows that the ring $S/f(I)S$ is finite as well. It follows that for
any ideal $I$ such that $R/I$ is finite, the map $R/I\to S/f(I)S$ is
an isomorphism.
On the other hand suppose $J$ is any ideal in $S$ such that $S/J$ is
finite and let $I=f^{-1}(J)$; then $R/I$ is a subring of $S/J$ and
thus also finite. We have seen above that this implies that $R/I\to
S/f(I)S$ is an isomorphism. But the inverse image of $J/f(I)S$ under
this is the zero ideal in $R/I$. Thus we have $J=f(I)S$. To
summarise,
\begin{lemma}\label{ideal}
Let $f:R\to S$ be as in the conditions of the theorem. The map
$I\mapsto f(I)S$ is a one-one correspondence between ideals of
finite index in $R$ and ideals of finite index in $S$. The map
$J\mapsto f^{-1}(J)$ is the inverse correspondence from ideals $J$
in $S$ to ideals in $R$. Moreover, the natural homomorphism $R/I\to
S/f(I)S$ is an isomorphism for such ideals.
\end{lemma}
Thus the original condition has been re-stated intrinsically in terms
of ideals. Next we wish to prove that the given homomorphism is
``closed''. That is to say given a prime ideal $Q$ in $S$, let
$m$ be a maximal ideal in $R$ that contains the prime ideal
$P=f^{-1}(Q)$. We wish to prove that there is a maximal
ideal $n$ in $S$ which contains $Q$ and satisfies $f^{-1}(n)=m$.
To do this we can restrict our attention to $R/P\to
S/f(Q)S$. Since $f^{-1}(f(P)S)\subset
f^{-1}(Q)=P$, the latter homomorphism is also injective.
\begin{lemma}
Let $f:R\to S$ be an injective homomorphism of finitely generated
rings with $R$ a domain. We have a factoring of $f$ as follows
\[ R \to R[X_1,\dots,X_a] =R_1 \to R_1[t_1,\dots,t_b] =R_2 \to S \]
where
\begin{enumerate}
\item $R_1$ is a polynomial ring over $R$.
\item There is a non-zero element $r$ of $R_1$ such that for each
$i$, the element $rt_i\in R_2$ satisfies a monic polynomial over
$R_1$. Other than this relation there are no further relations
among the $t_i$ in $R_2$.
\item $R_2\to S$ is the quotient by an ideal that intersects $R_1$
in the zero ideal.
\end{enumerate}
\end{lemma}
\begin{proof}
Since $S$ is finitely generated we can choose a maximal collection
of elements $X_1$, \dots, $X_a$ of $S$ that are algebraically
independent over (the quotient field of) $R$. Then
$R_1=R[X_1,\dots,X_a]$ is the polynomial ring over $R$ and is a
subring of $S$. The remaining generators of $S$ are algebraically
dependent on the $X_i$'s. Thus each of them satisfies an equation of
the form $r_0T^d+r_1T^{d-1}+\dots+r_d$ for some elements $r_j$ in
$R_1$. Moreover, we can assume that $r_0$ is non-zero in such an
equation. Let $r$ be the product in $R_1$ of polynomials $r_0$
corresponding to different generators of $S$. Since $R$ is a
domain, so is $R_1$ and the polynomial $r$ is non-zero. For each
generator $S$ choose a polynomial of the above form with leading
coefficient $r$ (one such such clearly exists) and let $R_2$ be the
ring obtained from $R_1$ by adjoining the roots of these equations.
We have a natural map $R_2\to S$; let $\gotha$ be the kernel. Since
$R_1\to S$ factors through $R_2$ and is injective, it follows that
$\gotha$ intersects $R_1$ in the zero ideal.
\end{proof}
Let $Q_1$, \dots, $Q_r$ be the minimal primes in $S$ or equivalently a
minimal primes in $R_2$ that contains the kernel of $R_2\to S$. Since
$R_1$ meets this kernel in the zero ideal, the intersection of the
prime ideals $Q_i\cap R_1$ in $R_1$ is a nilpotent ideal. Since $R_1$
is a domain there is an index $i$ such that $Q_i\cap R_1=(0)$. Let $Q$
denote the prime ideal $Q_i$ for any such index $i$.
Let $m$ be a maximal ideal in $R$ such that $r$ is not contained in
the prime ideal $m[X_1,\dots,X_a]$ of $R_1$. Since $R_1$ is a domain
we see that $Q_r\cap (R_1)_r$ is the zero ideal. Now, $(R_2)_r$ is a
finite free module over $(R_1)_r$ and so (by the going up theorem)
there is a prime ideal $Q'$ in $R_2$ which contains $Q$ and restricts
to $m[X_1,\dots,X_a]$ in $R_1$. Similarly, for any maximal ideal $n'$
in $R_1$ that contains $m[X_1,\dots,X_a]$ and does not contain $r$,
there is a maximal ideal $n$ in $R_2$ that contains $Q'$ (and hence
$Q$) that lies over $n'$.
Now, if $a>0$ (i.~e.\ $R\neq R_1$) then there are at least two (in
fact infinitely many) such maximal ideals $n'$. But then we see that
we have at least two maximal ideals in $S$ that lie over a given
maximal ideal $m$ in $R$ contradicting lemma~\ref{ideal}. Thus we must
have $R=R_1$.
Again, if $\tilde{Q}$ is another minimal prime in $R_2$ that contains
the kernel of $R_2\to S$ and such that $\tilde{Q}\cap R_1=(0)$, then
as above we can find a prime ideal $\tilde{Q'}$ which contains
$\tilde{Q}$ and lies over $m$ and is distinct from $Q'$. Now there are
distinct maximal ideals $n'$ and $\tilde{n'}$ in $R_2$, that contain
$Q'$ and $\tilde{Q'}$ respectively. This again contradicts
lemma~\ref{ideal}. It follows that there is a {\em unique} minimal
prime $Q$ containing the kernel of $R_2\to S$ such that $Q\cap
R=(0)$.
Now suppose that $Q_0$ is another miminal prime in $S$, or
equivalently a minimal prime in $R_2$ that contains the kernel of
$R_2\to S$. We must have $Q_0\cap R\neq(0)$. However, we have the
lemma
\begin{lemma}
Let $f:R\to S$ be a homomorphism of finitely generated rings with
$R$ a domain. Let $Q$ be a minimal prime in $S$ such that
$f^{-1}(Q)$ is non-zero. Then there is a maximal ideal $n$ in $S$
and an integer $k$ such that if $m=f^{-1}(n)$, then $R/m^k\to S/n^k$
is not an isomorphism.
\end{lemma}
\begin{proof}
Let $x$ be an element of all the minimal primes of $S$ other than
$Q$. Replacing $S$ by its localisation $S_x$ at $x$, we can
assume that $Q$ is the unique minimal prime in $S$. Then
$Q$ consists of nilpotent elements. Since $f^{-1}(Q)$ is
non-zero and $R$ is a domain it follows that $R\to S$ has a non-zero
kernel. Now let $n$ be {\em any} maximal ideal in $S$ and
$m=f^{-1}(m)$. The homomorphism of local rings $R_m\to S_n$ has a
non-zero kernel. The result follows by the Artin-Rees lemma.
\end{proof}
On the other hand, for our given homomorphism $R\to S$ we know that
$R/m^k\to S/n^k$ must be an isomorphism for all $k$. It follows that
there is no such prime ideal $Q_0$ in $S$.
We have thus proved that there is a unique prime ideal $Q$ in $S$ that
lies over a given prime ideal $P$ in $R$ and $f^{-1}(Q)=P$. The
``closed''-ness condition is an immediate corollary.
Let us note that if $R[X]$ is the polynomial ring over a ring $R$,
then $\Hom(R[X],A)$ is naturally identified with $\Hom(R,A)\times
A$. Thus, if $g:R[X]\to S[X]$ denotes the natural extension of the
above homomorphism to the corresponding polynomial rings then, for any
finite ring the induced map $\Hom(S[X],A)\to\Hom(R[X],A)$ is a
bijection whenever $\Hom(S,A)\to\Hom(R,A)$ is a bijection. In
particular, we can apply the above lemmas to the homomorphism $g$ as
well.
\begin{lemma}
Let $f:R\to S$ be as in the theorem and $g:R[X]\to S[X]$ be the
induced homomorphism on polynomial rings in one variable. Let
$\alpha$ be any element of $S$ and $\gothb$ be the ideal
$(X-\alpha)S[X]$ in $S[X]$. Let $\gotha$ be the ideal
$g^{-1}((X-a)S[X])$. Then $\gotha$ contains a {\em monic}
polynomial.
\end{lemma}
\begin{proof}
Let $A$ be any ring and $\gotha$ be an ideal in the polynomial ring
$A[X]$. Let $\gotha_1$ denote the ideal $\gotha\cdot A[X,X^{-1}]$
in the ring $A[X,X^{-1}]$. We have
\[
\gotha_1 = \{ P(X)\cdot X^{-n} | P(X) \in \gotha \text{~and~}
n \geq 0 \text{~an integer~}
\}
\]
Let $\gotha_2$ be the restriction $\gotha_1\cap A[X^{-1}]$ of this
ideal to $A[X^{-1}]$. We have
\[
\gotha_2 = \{ P(X)\cdot X^{-d} | P(X) \in \gotha \text{~and~}
d = \deg(P(X))
\}
\]
The {\em content} $c(\gotha)$ of the ideal $\gotha$ is defined as
the image of $\gotha_2$ in $A[X^{-1}]/(X^{-1})=A$. Clearly,
\[
c(\gotha) = \{ a \in A | \exists P(X) \in \gotha
\text{~such that~}
P(X) = aX^d + \text{~lower degree terms~}
\}
\]
Returning to the rings $R$ and $S$ let us use the subscripts 1 and 2
to denote the above constructions applied to ideals in $R[X]$ and
$S[X]$; specifically to the ideals $\gotha$ and $\gothb$.
We want to show that the content $c(\gotha)$ of the ideal $\gotha$
in $R[X]$ is the unit ideal. Suppose that $c(\gotha)\subset m$ for
some maximal ideal $m$ in $R$. The ideal
$\tilde{m}=m[X^{-1}]+X^{-1}R[X^-1]$ is then a maximal ideal in
$R[X^{-1}]$ which contains $\gotha_2$. Moreover, by the above
description of $\gotha_2$ it is clear that
$\gotha_2=g_2^{-1}(\gothb_2)$, where $g_2:R[X^{-1}]\to S[X^{-1}]$ is
the natural homomorphism. Applying the ``going-up'' which has been
proved above, it follows that there should exist a prime ideal
$\tilde{p}$ containing $\gothb_2$ such that
$g_2^{-1}(\tilde{p})=\tilde{m}$. But $\gothb_2$ is the ideal
generated by $1-\alpha X^{-1}$ and $X^{-1}$ lies in
$\tilde{m}$. Thus $\tilde{p}$ would have to be the unit ideal which
contradicts its primality. It follows that $c(\gotha)$ is the unit
ideal.
\end{proof}
From this lemma we see that $S$ is integral over $R$. Now the
result that $R/m\to S/f(m)S$ is an isomorphism for all maximal ideals
$m$ implies theorem~\ref{main} by Nakayama's lemma.
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