books/cring/integrality.tex

\chapter{Integrality and valuation rings}
\label{intchapter}

The notion of integrality is familiar from number theory: it is similar to  
``algebraic'' but with the polynomials involved are required to be monic. In algebraic geometry, integral
extensions of rings correspond to correspondingly nice morphisms on the
$\spec$'s---when the extension is finitely generated, it turns out that the
fibers are finite. That is, there are only finitely many ways to lift a prime
ideal to the extension: if $A \to B$ is integral and finitely generated, then
$\spec B \to \spec A$ has finite fibers.

Integral domains that are \emph{integrally closed} in their quotient field will play an
important role for us. Such ``normal domains'' are, for example, regular in
codimension one, which means that the theory of Weil divisors
(see \cref{weildivsec}) applies
to them. It is particularly nice because Weil divisors are sufficient to
determine whether a function is regular on a normal variety.

A canonical example of an integrally closed ring is a valuation ring; we shall
see in this chapter that any integrally closed ring is an intersection of such.

\section{Integrality}

\subsection{Fundamentals}


As stated in the introduction to the chapter, integrality is a condition on
rings parallel to that of algebraicity for field extensions.
\begin{definition} \label{intdefn}
Let $R$ be a ring, and $R'$ an $R$-algebra.  An element $x \in R'$
is said to be \textbf{integral} over $R$ if $x$ satisfies a monic polynomial
equation in $R[X]$, say
\[ x^n + r_1 x^{n-1} + \dots + r_n = 0, \quad r_1, \dots, r_n \in R.  \]
We can say that $R'$ is \textbf{integral} over $R$ if every $x \in R'$ is
integral over $R$.
\end{definition}

Note that in the definition, we are not requiring $R$ to be a \emph{subring} of
$R'$.

\begin{example} \label{sixthroot}
$\frac{1+\sqrt{-3}}{2}$ is integral over $\mathbb{Z}$; it is in fact a sixth
root of unity, thus satisfying the equation $X^6  -1 = 0$.
However, $\frac{1+\sqrt{5}}{2}$ is not integral over $\mathbb{Z}$. To explain this,
however, we will
need to work a bit more (see \cref{onegeneratorintegral} below).
\end{example}


\begin{example} 
Let $L/K$ be a field extension. Then $L/K$ is integral if and only if it is
algebraic, since $K$ is a field and we can divide polynomial equations by the
leading coefficient to make them monic.
\end{example} 

\begin{example} 
Let $R$ be a graded ring. Then the subring $R^{(d)} \subset R$ was defined in
\cref{dpowerofring}; recall that this consists of elements of $R$ all of whose
nonzero homogeneous components live in degrees that are multiples of $d$.
Then the $d$th power of any homogeneous
element in $R$ is in $R^{(d)}$. As a result, every homogeneous element of $R$
is integral over $R^{(d)}$.
\end{example} 

We shall now interpret the condition of integrality in terms of finite
generation of certain modules.
Suppose $R$ is a ring, and $R'$ an $R$-algebra.  Let $x \in R'$.

\begin{proposition}  \label{onegeneratorintegral}
$x \in R'$ is integral over $R$ if and only if the subalgebra $R[x] \subset R'$
(generated by $R, x$) is a finitely generated
$R$-module.
\end{proposition}

This notation is an abuse of notation (usually $R[x]$ refers to a polynomial
ring), but it should not cause confusion. 

This result for instance lets us show that $\frac{1+\sqrt{-5}}{2}$ is not integral
over $\mathbb{Z}$, because when you keep taking powers, you get arbitrarily
large denominators: the arbitrarily large denominators imply that it cannot be
integral.

\begin{proof}
If $x  \in R'$ is integral, then $x$ satisfies
\[ x^n + r_1 x^{n-1}+\dots+r_n = 0, \quad r_i \in R.  \]
Then $R[x]$ is generated as an $R$-module by $1, x, \dots, x^{n-1}$.  This is
because the submodule of $R'$ generated by $1, x ,\dots, x^{n-1}$ is closed under
multiplication by $R$ and by multiplication by $x$ (by the above equation).

Now suppose $x$ generates a subalgebra $R[x] \subset R'$ which is a finitely
generated $R$-module.  Then the increasing sequence
of $R$-modules generated by $\{1\}, \left\{1, x\right\}, \left\{1, x,
x^2\right\}
, \dots$ must stabilize, since the union is $R[x]$.\footnote{As an easy
exercise, one may see that if a finitely generated module $M$ is the union of
an increasing sequence of submodules $M_1 \subset M_2 \subset M_3 \subset
\dots$, then $M  = M_n$ for some $n$; we just need to take $n$ large enough
such that $M_n$ contains each of the finitely many generators of $M$.}  It follows that some $x^n$
can be expressed as a linear combination of smaller powers of $x$. Thus $x$ is
integral over $R$.
\end{proof}

So, if $R'$ is an $R$-module, we can say that
an element $x \in R'$ is \textbf{integral} over $R$ if either of the following
equivalent conditions are satisfied:

\begin{enumerate}
\item There is a monic polynomial in $R[X]$ which vanishes on $x$.
\item $R[x] \subset R'$ is a finitely generated $R$-module.
\end{enumerate}

\begin{example}
Let $F$ be a field, $V$ a finite-dimensional $F$-vector space, $T: V \to V$ a
linear transformation. Then the ring generated by $T$ and $F$ inside
$\mathrm{End}_F(V)$ (which is a noncommutative ring) is finite-dimensional
over $F$.
Thus, by similar reasoning, $T$ must satisfy a polynomial equation with
coefficients in $F$ (e.g. the characteristic polynomial).
\end{example}

Of course, if $R'$ is integral over $R$, $R'$ may not be a finitely generated
$R$-module. For instance, $\overline{\mathbb{Q}}$ is not a finitely generated
$\mathbb{Q}$-module, although the extension is integral. As we shall see in
the next section, this is always the case if $R'$ is a finitely
generated $R$-\emph{algebra}.


We now will add a third equivalent condition to this idea of ``integrality,''
at least in the case where the structure map is an injection.

\begin{proposition} \label{thirdintegralitycriterion}
Let $R$ be a ring, and suppose $R$ is a subring of $R'$.
$x \in R'$ is integral if and only if there exists a
 finitely generated faithful $R$-module $M \subset R'$ such that $R \subset M$ and
 $xM \subset M$.
\end{proposition}
A module $M$ is \emph{faithful} if $x M = 0$ implies $x=0$. That is, the map
from $R$ into the $\mathbb{Z}$-endomorphisms of $M$ is injective.
If $R$ is a \emph{subring} of $R'$ (i.e. the structure map $R \to R'$ is
injective), then $R'$ for instance is a faithful $R$-module.
\begin{proof}
It's obvious that the second condition above (equivalent to integrality)
implies the condition of this
proposition. Indeed, one could just take $M = R[x]$.

Now let us prove that if there exists such an $M$ which is finitely generated,
then $x$ is integral. Just because $M$ is finitely generated, the
submodule $R[x]$ is not obviously finitely generated. In particular, this
implication  requires a bit of proof.


We shall prove that the condition of this proposition implies integrality.
Suppose $y_1, \dots, y_k \in M$ generate $M$ as $R$-module. Then multiplication
by $x$ gives an $R$-module map $M \to M$. In particular, we can write
\[ xy_i = \sum a_{ij} y_j  \]
where each $a_{ij} \in R$.
These $\left\{a_{ij}\right\}$ may not be unique, but let us make some choices;
we get a $k$-by-$k$ matrix $A \in M_k(R)$. The claim is that $x$ satisfies the
characteristic polynomial of $A$.

Consider the matrix
\[ (x 1 - A) \in M_n(R').  \]
Note that $(x1-A)$ annihilates each $y_i$, by the choice of $A$.
We can consider the adjoint $B = (x1  -A)^{adj}$.  Then
\[ B(x1 - A) = \det(x1 - A) 1.  \]
This product of matrices obviously annihilates each vector $y_i$.  It follows
that
\[ (\det(x1 - A) y_i = 0, \quad \forall i,  \]
which implies that $\det (x1-A)$ kills $M$. This implies that $\det (x1 -
A)=0$ since $M$ is faithful.

As a result, $x$ satisfies the characteristic polynomial.
\end{proof}

\begin{exercise}
Let $R$ be a noetherian
local domain with maximal ideal $\mathfrak{m}$. As we will define shortly, $R$
is \emph{integrally closed} if every element of the quotient field $K=K(R)$
integral over $R$ belongs to $R$ itself. Then if $x \in K$ and $x \mathfrak{m}
\subset \mathfrak{m}$, we have $x \in R$.
\end{exercise}
\begin{exercise} Let us say that an $A$-module is \emph{$n$-generated}
if it is generated by at most $n$ elements. 

Let $A$ and $B$ be two rings such that $A\subset B$, so that $B$ is an
$A$-module.

Let
$n\in\mathbb{N}$. Let $u\in B$. Then, the following four assertions
 are equivalent:

\begin{enumerate}
\item   There exists a monic polynomial
$P\in A\left[  X\right]  $ with $\deg P=n$ and $P\left(  u\right)  =0$.
\item  There exist a $B$-module $C$ and an
$n$-generated $A$-submodule $U$ of $C$ such that $uU\subset U$ and such that
every $v\in B$ satisfying $vU=0$ satisfies $v=0$. (Here, $C$ is an $A$-module,
since $C$ is a $B$-module and $A\subset B$.)
\item  There exists an $n$-generated
$A$-submodule $U$ of $B$ such that $1\in U$ and $uU\subset U$.
\item  As an $A$-module, $A[u]$ is spanned by $1, u, \dots,
u^{n-1}$.\end{enumerate}
\end{exercise} 


We proved this to show that the set of integral elements is well behaved.

\begin{proposition}
Let $R \subset R'$. Let $S = \left\{x \in R': x \text{ is integral over }
R\right\}$. Then $S$ is a subring of $R'$. In particular, it is closed under
addition and multiplication.
\end{proposition}
\begin{proof}
Suppose $x,y \in S$.
We can consider the finitely generated modules $R[x], R[y] \subset R'$
generated (as algebras) by $x$ over $R$. By assumption, these are finitely
generated $R$-modules. In particular, the tensor product
\[ R[x] \otimes_R R[y]  \]
is a finitely generated $R$-module (by \cref{fingentensor}).

We have a ring-homomorphism $R[x]\otimes_R R[y] \to R'$
which comes from the inclusions $R[x], R[y] \rightarrowtail R'$.
Let $M$ be the image of $R[x] \otimes_R R[y]$ in $R'$. Then $M$ is an
$R$-submodule of $R'$, indeed an $R$-subalgebra containing $x,y$.  Also, $M$ is
finitely generated. Since $x+y, xy\in M$ and $M$ is a subalgebra, it
follows that
\[ (x+y) M \subset M, \quad xy M \subset M.  \]
Thus $x+y, xy$ are integral over $R$.
\end{proof}

Let us consider the ring $\mathbb{Z}[\sqrt{-5}]$; this is the canonical
example of a ring where unique factorization fails. This is because \( 6 = 2
\times 3 = (1+\sqrt{-5})(1-\sqrt{-5}).  \)
One might ask: what about $\mathbb{Z}[\sqrt{-3}]$? It turns out that
$\mathbb{Z}[\sqrt{-3}]$ lacks unique factorization as well. Indeed, here we have
\[ (1 - \sqrt{-3})(1+\sqrt{-3}) = 4 = 2 \times 2.  \]
These elements can be factored no more, and $1 - \sqrt{-3}$ and $2$ do not
differ by units.
So in this ring, we have a failure of unique factorization. Nonetheless, the
failure of unique factorization in $\mathbb{Z}[\sqrt{-3}]$ is less
noteworthy, because $\mathbb{Z}[\sqrt{-3}]$ is not \emph{integrally closed}. Indeed, it turns out that $\mathbb{Z}[\sqrt{-3}]$ is
contained in the larger ring
\( \mathbb{Z}\left[ \frac{1 + \sqrt{-3}}{2}\right],  \)
which does have unique factorization, and this larger ring is finite over
$\mathbb{Z}[\sqrt{-3}]$.\footnote{In fact, $\mathbb{Z}[\sqrt{-3}]$ is an index two subgroup of $\mathbb{Z}\left[
\frac{1 + \sqrt{-3}}{2}\right]$, as the ring $\mathbb{Z}[ \frac{1 + \sqrt{-3}}{2}]$
can be described as the set of elements $a +
b\sqrt{-3}$ where $a,b$ are either both integers or both integers plus
$\frac{1}{2}$, as is easily seen: this set is closed under addition and
multiplication.}
 Since being integrally closed is a
prerequisite for having unique factorization (see \cref{} below), the failure in
$\mathbb{Z}[\sqrt{-3}]$ is not particularly surprising.

Note that, by contrast, $\mathbb{Z}[ \frac{1 + \sqrt{-5}}{2}]$ does not
contain $\mathbb{Z}[\sqrt{-5}]$ as a finite index subgroup---it cannot be
slightly enlarged in the same sense. When one enlarges $\mathbb{Z}[\sqrt{-5}]$,
one has to add a lot of stuff.
We will see more formally that $\mathbb{Z}[\sqrt{-5}]$ is \emph{integrally
closed} in its quotient field, while $\mathbb{Z}[\sqrt{-3}]$ is not. Since
unique factorization domains are automatically integrally closed, the failure
of $\mathbb{Z}[\sqrt{-5}]$ to be a UFD is much more significant than that of
$\mathbb{Z}[\sqrt{-3}]$.


\subsection{Le sorite for integral extensions}

In commutative algebra and algebraic geometry, there are a lot of standard
properties that a \emph{morphism} of rings $\phi: R \to S$ can have: it could
be of \emph{finite type} (that is, $S$ is finitely generated over $\phi(R)$),
it could be \emph{finite} (that is, $S$ is a finite $R$-module), or it could
be \emph{integral} (which we have defined in \cref{intdefn}). There are many more examples
that we will encounter as we dive deeper into commutative algebra.
In algebraic geometry, there are corresponding properties of morphisms of
\emph{schemes,} and there are many more interesting ones here.

In these cases, there is usually---for any reasonable property---a standard
and familiar list of
properties that one proves about them. We will refer to such lists as
``sorites,'' and prove our first one now.

\begin{proposition}[Le sorite for integral morphisms]
\begin{enumerate}
\item For any ring $R$ and any ideal $I \subset R$, the map $R \to R/I$ is
integral.
\item If $\phi: R \to S$ and $\psi: S \to T$ are integral morphisms, then so
is $\psi \circ \phi: R \to T$.
\item If $\phi: R \to S$ is an integral morphism and $R'$ is an $R$-algebra,
then the base-change
$R' \to R' \otimes_R S$ is integral.
\end{enumerate}
\end{proposition}

\begin{proof}
The first property is obvious. For the second, the condition of
integrality in a morphism of rings depends on the inclusion of the image
in the codomain. So we can suppose that $R \subset S \subset T$. Suppose $t
\in T$. By assumption, there is a monic polynomial equation
\[ t^n + s_1 t^{n-1} + \dots + s_n = 0  \]
that $t$ satisfies, where each $s_i \in S$.

In particular, we find that $t$ is integral over $R[s_1, \dots, s_n]$.
As a result, the module $R[s_1, \dots, s_n, t]$ is finitely generated over the
ring $R'=R[s_1, \dots, s_n]$.
By the following \cref{finitelygeneratedintegral}, $R'$ is a finitely generated
$R$-module. In
particular, $R[s_1, \dots, s_n,t]$ is a finitely generated $R$-module (not
just a
finitely generated $R'$-module).

Thus the $R$-module $R[s_1, \dots, s_n,t]$  is a faithful
$R'$ module, finitely generated over $R$, which is preserved under
multiplication by $t$.
\end{proof}

We now prove a result that can equivalently be phrased as ``finite type plus
integral implies finite'' for a map of rings.

\begin{proposition} \label{finitelygeneratedintegral}
Let $R'$ be a finitely generated, integral $R$-algebra. Then $R'$ is a
finitely generated $R$-module: that is, the map $R \to R'$ is finite.
\end{proposition}
\begin{proof}
Induction on the number of generators of $R'$ as $R$-algebra. For one
generator, this follows from \rref{onegeneratorintegral}.
In general, we will have $R' = R[\alpha_1 ,\dots, \alpha_n]$ for some
$\alpha_i \in R'$.
By the inductive hypothesis, $R[\alpha_1 , \dots, \alpha_{n-1}]$ is a finite
$R$-module; by the case of one generator, $R'$ is a finite $R[\alpha_1, \dots,
\alpha_{n-1}]$-module. This establishes the result by the next exercise.
\end{proof}

\begin{exercise}
Let $R \to S, S \to T$ be morphisms of rings. Suppose $S$ is a finite
$R$-module and $T$ a finite $T$-module. Then $T$ is a finite $R$-module.
\end{exercise}




\subsection{Integral closure}

Let $R, R'$ be rings.
\begin{definition}
If $R \subset R'$, then the set $S = \left\{x \in R': x \ \mathrm{is \
integral  }\right\}$ is called the \textbf{integral closure} of $R$ in $R'$. We
say that $R$ is \textbf{integrally closed in $R'$} if $S = R'$.


When $R$ is a domain, and $K$ is the quotient field,  we shall simply
say that $R$ is \textbf{integrally closed} if it is integrally closed in
$K$.
Alternatively, some people say that $R$ is \textbf{normal} in this case.
\end{definition}

Integral closure (in, say, the latter sense) is thus an operation that maps
integral domains to integral domains. It is easy to see that the operation is
\emph{idempotent:} the integral closure of the integral closure is the integral
closure.


\begin{example}
The integers $\mathbb{Z} \subset \mathbb{C}$ have as integral closure (in
$\mathbb{C}$) the set
of complex numbers  satisfying a monic polynomial with integral
coefficients.  This set is called the set of \textbf{algebraic integers}.

For instance, $i$ is an algebraic integer because it satisfies the equation $X^2 +1 = 0$.
$\frac{1 - \sqrt{-3}}{2}$ is an algebraic integer, as we talked about last
time; it is a sixth root of unity.  On the other hand, $\frac{1+\sqrt{-5}}{2}$
is not an algebraic integer.
\end{example}

\begin{example}
Take $\mathbb{Z} \subset \mathbb{Q}$. The claim is that $\mathbb{Z}$ is
integrally closed in its quotient field $\mathbb{Q}$, or simply---integrally
closed.
\end{example}
\begin{proof}
We will build on this proof later. Here is the point. Suppose $\frac{a}{b}
\in \mathbb{Q}$ satisfying an equation
\[ P(a/b) = 0, \quad P(t) = t^n + c_1 t^{n-1} + \dots + c_0 , \ \forall c_i \in
\mathbb{Z}.\]
Assume that $a,b$ have no common factors; we must prove that $b$ has no prime
factors, so is $\pm 1$.
If $b$ had a prime factor, say $q$, then we must obtain a contradiction.

We interrupt with a definition.
\begin{definition}
The \textbf{valuation at $q$} (or \textbf{$q$-adic valuation}) is the map
$v_q: \mathbb{Q}^* \to \mathbb{Z}$ is the
function sending $q^k (a/b)$ to $k$ if $q \nmid a,b$. We extend this to all
rational numbers via $v(0) = \infty$.
\end{definition}
In general, this just counts the number of factors of $q$ in the expression.


Note the general property that
\begin{equation} \label{val1} v_q(x+y) \geq \min( v_q(x), v_q(y)) .
\end{equation}
If $x,y$ are both divisible by some power of $q$, so is $x+y$; this is the
statement above. We also have the useful property
\begin{equation} \label{val2} v_q(xy) = v_q(x) + v_q(y).  \end{equation}




Now return to the proof that $\mathbb{Z}$ is normal. We would like to show that
\( v_q(a/b) \geq 0.  \)
This will prove that $b$ is not divisible by $q$. When we show this for all
$q$, it will follow that $a/b \in \mathbb{Z}$.

We are assuming that $P(a/b) = 0$. In particular,
\[ \left( \frac{a}{b}  \right)^n = -c_1 \left( \frac{a}{b}  \right)^{n-1} -
\dots - c_0.  \]
Apply $v_q$ to both sides:
\[ n v_q ( a/b) \geq \min_{i>0} v_q( c_i (a/b)^{n-i}).  \]
Since the $c_i \in \mathbb{Z}$, their valuations are nonnegative. In
particular, the right hand side is at least
\[ \min_{i>0}  (n-i) v_q(a/b). \]
This cannot happen if $v_q(a/b)<0$, because $n-i < n$ for each $i>0$.
\end{proof}

This argument applies more generally. If  $K$ is a field, and $R \subset K$ is
a subring  ``defined
by valuations,'' such as the $v_q$, then $R$ is integrally closed in its
quotient field.
More precisely, 
note the reasoning of the previous example: the key idea was that $\mathbb{Z}
\subset \mathbb{Q}$ was characterized by the rational numbers $x$ such that
$v_q(x) \geq 0$ for all primes $q$.
We can abstract this idea as follows. If there exists a family of functions $\mathcal{V}$ from $K^*
\to \mathbb{Z}$ (such as
$\left\{v_q: \mathbb{Q}^* \to \mathbb{Z}\right\}$) satisfying \eqref{val1} and
\eqref{val2} above such that $R$ is
the set of elements such that $v(x) \geq 0, v \in \mathcal{V}$ (along with
$0$), then $R$ is integrally closed in $K$.
We will talk more about
this, and about valuation rings, below.

\begin{example} 
We saw earlier (\cref{sixthroot}) that $\mathbb{Z}[\sqrt{-3}]$ is not
integrally closed, as $\frac{1 + \sqrt{-3}}{2}$ is integral over this ring and
in the quotient field, but not in the ring.
\end{example} 

We shall give more examples in the next subsection.


\subsection{Geometric examples}

Let us now describe the geometry of a non-integrally closed ring.
Recall that finitely generated (reduced) $\mathbb{C}$-algebras are supposed to
correspond to affine algebraic varieties. A \emph{smooth} variety (i.e., one
that is a complex manifold) will always correspond to an integrally closed
ring (though this relies on a deep result that a regular local ring is a
factorization domain, and consequently integrally closed): non-normality is a sign of singularities.
\begin{example}
Here is a ring which is not integrally closed. Take  $\mathbb{C}[x, y]/(x^2
- y^3)$. Algebraically, this is the subring of the polynomial ring
  $\mathbb{C}[t]$ generated by $t^2$ and $t^3$.

In the complex plane, $\mathbb{C}^2$, this corresponds to the subvariety $C
\subset \mathbb{C}^2$ defined by $x^2 =
y^3$.  In $\mathbb{R}^2$, this can be drawn: it has a singularity at $(x,y)=0$.

Note that $x^2 = y^3$ if and only if there is a complex number $z$ such that $x
= z^3, y = z^2$. This complex number $z$ can be recovered via $x/y$ when $x,y
\neq 0$. In particular, there is a map $\mathbb{C} \to C$ which sends $z \to
(z^3, z^2)$.  At every point other than the origin, the inverse can be
recovered using rational functions. But this does not work at the origin.

We can think of $\mathbb{C}[x,y]/(x^2 - y^3)$ as the subring $R'$ of
$\mathbb{C}[z]$
generated by $\left\{z^n, n \neq 1\right\}$.  There is a map from
$\mathbb{C}[x,y]/(x^2 - y^3)$ sending $x \to z^3, y \to z^2$.  Since these two
domains are isomorphic, and $R'$ is not integrally closed, it follows that
$\mathbb{C}[x,y]/(x^2 - y^3)$ is not integrally closed.
The element $z$ can be thought of as an element of the fraction field of $R'$
or of $\mathbb{C}[x,y]/(x^2 - y^3)$.
It is integral, though.

The failure of the ring to be integrally closed has to do with the singularity
at the
origin.
\end{example}


We now give a generalization of the above example.

\begin{example}
This example is outside the scope of the present course.  Say that $X \subset
\mathbb{C}^n$ is given as the zero locus of some holomorphic functions
$\left\{f_i: \mathbb{C}^{n} \to \mathbb{C}\right\}$.  We just gave an example
when $n=2$.
Assume that $0 \in X$, i.e. each $f_i$ vanishes at the origin.

Let $R$ be the ring of germs of holomorphic functions $0$, in other words
holomorphic functions from small open neighborhoods of zero.  Each of these
$f_i$ becomes an  element of $R$.  The ring
\( R/(\left\{f_i\right\} ) \)
is called the ring of germs of holomorphic functions on $X$ at zero.

Assume that $R$ is a domain.  This assumption, geometrically, means that near
the point zero in $X$, $X$ can't be broken into two smaller closed analytic
pieces.  The fraction field of $R$ is to be thought of as the ring of
germs of meromorphic functions on $X$ at zero.

We state the following without proof:

\begin{theorem}
Let $g/g'$ be an element of the fraction field, i.e. $g, g' \in R$. Then $g/g'$
is integral over $R$ if and only if $g/g'$ is bounded near zero.
\end{theorem}

In the previous example of $X$ defined by $x^2 = y^3$, the function $x/y$
(defined near the origin on the curve) is
bounded near the origin, so it is integral over the ring of germs of regular
functions. The reason it is not defined near the origin is \emph{not} that it
blows up. In fact, it extends continuously, but not holomorphically, to the
rest of the variety $X$.
\end{example}

\section{Lying over and going up}
We now interpret integrality in terms of the geometry of $\spec$.
In general, for $R \to S$ a ring-homomorphism, the induced map $\spec S \to
\spec R$ need not be topologically nice; for instance, even if $S$ is a
finitely generated $R$-algebra, the image of $\spec S$ in $\spec R$ need not
be either open or closed.\footnote{It is, however, true that if $R$ is
\emph{noetherian} (see \rref{noetherian}) and $S$ finitely generated over
$R$, then the image of $\spec S$ is \emph{constructible,} that is, a finite
union of locally closed subsets. \add{this result should be added sometime}.}

We shall see that under conditions of integrality, more can be said.



\subsection{Lying over}

In general, given a morphism of algebraic varieties $f: X \to Y$, the image of
a closed subset $Z \subset X$ is far from closed.  For instance, a regular function $f: X \to \mathbb{C}$ 
that is a closed map would have to be either surjective or constant (if $X$ is
connected, say).
Nonetheless, under integrality hypotheses, we can say more.

\begin{proposition}[Lying over] \label{lyingover}
If $\phi: R \to R'$ is an integral morphism, then the induced map
\[  \spec R' \to \spec R  \]
is a closed map; it is surjective if $\phi$ is injective.
 \end{proposition}

Another way to state the last claim, without mentioning $\spec R'$, is the
following. Assume $\phi$ is injective and integral. Then if
$\mathfrak{p} \subset R$ is prime, then there exists $\mathfrak{q} \subset R'$
such that $\mathfrak{p}$ is the inverse image $\phi^{-1}(\mathfrak{q})$.

\begin{proof} First suppose $\phi$ injective, in which case we must prove the
map $\spec R' \to \spec R$ surjective.
Let us reduce to the case of a local ring.
For a prime $\mathfrak{p} \in \spec R$, we must show that $\mathfrak{p}$
arises as the inverse image of an element of $\spec R'$.
So we replace $R$ with $R_{\mathfrak{p}}$.  We get a map
\[ \phi_{\mathfrak{p}}: R_{\mathfrak{p}} \to (R- \mathfrak{p})^{-1} R'  \]
which is injective if $\phi$ is, since localization is an exact functor. Here
we
have localized both $R, R'$ at the multiplicative subset $R - \mathfrak{p}$.

Note that $\phi_{\mathfrak{p}}$ is an integral extension too. This follows
because integrality is preserved by base-change.
We will now prove the result for $\phi_{\mathfrak{p}}$; in particular, we
will show
that there is a prime ideal of $(R- \mathfrak{p})^{-1} R'$ that pulls back to
$\mathfrak{p}R_{\mathfrak{p}}$. These will imply that if we pull this prime
ideal back to $R'$, it will pull back to $\mathfrak{p}$ in $R$. In detail, we
can consider the diagram
\[ \xymatrix{
\spec (R-\mathfrak{p})^{-1} R'\ar[d]  \ar[r] & \spec R_{\mathfrak{p}} \ar[d] \\
\spec R' \ar[r] &  \spec R
}\]
which shows that if $\mathfrak{p} R_{\mathfrak{p}}$ appears in the image of the top map, then
$\mathfrak{p}$
arises as the image of something in $\spec R'$.
So it is sufficient for the proposition (that is, the case of $\phi$
injective) to handle the case of $R$ local, and
$\mathfrak{p}$ the maximal ideal.


In other words, we need to show that:
\begin{quote}
If $R$ is a \emph{local} ring, $\phi: R \hookrightarrow R'$ an injective
integral morphism, then the maximal ideal of $R$ is the inverse image of
something in $\spec R'$.
\end{quote}

Assume $R$ is local with maximal ideal $\mathfrak{p}$. We want to find a prime ideal $\mathfrak{q} \subset R'$ such that
$\mathfrak{p}  = \phi^{-1}(\mathfrak{q})$. Since $\mathfrak{p}$ is already
maximal, it will suffice to show that $\mathfrak{p} \subset
\phi^{-1}(\mathfrak{q})$. In particular, we need to show that there is a prime
ideal $\mathfrak{q}$ such that
\( \mathfrak{p} R' \subset \mathfrak{q}.  \)
The pull-back of this will be $\mathfrak{p}$.

If $\mathfrak{p}R' \neq R'$, then
$\mathfrak{q}$ exists, since every proper ideal of a ring is contained in a
maximal ideal.  We will in fact show
\begin{equation} \label{thingdoesn'tgenerate} \mathfrak{p} R' \neq R',
\end{equation}
or that $\mathfrak{p}$ does not generate the unit ideal in $R'$.
If we prove \eqref{thingdoesn'tgenerate}, we will thus be able to find our
$\mathfrak{q}$, and we will be done.

Suppose the
contrary, i.e. $\mathfrak{p}R' = R'$. We will derive a contradiction using
Nakayama's lemma (\cref{nakayama}). 
Right now, we cannot apply Nakayama's lemma directly because $R'$ is not a
finite $R$-module. The idea is that we will ``descend'' the ``evidence'' that 
\eqref{thingdoesn'tgenerate} fails to a small subalgebra of $R'$, and then
obtain a contradiction.
To do this, note that $1 \in \mathfrak{p}R'$, and we can write
\[ 1 = \sum x_i \phi(y_i)  \]
where $x_i \in R', y_i \in \mathfrak{p}$.
This is the ``evidence'' that \eqref{thingdoesn'tgenerate} fails, and it
involves only a finite amount of data. 

Let $R''$ be the subalgebra of $R'$ generated by $\phi(R)$ and the $x_i$. Then
$R'' \subset R'$ and is finitely generated as an $R$-algebra, because it is generated by
the $x_i$. However, $R''$ is integral over $R$ and thus finitely generated as an
$R$-module, by \cref{finitelygeneratedintegral}. This
is where integrality comes in.

So $R''$ is a finitely generated $R$-module. Also, the expression
$1 = \sum x_i \phi(y_i)$ shows that $\mathfrak{p}R'' = R''$. However, this
contradicts Nakayama's lemma. That brings the contradiction, showing that
$\mathfrak{p}$ cannot generate $(1)$ in $R'$, and proving the surjectivity
part of  lying over theorem.

Finally, we need to show that if $\phi: R \to R'$ is \emph{any} integral
morphism, then $\spec R' \to \spec R$ is a closed map. Let $X = V(I) $ be a
closed subset of $\spec R'$. Then the image of $X$ in $\spec R$ is the image
of the map
\[ \spec R'/I \to \spec R   \]
obtained from the morphism $R \to R' \to  R'/I$, which is integral; thus we are
reduced to showing that any integral morphism $\phi$ has closed image on the
$\spec$. 
Thus we are reduced to $X = \spec R'$, if we throw out $R'$ and replace it by
$R'/I$.

In other words, we must prove the following statement. Let $\phi: R \to R'$ be
an integral morphism; then the image of $\spec R'$ in $\spec R$ is closed.
But, quotienting by $\ker \phi$ and taking the map $R/\ker \phi \to R'$, we
may reduce to the case of $\phi$ injective; however, then this follows from
the surjectivity result already proved.
\end{proof}

In general, there will be \emph{many} lifts of a given prime ideal.
Consider for instance the inclusion $\mathbb{Z} \subset \mathbb{Z}[i]$.
Then the prime ideal $(5) \in \spec \mathbb{Z}$ can be lifted either to $(2+i)
\in \spec \mathbb{Z}[i]$
or $(2-i) \in \spec \mathbb{Z}[i]$.
These are distinct prime ideals: $\frac{2+i}{2-i} \notin \mathbb{Z}[i]$. But
note that any element of $\mathbb{Z}$ divisible by $2+i$ is automatically
divisible by its conjugate $2-i$, and consequently by their product $5$
(because $\mathbb{Z}[i]$ is  a UFD, being a euclidean domain).

Nonetheless, the different lifts are incomparable.

\begin{proposition} \label{incomparableintegral}
Let $\phi: R \to R'$ be an integral morphism. Then given $\mathfrak{p} \in
\spec R$, there are no inclusions among the elements $\mathfrak{q} \in \spec
R'$ lifting $\mathfrak{p}$.
\end{proposition} 
In other words, if $\mathfrak{q}, \mathfrak{q}' \in \spec R'$ lift
$\mathfrak{p}$, then $\mathfrak{q} \not\subset \mathfrak{q}'$.
\begin{proof} 
We will give a ``slick'' proof by various reductions.
Note that the operations of localization and quotienting only shrink the
$\spec$'s: they do not ``merge'' heretofore distinct prime ideals into one.
Thus, by quotienting $R$ by $\mathfrak{p}$, we may assume $R$ is a
\emph{domain} and that $\mathfrak{p} = 0$.
Suppose we had two primes $\mathfrak{q} \subsetneq \mathfrak{q}'$ of $R'$
lifting $(0) \in \spec R$.
Quotienting $R'$ by $\mathfrak{q}$, we may assume that $\mathfrak{q}  = 0$.
We could even assume $R \subset R'$, by quotienting by the kernel of $\phi$.
The next lemma thus completes the proof, because it shows that $\mathfrak{q}'
= 0$, contradiction.
\end{proof} 

\begin{lemma} 
Let $R \subset R'$ be an inclusion of integral domains, which is an integral
morphism. If $\mathfrak{q} \in \spec R'$ is a nonzero prime ideal, then
$\mathfrak{q} \cap R$ is nonzero.
\end{lemma} 
\begin{proof} 
Let $x \in \mathfrak{q}'$ be nonzero. There is an equation
\[ x^n + r_1 x^{n-1} + \dots + r_n = 0,  \quad r_i \in R, \]
that $x$ satisfies, by assumption. 
Here we can assume $r_n \neq 0$; then $r_n \in \mathfrak{q}' \cap R$ by
inspection, though. So this intersection is nonzero.

\end{proof} 

\begin{corollary} 
Let $R \subset R'$ be an inclusion of integral domains, such that $R'$ is
integral over $R$. Then if one of $R, R'$ is a field, so is the other.
\end{corollary} 
\begin{proof} 
Indeed, $\spec R' \to \spec R$ is surjective by \cref{lyingover}: so if $\spec
R'$ has one element (i.e., $R'$ is a field), the same holds for $\spec R$
(i.e., $R$ is a field). Conversely, suppose $R$ a field. 
Then any two prime ideals in $\spec R'$ pull back to the same element of $\spec
R$. So, by \cref{incomparableintegral}, there can be no inclusions among the prime ideals of $\spec
R'$. But $R'$ is a domain, so it must then be a field.
\end{proof} 


\begin{exercise} Let $k$ be a field.
Show that $k[\mathbb{Q}_{\geq 0}]$ is integral over the polynomial ring $k[T]$.
Although this is a \emph{huge} extension,  the prime ideal $(T)$ lifts in only
one way to $\spec k[\mathbb{Q}_{\geq 0}]$.
\end{exercise} 

\begin{exercise} 
Suppose $A \subset B$ is an inclusion of rings over a field of characteristic
$p$. Suppose $B^p \subset A$, so that $B/A$ is integral in a very strong sense. 
Show that the map $\spec B \to \spec A$ is a \emph{homeomorphism.} 
\end{exercise} 



\subsection{Going up}
Let $R \subset R'$ be an inclusion of rings with $R'$ integral over $R$. We saw in the
lying over theorem (\cref{lyingover}) that any prime $\mathfrak{p} \in \spec R$ 
has a prime $\mathfrak{q} \in \spec R'$ ``lying over'' $\mathfrak{p}$, i.e.
such that $R \cap \mathfrak{q} = \mathfrak{p}$.
We now want to show that we can lift finite \emph{inclusions} of primes to $R'$.

\begin{proposition}[Going up]
Let $R \subset R'$ be an integral inclusion of rings.
Suppose $\mathfrak{p}_1 \subset \mathfrak{p}_2 \subset \dots \subset
\mathfrak{p}_n \subset R$ is a 
finite ascending chain of prime ideals in $R$.
Then there is an ascending chain $\mathfrak{q}_1 \subset \mathfrak{q}_2 \subset
\dots \subset \mathfrak{q}_n$ in $\spec R'$ lifting this chain. 

Moreover, $\mathfrak{q}_1$ can be chosen arbitrarily so as to lift
$\mathfrak{p}_1$.
\end{proposition} 
\begin{proof} 
By induction and lying over (\cref{lyingover}), it suffices to show:
\begin{quote}
Let $\mathfrak{p}_1 \subset \mathfrak{p}_2$ be an inclusion of primes in $\spec
R$. Let $\mathfrak{q}_1 \in \spec R'$ lift $\mathfrak{p}_1$. Then there is
$\mathfrak{q}_2 \in \spec R'$, which satisfies the dual conditions of lifting
$\mathfrak{p}_2$ and containing $\mathfrak{q}_1$.
\end{quote}
To show that this is true, we apply \cref{lyingover} to the inclusion
$R/\mathfrak{p}_1 \hookrightarrow R'/\mathfrak{q}_1$. There is an element of
$\spec R'/\mathfrak{q}_1$ lifting $\mathfrak{p}_2/\mathfrak{p}_1$; the
corresponding element of $\spec R'$ will do for $\mathfrak{q}_2$.
\end{proof} 

\section{Valuation rings}

A valuation ring is a special type of local ring. Its distinguishing
characteristic is that divisibility is a ``total preorder.'' That is, two
elements of the quotient field are never incompatible under divisibility.
 We shall see in this section that integrality can be detected using
valuation rings only.

Geometrically, the valuation ring is something like a local piece of a smooth
curve. In fact, in algebraic geometry, a more compelling reason to study
valuation rings is provided by the valuative criteria for separatedness and
properness (cf. \cite{EGA} or \cite{Ha77}).  One key observation about
valuation rings that leads the last results is that any local domain can be
``dominated'' by a valuation ring with the same quotient field (i.e. mapped
into a valuation ring via local
homomorphism), but valuation rings are the maximal elements in this relation
of domination.

\subsection{Definition}

\begin{definition}
A \textbf{valuation ring} is a domain $R$ such that for every pair of elements
$a,b \in R$, either $a \mid b$ or $b \mid a$.
\end{definition}

\begin{example}
$\mathbb{Z}$ is not a valuation ring. It is neither true that  2 divides 3
nor that 3 divides 2.
\end{example}

\begin{example}
$\mathbb{Z}_{(p)}$, which is the set of all fractions of the form $a/b \in
\mathbb{Q}$ where $p \nmid b$, is a valuation ring. To check whether $a/b$
divides $a'/b'$ or vice versa, one  just has to check which is divisible by
the larger power of $p$.
\end{example}

\begin{proposition}
Let $R$ be a domain with quotient field $K$. Then $R$ is a valuation ring if
and only if for every $x \in K$, either $x$ or $x^{-1}$ lies in $R$.
\end{proposition}

\begin{proof} Indeed, if $x=a/b , \ a,b \in R$, then either $a \mid
b$ or $b \mid a$, so either $x$ or $x^{-1} \in R$. This condition is equivalent
to $R$'s being a valuation ring.
\end{proof}


\subsection{Valuations}
The reason for the name ``valuation ring'' is provided by the next definition.
As we shall see, any valuation ring comes from a ``valuation.''

By definition, an \emph{ordered abelian group} is an abelian group $A$
together with a set of \emph{positive elements} $A_+ \subset A$. This set is
required to be closed under addition and satisfy the property that if $x \in
A$, then precisely one of the following is true: $x \in A_+$, $-x \in A_+$,
and $x = 0$. This allows one to define an ordering $<$ on $A$ by writing $x<y$
if $y-x \in A_+$.
Given $A$, we often formally adjoin an element $\infty$ which is bigger than
every element in $A$.


\begin{definition}
Let $K$ be a field. A \textbf{valuation} on $K$ is a map $v: K \to A \cup
\left\{\infty\right\}$ for some
 ordered abelian group $A$  satisfying:
\begin{enumerate}
\item  $v(0) = \infty$ and $v(K^*) \subset A$.
\item For $x,y \in K^*$, $v(xy) = v(x) + v(y)$. That is, $v|_{K^*}$ is
a homomorphism.
\item For $x,y \in K$, $v(x+y) \geq \min (v(x), v(y))$.
\end{enumerate}

\end{definition}
Suppose that $K$ is a field and $v: K \to A \cup \left\{\infty\right\}$ is a
valuation (i.e. $v(0) = \infty$). Define $R = \left\{x \in K: v(x) \geq
0\right\}$.
\begin{proposition}
$R$ as just defined is a valuation ring.
\end{proposition}
\begin{proof}  First, we prove that $R$ is a ring.
$R$ is closed under addition and multiplication by the two conditions
\[ v(xy)  = v(x) + v(y)  \]
and
\[ v(x+y) \geq \min v(x), v(y) , \]
so if $x,y \in R$, then $x+y, xy$ have nonnegative valuations.

Note that $0 \in R$ because $v(0) = \infty$. Also $v(1) = 0$ since $v: K^*
\to A$
is a homomorphism. So $1 \in R$ too.
Finally, $-1 \in R$ because $v(-1) =0$ since $A$ is totally ordered.  It
follows that $R$ is also a group.

Let us now show that $R$ is a valuation ring. If $x \in K^*$, either $v(x) \geq
0$ or $v(x^{-1}) \geq 0$ since $A$ is totally ordered.\footnote{Otherwise $0
=v(x)+v(x^{-1}) < 0$, contradiction.} So either $x, x^{-1} \in R$.
\end{proof}

In particular, the set of elements with nonnegative valuation is a valuation
ring.
The converse also holds. Whenever you have a valuation ring, it comes about in
this manner.

\begin{proposition}
Let $R$ be a valuation ring with quotient field $K$. There is an ordered
abelian group
$A$ and a valuation $v: K^* \to A$ such that $R$ is the set of elements with
nonnegative valuation.
\end{proposition}
\begin{proof}
First, we construct $A$. In fact, it is the quotient of $K^*$ by the subgroup
of units  $R^*$ of $R$.
We define an ordering by saying that $x \leq y$ if $y/x \in R$---this doesn't
depend on the representatives in $K^*$ chosen.  Note that either $x \leq y$ or
$y \leq x$ must hold, since $R$ is a valuation ring.
The combination of $ x \leq y$ and $y \leq x$ implies that $x,y$ are equivalent
classes.
The nonnegative elements in this group are those whose representatives in $K^*$
belong to $R$.

It is easy to see that $K^*/R^*$ in this way is a totally ordered abelian
group with
the image of 1 as the unit. The
reduction map $K^* \to K^*/R^*$  defines a valuation whose corresponding ring
is just $R$. We have omitted some details; for instance, it should be checked
that the valuation of $x+y$ is at least the minimum of $v(x), v(y)$.
\end{proof}



To summarize:
\begin{quote}
Every valuation ring $R$ determines a valuation $v$ from the fraction field of
$R$ into $A \cup \left\{\infty\right\}$ for $A$ a totally ordered abelian group
such that $R$ is just the set of elements of $K$ with nonnegative valuation. As
long as we require that $v: K^* \to A$ is surjective, then $A$ is uniquely
determined as well.
\end{quote}

\begin{definition}
A valuation ring $R$ is \textbf{discrete} if we can choose $A$ to be
$\mathbb{Z}$.
\end{definition}

\begin{example}
$\mathbb{Z}_{(p)}$ is a discrete valuation ring.
\end{example}

The notion of a valuation ring is a useful one.

\subsection{General remarks}
Let $R$ be a commutative ring. Then $\spec R$ is the set of primes of $R$,
equipped
with a certain topology. The space $\spec R$ is almost never Hausdorff. It is
almost always a bad idea to apply the familiar ideas from elementary topology
(e.g. the fundamental group) to $\spec R$. Nonetheless, it has some other nice
features that substitute for its non-Hausdorffness.

For instance, if $R = \mathbb{C}[x,y]$, then $\spec R$ corresponds to
$\mathbb{C}^2$ with some additional nonclosed points.  The injection of
$\mathbb{C}^2$ with its usual topology into $\spec R$ is continuous. While in
$\spec R$ you don't want to think of continuous paths, you can in
$\mathbb{C}^2$.

Suppose you had two points $x,y \in \mathbb{C}^2$ and their images in $\spec
R$.  Algebraically, you can still think about algebraic curves passing
through $x,y$.
This is a subset of $x,y$ defined by a single polynomial equation.
This curve will have what's called a ``generic point,'' since the ideal
generated by this curve will be a prime ideal.
The closure of this generic point will be precisely this algebraic
curve---including $x,y$.

\begin{remark}
If $ \mathfrak{p}, \mathfrak{p}' \in \spec R$, then
\[ \mathfrak{p}' \in \overline{\left\{\mathfrak{p}\right\}}  \]
iff
\[ \mathfrak{p}' \supset \mathfrak{p}.  \]
Why is this? Well, the closure of $\left\{\mathfrak{p}\right\}$ is just
$V(\mathfrak{p})$, since this is the smallest closed subset of $\spec R$
containing $\mathfrak{p}$.
\end{remark}

The point of this discussion is that instead of paths, one can transmit
information from point to point in $\spec R$ by having one point be in a
closure of another.
However, we will show that this relation is contained by the theory of
valuation rings.

\begin{theorem}
Let $R$ be a domain containing a prime ideal $\mathfrak{p}$.  Let $K$ be the
fraction field of $R$.

Then there is a valuation  $v$ on $K$ defining a valuation ring $R' \subset
K$  such that
\begin{enumerate}
\item $R \subset R'$.
\item $\mathfrak{p} = \left\{x \in R: v(x) > 0\right\}$.
\end{enumerate}

\end{theorem}

Let us motivate this by the remark:
\begin{remark}
A valuation ring is automatically a local ring. A local ring is a ring where
either $x, 1-x$ is invertible for all $x$ in the ring. Let us show that this is
true for a valuation ring.

If $x $ belongs to a valuation ring $R$ with valuation $v$, it is invertible if
$v(x)=0$.  So if $x, 1-x$ were both noninvertible, then both would have
positive valuation.  However, that would imply that $v(1) \geq \min v(x),
v(1-x)$ is positive, contradiction.
\end{remark}

\begin{quote}
If $R'$ is any valuation ring (say defined by a valuation $v$), then $R'$ is
local with maximal ideal consisting of elements with positive valuation.
\end{quote}

The theorem above says that there's a good supply of valuation rings.
In particular, if $R$ is any domain, $\mathfrak{p} \subset R$ a prime ideal,
then we can choose a valuation ring $R' \supset R$ such that $\mathfrak{p}$ is
the intersection of the maximal ideal of $R'$ intersected with $R$.
So the map $\spec R' \to \spec R$ contains $\mathfrak{p}$.

\begin{proof}
Without loss of generality, replace $R$ by $R_{\mathfrak{p}}$, which is a local
ring with maximal ideal $\mathfrak{p}R_{\mathfrak{p}}$. The maximal ideal
intersects $R$ only in $\mathfrak{p}$.

So, we can assume without loss of generality that
\begin{enumerate}
\item $R$ is local.
\item $\mathfrak{p}$ is maximal.
\end{enumerate}

Let $P$ be the collection of all subrings $R' \subset K$ such that $R' \supset
R$ but $\mathfrak{p}R' \neq R'$.  Then $P$ is a poset under inclusion. The
poset is nonempty, since $R \in P$.  Every totally ordered chain in $P$ has an
upper bound.  If you have a totally ordered subring of elements in $P$, then
you can take the union.
We invoke:
\begin{lemma}
Let $R_\alpha$ be a chain in $P$ and $R' = \bigcup R_\alpha$. Then $R' \in P$.
\end{lemma}
\begin{proof}
Indeed, it is easy to see that this is a subalgebra of $K$ containing $R$. The
thing to observe is that
\[ \mathfrak{p}R' = \bigcup_\alpha \mathfrak{p} R_\alpha  ;\]
since by assumption, $1 \notin \mathfrak{p}R_\alpha$ (because each $R_\alpha
\in P$), $1 \notin \mathfrak{p}R'$. In particular, $R' \notin P$.
\end{proof}

By the lemma, Zorn's lemma to the poset $P$. In particular, $P$ has a maximal
element $R'$. By construction, $R'$ is some subalgebra of $K$ and
$\mathfrak{p}R' \neq R'$. Also, $R'$ is maximal with respect to these
properties.

We show first that $R'$ is local, with maximal ideal $\mathfrak{m}$ satisfying
\[ \mathfrak{m} \cap R = \mathfrak{p}.  \]
The second part is evident from locality of $R'$, since $\mathfrak{m} $
must contain
the proper ideal $\mathfrak{p}R'$, and $\mathfrak{p} \subset R$ is a maximal
ideal.

Suppose that $x \in R'$; we show that either $x, 1-x$ belongs to $R'^*$ (i.e.
is invertible). Take the ring $R'[x^{-1}]$.  If $x$ is noninvertible, this
properly contains $R'$.  By maximality, it follows that $\mathfrak{p}R'[x^{-1}]
= R'[x^{-1}]$.

And we're out of time. We'll pick this up on Monday.

\end{proof}

Let us set a goal.


First, recall the notion introduced last time. A \textbf{valuation ring} is a
domain $R$ where for all $x$ in the fraction field of $R$, either $x$ or
$x^{-1}$ lies in $R$. We saw that if $R$ is a valuation ring, then $R$ is
local. That is, there is a unique maximal ideal $\mathfrak{m} \subset R$,
automatically prime.  Moreover, the zero ideal $(0)$ is prime, as $R$ is a
domain. So if you look at the spectrum $\spec R$ of a valuation ring $R$, there
is a unique closed point $\mathfrak{m}$, and a unique generic point
$(0)$.  There might be some other prime ideals in $\spec R$; this depends on
where the additional valuation lives.

\begin{example}
Suppose the valuation defining the valuation ring $R$ takes values in
$\mathbb{R}$. Then the only primes are $\mathfrak{m}$ and zero.
\end{example}

Let $R$ now be any ring, with $\spec R$ containing prime ideals
$\mathfrak{p} \subset \mathfrak{q}$.  In particular, $\mathfrak{q}$ lies in
the closure of $\mathfrak{p}$.
As we will see, this implies that there is a map
\[  \phi: R \to R'  \]
such that $\mathfrak{p} = \phi^{-1}(0)$ and $\mathfrak{q} =
\phi^{-1}(\mathfrak{m})$, where $\mathfrak{m}$ is the maximal ideal of $R'$.
This statement says that the relation of closure in $\spec R$ is always
controlled by valuation rings.
In yet another phrasing, in the map
\[ \spec R' \to \spec R  \]
the closed point goes to $\mathfrak{q}$ and the generic point to
$\mathfrak{p}$. This is our eventual goal.

To carry out this goal, we need some more elementary facts. Let us discuss
things that don't have any obvious relation to it.

\subsection{Back to the goal} Now we return to the goal of the  lecture. Again,
$R$
was any ring, and we had primes $\mathfrak{p} \subset \mathfrak{q} \subset
R$. We
wanted a valuation ring $R'$ and a map $\phi: R \to R'$ such that zero pulled
back to $\mathfrak{p}$ and the maximal ideal pulled back to $\mathfrak{q}$.

What does it mean for $\mathfrak{p}$ to be the inverse image of $(0) \subset
R'$? This means that $\mathfrak{p} = \ker \phi$. So we get an injection
\[ R/\mathfrak{p} \rightarrowtail R'.  \]
We will let $R'$ be a subring of the quotient field $K$ of the domain
$R/\mathfrak{p}$. Of course, this subring will contain $R/\mathfrak{p}$.

In this case, we will get  a map $R \to R'$ such that the pull-back of zero is
$\mathfrak{p}$.  What we want, further, to be true is that $R'$ is a valuation
ring and the pull-back of the maximal ideal is $\mathfrak{q}$.



This is starting to look at the problem we discussed last time.
Namely, let's throw out $R$, and replace it with  $R/\mathfrak{p}$.
Moreover, we can replace $R$ with $R_{\mathfrak{q}}$ and assume that $R$ is
local with maximal ideal $\mathfrak{q}$.
What we need to show is that a valuation ring $R' $ contained in the fraction
field of $R$, containing $R$, such that the intersection of the maximal
ideal of
$R'$ with $R$ is equal to $\mathfrak{q} \subset R$.
If we do this, then we will have accomplished our goal.

\begin{lemma}
Let $R$ be a local domain. Then there is a valuation subring $R'$ of the
quotient
field of $R$ that \emph{dominates} $R$, i.e .the map $R \to R'$ is a
\emph{local} homomorphism.
\end{lemma}

Let's find $R'$ now.

Choose $R'$ maximal such that $\mathfrak{q} R' \neq R'$. Such a ring exists, by
Zorn's lemma. We gave this argument at the end last time.

\begin{lemma}
$R'$ as described is local.
\end{lemma}
\begin{proof}
Look at $\mathfrak{q}R' \subset R'$; it is a proper subset, too, by assumption.
In particular, $\mathfrak{q}R'$ is contained in some maximal ideal
$\mathfrak{m}\subset R'$.   Replace $R'$ by $R'' = R'_{\mathfrak{m}}$.
Note that
\[ R' \subset R''  \]
and
\[ \mathfrak{q}R'' \neq R''  \]
because $\mathfrak{m}R'' \neq R''$.  But $R'$ is maximal, so $R' = R''$, and
$R''$ is a local ring. So $R'$ is a local ring.
\end{proof}

Let $\mathfrak{m}$ be the maximal ideal of $R'$. Then $\mathfrak{m} \supset
\mathfrak{q}R$, so $\mathfrak{m} \cap R = \mathfrak{q}$.
All that is left to prove now is that $R'$ is a valuation ring.

\begin{lemma}
$R'$ is integrally closed.
\end{lemma}

\begin{proof}
Let $R''$ be its integral closure. Then $\mathfrak{m} R'' \neq R''$ by lying
over, since $\mathfrak{m}$ (the maximal ideal of $R'$) lifts up to $R''$. So
$R''$ satisfies
\[ \mathfrak{q}R'' \neq R''  \]
and by maximality, we have $R'' = R'$.
\end{proof}

To summarize, we know that $R'$ is a  local, integrally closed subring of the
quotient field of $R$, such that the maximal ideal of $R'$ pulls back to
$\mathfrak{q}$ in $R$.
All we now need is:

\begin{lemma}
$R'$ is a valuation ring.
\end{lemma}
\begin{proof}
Let $x$ lie in the fraction field. We must show that either $x$ or $x^{-1} \in
R'$.  Say $x \notin R'$.  This means by maximality of $R'$ that $R'' =
R'[x]$ satisfies
\[ \mathfrak{q}R'' = R''.  \]
In particular, we can write
\[ 1 = \sum q_i x^i, \quad q_i \in \mathfrak{q}R' \subset R'.  \]
This implies that
\[ (1-q_0) + \sum_{i > 0} -q_i x^i  = 0.  \]
But $1-q_0$ is invertible in $R'$, since $R'$ is local.  We can divide by the
highest power of $x$:
\[  x^{-N} + \sum_{i>0} \frac{-q_i}{1-q_0} x^{-N+i} = 0. \]
In particular, $1/x$ is integral over $R'$; this implies that $1/x \in
R'$ since
$R'$ is integrally closed and $q_0$ is a nonunit. So
$R'$ is a valuation ring.
\end{proof}

We can state the result formally.
\begin{theorem}
Let $R$ be a ring, $\mathfrak{p} \subset \mathfrak{q}$ prime ideals. Then there
is a homomorphism $\phi: R \to R'$ into a valuation ring $R'$ with maximal
ideal
$\mathfrak{m}$ such that
\[ \phi^{-1}(0) = \mathfrak{p}  \]
and
\[ \phi^{-1}(\mathfrak{m} ) = \mathfrak{q} .\]
\end{theorem}

There is a related fact which we now state.
\begin{theorem}
Let $R$ be any domain. Then the integral closure of $R$ in the quotient field
$K$ is the intersection
\[ \bigcap R_{\alpha}  \]
of all valuation rings $R_{\alpha} \subset K$ containing $R$.
\end{theorem}
So an element of the quotient field is integral over $R$ if and only if its
valuation is nonnegative at every valuation which is nonnegative on $R$.

\begin{proof}
The $\subset$ argument is easy, because one can check that a valuation ring is
integrally closed. (Exercise.)
The interesting direction is to assume that $v(x) \geq 0$ for all $v$
nonnegative
on $R$.

Let us suppose $x$ is nonintegral. Suppose $R' = R[1/x]$ and $I$ be the ideal
$(x^{-1}) \subset R'$. There are two cases:
\begin{enumerate}
\item $I = R'$. Then in the ring $R'$, $x^{-1} $ is invertible. In particular,
$x^{-1}P(x^{-1}) = 1$. Multiplying by a high power of $x$ shows that $x$ is
integral over $R$.  Contradiction.
\item  Suppose $I \subsetneq R'$. Then $I$ is contained in a maximal ideal
$\mathfrak{q} \subset R'$.  There is a valuation subring $R'' \subset K$ ,
containing $R'$, such that the corresponding valuation is positive on
$\mathfrak{q}$.  In particular, this valuation is positive on $x^{-1}$,
so it is
negative on $x$, contradiction.
\end{enumerate}
\end{proof}

So the integral closure has this nice characterization via valuation rings. In
some sense, the proof that $\mathbb{Z}$ is integrally closed has the property
that every integrally closed ring is integrally closed for that reason:
it's the
common nonnegative locus for some valuations.

\section{The Hilbert Nullstellensatz}

The Nullstellensatz is the basic algebraic fact, which we have invoked in the
past to justify various examples, that connects the idea of
the $\spec$ of a ring to classical algebraic geometry.

\subsection{Statement and initial proof of the Nullstellensatz}

There are several ways in which the Nullstellensatz can be stated. Let us
start with the following very concrete version.

\begin{theorem} \label{nullstellensatzoverC}
All maximal ideals in the polynomial ring $R=\mathbb{C}[x_1, \dots, x_n]$ come
from points in $\mathbb{C}^n$. In other words, if $\mathfrak{m} \subset R$ is
maximal, then there exist $a_1, \dots, a_n \in \mathbb{C}$ such that
$\mathfrak{m} = (x_1 - a_1, \dots, x_n - a_n)$.
\end{theorem}


The maximal spectrum of $R=\mathbb{C}[x_1, \dots, x_n]$ is thus identified with
$\mathbb{C}^n$.

We shall now reduce \rref{nullstellensatzoverC} to an easier claim.
Let
$\mathfrak{m}\subset R$ be a maximal ideal. Then there is a map
\[ \mathbb{C} \to R \to R/\mathfrak{m}  \]
where $R/\mathfrak{m}$ is thus a finitely generated $\mathbb{C}$-algebra, as
$R$ is.  The ring $R/\mathfrak{m}$ is also a field by maximality.

We would like to show that $R/\mathfrak{m}$ is a finitely generated
$\mathbb{C}$-vector
space. This would imply that $R/\mathfrak{m}$ is integral over $\mathbb{C}$,
and there are no proper algebraic extensions of $\mathbb{C}$. Thus, if we
prove this, it will follow that the map $\mathbb{C} \to R/\mathfrak{m}$ is an
isomorphism. If $a_i \in \mathbb{C}$ ($1 \leq i \leq n$) is the image of
$x_i $ in $R/\mathfrak{m} =
\mathbb{C}$, it will follow that $(x_1 - a_1, \dots, x_n - a_n) \subset
\mathfrak{m}$, so $(x_1 - a_1, \dots, x_n - a_n)=
\mathfrak{m}$.


Consequently, the Nullstellensatz in this form would follow from the next
claim:

\begin{proposition}
Let $k$ be a field, $L/k$ an extension of fields. Suppose $L$ is a finitely
generated $k$-algebra. Then $L$ is a finite $k$-vector space.
\end{proposition}
This is what we will prove.

We start with an easy proof in the special case:
\begin{lemma}
Assume $k$ is uncountable (e.g. $\mathbb{C}$, the original case of interest).
Then the above proposition is true.
\end{lemma}
\begin{proof}
Since $L$ is a finitely generated $k$-algebra, it suffices to show that $L/k$
is algebraic.
If not, there exists $x \in L$ which isn't algebraic over $k$. So $x$ satisfies
no nontrivial polynomials.
I claim now that the uncountably many elements $\frac{1}{x-\lambda}, \lambda
\in K$ are linearly
independent over $K$. This will be a contradiction as $L$ is a finitely
generated $k$-algebra, hence at most countably dimensional over $k$. (Note that
the polynomial ring is countably dimensional over $k$, and $L$ is a quotient.)

So let's prove this. Suppose not. Then there is a nontrivial linear dependence
\[ \sum \frac{c_i}{x - \lambda_i}  = 0, \quad c_i, \lambda_i \in K. \]
Here the $\lambda_j$ are all distinct to make this nontrivial. Clearing
denominators, we find
\[ \sum_i c_i \prod_{j \neq i } (x- \lambda_j) = 0. \]
Without loss of generality, $c_1 \neq 0$.
This equality was in the field $L$. But $x$ is transcendental over $k$. So we
can think of this as a polynomial ring relation.
Since we can think of this as a relation in the polynomial ring, we see that
doing so, all but the $i =1$ term in the sum is divisible by $x - \lambda_1$
as a polynomial.
It follows that, as polynomials in the indeterminate $x$,
\[ x - \lambda_1 \mid c_1 \prod_{j \neq 1} (x - \lambda_j).  \]
This is a contradiction since all the $\lambda_i$ are distinct.
\end{proof}

This is kind of a strange proof, as it exploits the fact that $\mathbb{C}$ is
uncountable.
This shouldn't be relevant.

\subsection{The normalization lemma}

Let's now give a more algebraic proof.
We shall exploit the following highly useful fact in commutative algebra:

\begin{theorem}[Noether normalization lemma] Let $k$ be a field, and $R =
k[x_1, \dots, x_n]/\mathfrak{p}$ be a finitely generated domain over $k$ (where
$\mathfrak{p}$ is a prime ideal in the polynomial ring).

Then there exists a polynomial subalgebra $k[y_1, \dots, y_m] \subset R$ such
that $R$ is integral over $k[y_1, \dots, y_m]$.
\end{theorem}

Later we will see that $m$ is the \emph{dimension} of $R$.

There is a geometric picture here. Then $\spec R$ is some irreducible algebraic
variety in $k^n$ (plus some additional points), with a smaller dimension than
$n$ if $\mathfrak{p} \neq 0$. Then there exists a \emph{finite map} to $k^m$.
In particular, we can map surjectively $\spec R \to k^m$ which is integral.
The fibers are in fact finite, because integrality implies finite fibers.  (We
have not actually proved this yet.)

How do we actually find such a finite projection? In fact, in characteristic
zero, we just take a
vector space projection $\mathbb{C}^n \to \mathbb{C}^m$. For a ``generic''
projection onto a subspace of the appropriate dimension, the projection will
will do as our finite map. In characteristic $p$, this may not work.

\begin{proof}
First, note that $m$ is uniquely determined as the transcendence degree of the
quotient field of $R$ over $k$.

Among the variables $x_1, \dots, x_n \in R$ (which we think of as in $R$ by an
abuse of notation), choose a maximal subset which is algebraically independent.
This subset has no nontrivial polynomial relations. In particular, the ring
generated by that subset is just the polynomial ring on that subset.
We can permute these variables and assume that
$$\left\{x_1,\dots, x_m\right\}$$ is the maximal subset. In particular, $R$
contains the \emph{polynomial ring} $k[x_1, \dots, x_m]$ and is generated by
the rest of the variables. The rest of the variables are not adjoined freely
though.

The strategy is as follows.  We will implement finitely many changes of
variable so that $R$ becomes integral over $k[x_1, \dots, x_m]$.

The essential case is where $m=n-1$. Let us handle this. So we have
\[ R_0 = k[x_1, \dots, x_m]  \subset R = R_0[x_n]/\mathfrak{p}.  \]
Since $x_n$ is not algebraically independent, there is a nonzero polynomial
$f(x_1, \dots, x_m, x_n) \in \mathfrak{p}$.

We want $f$ to be monic in $x_n$. This will buy us integrality. A priori, this
might not be true. We will modify the coordinate system to arrange that,
though. Choose $N \gg 0$. Define for $1 \leq i \leq m$,
\[ x_i' = x_i + x_n^{N^i}.  \]
Then the equation becomes:
\[ 0 = f(x_1, \dots, x_m, x_n) = f( \left\{x_i' - x_n^{N^i}\right\} , x_n). \]
Now $f(x_1, \dots, x_n, x_{n+1})$ looks like some sum
\[ \sum \lambda_{a_1 \dots b} x_1^{a_1} \dots x_m^{a_m} x_n^{b} , \quad
\lambda_{a_1 \dots b} \in k. \]
But $N$ is really really big. Let us expand this expression in the $x_i'$ and
pay attention to the largest power of $x_n$ we see.
We find that
\[ f(\left\{x_i' - x_n^{N_i}\right\},x_n)
\]
has the largest power of $x_n$ precisely where, in the expression for $f$,
$a_m$ is maximized first, then
$a_{m-1}, $ and so on. The largest exponent would have the form
\[ x_n^{a_m N^m + a_{m-1}N^{m-1} + \dots + b}.	\]
We can't, however, get any exponents of $x_n$ in the expression
\( f(\left\{x_i' - x_n^{N_i}\right\},x_n)\) other than these. If $N$ is super
large, then all these exponents will be different from each other.
In particular, each power of $x_n$ appears precisely once in the expansion of
$f$. We see in particular that $x_n$ is integral over $x_1', \dots, x_n'$.
Thus each $x_i$ is as well.

So we find
\begin{quote}
$R$ is integral over $k[x_1', \dots, x_m']$.
\end{quote}

We have thus proved the normalization lemma in the codimension one case. What
about the general case? We repeat this.
Say we have
\[ k[x_1, \dots, x_m] \subset R.  \]
Let $R'$ be the subring of $R$ generated by $x_{1}, \dots,x_m, x_{m+1}$. The
argument we just gave implies that we can choose $x_1', \dots, x_m'$ such that
$R'$ is integral over $k[x_1', \dots, x_m']$, and the $x_i'$ are
algebraically independent.
We know in fact that $R' = k[x_1', \dots, x_m', x_{m+1}]$.

Let us try repeating the argument while thinking about $x_{m+2}$. Let $R'' =
k[x_1', \dots, x_m', x_{m+2}]$ modulo whatever relations that $x_{m+2}$ has to
satisfy. So this is a subring of $R$. The same argument shows that we can
change variables such that $x_1'', \dots, x_m''$ are algebraically independent
and $R''$ is integral over $k[x_1'', \dots, x_m'']$. We have furthermore that
$k[x_1'', \dots, x_m'', x_{m+2}] = R''$.

Having done this, let us give the argument where $m=n-2$. You will then see
how to do the general case. Then I claim that:
\begin{quote}
$R$ is integral over $k[x_1'', \dots, x_m'']$.
\end{quote}
For this, we need to check that $x_{m+1}, x_{m+2}$ are integral (because these
together with the $x''_i$ generate $R''[x_{m+2}][x_{m+2}]=R$.
But $x_{m+2}$ is integral over this by construction. The integral closure of
$k[x_1'', \dots, x_{m}'']$ in $R$ thus contains
\[ k[x_1'', \dots, x''_m, x_{m+2}] = R''.  \]
However, $R''$ contains the elements $x_1', \dots, x_m'$. But by construction,
$x_{m+1}$ is integral over the $x_1', \dots, x_m'$. The integral closure of
$k[x_1'', \dots, x_{m}'']$ must contain $x_{m+2}$. This completes the proof in
the case $m=n-2$. The general case is similar; we just make several changes of
variables, successively.

\end{proof}
\subsection{Back to the Nullstellensatz}

Consider a finitely generated $k$-algebra $R$ which is a field. We need to
show that $R$ is a
finite $k$-module. This will prove the proposition.
Well, note that $R$ is integral over a polynomial ring $k[x_1, \dots, x_m]$ for
some $m$.
If $m > 0$, then this polynomial ring has more than one prime.
For instance, $(0)$ and $(x_1, \dots, x_m)$. But these must lift to primes in
$R$. Indeed, we have seen that whenever you have an integral extension, the
induced map on spectra is surjective. So
\[ \spec R \to \spec k[x_1, \dots, x_m]  \]
is surjective. If $R$ is a field, this means $\spec k[x_1, \dots, x_m]$ has one
point and $m=0$. So $R$ is integral over $k$, thus algebraic. This implies that
$R$ is finite as it is finitely generated. This proves one version of the
Nullstellensatz.


 Another version of the Nullstellensatz, which is
more precise, says:

\begin{theorem} \label{gennullstellensatz}
Let $I \subset \mathbb{C}[x_1, \dots, x_n]$. Let $V \subset \mathbb{C}^n$ be
the subset of $\mathbb{C}^n$ defined by the ideal $I$ (i.e. the zero locus of
$I$).

Then $\rad(I)$ is precisely the collection of $f$ such that $f|_V = 0$. In
particular,
\[ \rad(I) = \bigcap_{\mathfrak{m} \supset I, \mathfrak{m} \
\mathrm{maximal}} \mathfrak{m}.  \]
\end{theorem}

In particular, there is a bijection between radical ideals and algebraic
subsets of $\mathbb{C}^n$.

The last form of the theorem, which follows from the expression of maximal
ideals in the polynomial ring, is very similar to the result
\[ \rad(I) = \bigcap_{\mathfrak{p} \supset I, \mathfrak{p} \
\mathrm{prime}} \mathfrak{p},  \]
true in any commutative ring. However, this general result is not necessarily
true.

\begin{example}
The intersection of all primes in a DVR is zero, but the intersection of all
maximals is nonzero.
\end{example}

\begin{proof}[Proof of \cref{gennullstellensatz}] 
It now suffices to show that for every $\mathfrak{p} \subset \mathbb{C}[x_1,
\dots, x_n]$ prime, we have
\[ \mathfrak{p} = \bigcap_{\mathfrak{m} \supset I \ \mathrm{maximal}}
\mathfrak{m}  \]
since every radical ideal is an intersection of primes.

Let $R = \mathbb{C}[x_1, \dots,
x_n]/\mathfrak{p}$. This is a domain finitely generated over $\mathbb{C}$. We
want to show that the intersection of maximal ideals in $R$ is zero. This is
equivalent to the above displayed equality.

So fix $f \in R - \left\{0\right\}$. Let $R'$ be the localization $R'= R_f$. Then $R'$ is also an
integral domain, finitely generated over $\mathbb{C}$.	$R'$ has a maximal
ideal $\mathfrak{m}$ (which a priori could be zero). If
we look at the map $R' \to R'/\mathfrak{m}$, we get a map into a field
finitely generated
over $\mathbb{C}$, which is thus $\mathbb{C}$.
The composite map
\[ R \to  R' \to R'/\mathfrak{m}   \]
is just given by an $n$-tuple of complex numbers, i.e. to a point in
$\mathbb{C}^n$ which is even in $V$ as it is a map out of $R$. This corresponds
to a maximal ideal in $R$.
This maximal ideal does not contain $f$ by construction.
\end{proof}

\begin{exercise} 
Prove the following result, known as ``Zariski's lemma'' (which easily implies
the Nullstellensatz): if $k$ is a field, $k'$ a field extension of $k$ which
is a finitely generated $k$-\emph{algebra}, then $k'$ is finite algebraic over
$k$. Use the following argument of McCabe (in \cite{Mc76}):
\begin{enumerate}
\item $k'$ contains a subring $S$ of the form $S= k[ x_1, \dots, x_t]$ where
the $x_1, \dots, x_t$ are algebraically independent over $k$, and $k'$ is
algebraic over the quotient field of $S$ (which is a polynomial ring). 
\item  If $k'$ is not algebraic over $k$, then $S \neq k$ is not a field.
\item Show that there is $y \in S$ such that $k'$ is integral over $S_y$.
Deduce that $S_y$ is a field.
\item Since $\spec (S_y) = \left\{0\right\}$, argue that $y$ lies in every
non-zero prime ideal of $\spec S$. Conclude that $1+y \in k$, and $S$ is a
field---contradiction.
\end{enumerate}
\end{exercise} 

\subsection{A little affine algebraic geometry}

In what follows, let $k$ be algebraically closed, and let $A$ be a finitely generated $k$-algebra.  Recall that $\Specm A$ denotes the set of maximal ideals in $A$.  Consider the natural $k$-algebra structure on $\mathrm{Funct}(\Specm A, k)$.  We have a map
$$A \rightarrow \mathrm{Funct}(\Specm A, k)$$
which comes from the Weak Nullstellensatz as follows.  Maximal ideals $\mathfrak{m}\subset A$ are in bijection with maps $\varphi_\mathfrak{m}:A\rightarrow k$ where $\ker(\varphi_\mathfrak{m})=\mathfrak{m}$, so we define $a\longmapsto [\mathfrak{m}\longmapsto \varphi_\mathfrak{m}(a)]$.  If $A$ is reduced, then this map is injective because if $a\in A$ maps to the zero function, then $a\in \cap\, \mathfrak{m}$ $\rightarrow$ $a$ is nilpotent $\rightarrow$ $a=0$.\\

\begin{definition} A function $f\in \mathrm{Funct}(\Specm A,k)$ is called {\bf algebraic} if it is in the image of $A$ under the above map.  (Alternate words for this are {\bf polynomial} and {\bf regular}.) \end{definition}

Let $A$ and $B$ be finitely generated $k$-algebras and $\phi:A\rightarrow B$ a homomorphism.  This yields a map $\Phi:\Specm B\rightarrow \Specm A$ given by taking pre-images.

\begin{definition} A map $\Phi:\Specm B\rightarrow \Specm A$ is called {\bf algebraic} if it comes from a homomorphism $\phi$ as above.\end{definition}

To demonstrate how these definitions relate to one another we have the following proposition.

\begin{proposition} A map $\Phi:\Specm B\rightarrow \Specm A$ is algebraic if and only if for any algebraic function $f\in \mathrm{Funct}(\Specm A,k)$, the pullback $f\circ \Phi\in \mathrm{Funct}(\Specm B,k)$ is algebraic.\end{proposition}

\begin{proof}
Suppose that $\Phi$ is algebraic.  It suffices to check that the following diagram is commutative:
\[ 
\xymatrix{
\mathrm{Funct}(\Specm A,k) \ar[r]^{-\circ\Phi} & \mathrm{Funct}(\Specm B,k) \\
A \ar[u] \ar[r]_{\phi} & B \ar[u]\\
}
\]
where $\phi:A\rightarrow B$ is the map that gives rise to $\Phi$.

[$\Leftarrow$] Suppose that for all algebraic functions $f\in \mathrm{Funct}(\Specm A,k)$, the pull-back $f\circ\Phi$ is algebraic.  Then we have an induced map, obtained by chasing the diagram counter-clockwise:
$$
\xymatrix{
\mathrm{Funct}(\Specm A,k) \ar[r]^{-\circ\Phi} & \mathrm{Funct}(\Specm B,k) \\
A \ar[u] \ar@{-->}[r]_{\phi} & B \ar[u]\\
}
$$
From $\phi$, we can construct the map $\Phi':\Specm B \rightarrow \Specm A$ given by $\Phi'(\mathfrak{m})=\phi^{-1}(\mathfrak{m})$.  I claim that $\Phi=\Phi'$.  If not, then for some $\mathfrak{m}\in \Specm B$ we have $\Phi(\mathfrak{m})\neq \Phi'(\mathfrak{m})$.  By definition, for all algebraic functions $f\in \mathrm{Funct}(\Specm A,k)$, $f\circ\Phi=f\circ\Phi'$ so to arrive at a contradiction we show the following lemma:\\
Given any two distinct points in $\Specm A=V(I)\subset k^n$, there exists some
algebraic $f$ that separates them.  This is trivial when we realize that any
polynomial function is algebraic, and such polynomials separate points.
\end{proof}

\section{Serre's criterion and its variants}

We are going to now prove a useful criterion for a noetherian ring to be a
product of 
normal domains, due to Serre: it states that a (noetherian) ring is normal if
and only if most of the localizations at prime ideals are discrete valuation
rings (this corresponds to the ring being \emph{regular} in codimension one,
though we have not defined regularity yet) and a more technical condition that
we will later interpret in terms of \emph{depth.} One advantage of this
criterion is that it does \emph{not} require the ring to be a product of
domains a priori.

\subsection{Reducedness}

There is a ``baby'' version of Serre's criterion for testing whether a ring is
reduced, which we star with.

Recall:

\begin{definition}
A ring $R$ is \textbf{reduced} if it has no nonzero nilpotents.
\end{definition}

\begin{proposition} \label{reducedcrit}
If $R$ is noetherian, then $R$ is reduced if and only if it satisfies the
following conditions:
\begin{enumerate}
\item Every associated prime of $R$ is	minimal (no embedded primes).
\item If $\mathfrak{p}$ is minimal, then $R_{\mathfrak{p}}$ is	a field.
\end{enumerate}
\end{proposition}
\begin{proof}
First, assume $R$ reduced. What can we say? Say $\mathfrak{p}$ is a minimal
prime; then $R_{\mathfrak{p}}$ has precisely one prime ideal (namely,
$\mathfrak{m}=\mathfrak{p}R_{\mathfrak{p}}$). It is in fact a local artinian
ring, though we
don't need that fact. The radical of $R_{\mathfrak{p}}$ is just $\mathfrak{m}$.
But $R$ was reduced, so $R_{\mathfrak{p}}$ was reduced; it's an easy argument
that localization preserves reducedness. So $\mathfrak{m}=0$. The fact that 0
is a maximal ideal in $R_{\mathfrak{p}}$ says that it is a field.

On the other hand, we still have to do part 1. $R$ is reduced, so $\rad(R) =
\bigcap_{\mathfrak{p} \in \spec R} \mathfrak{p} = 0$. In particular,
\[ \bigcap_{\mathfrak{p} \ \mathrm{minimal}}\mathfrak{p} = 0.  \]
The map
\[ R \to \prod_{\mathfrak{p} \ \mathrm{minimal}}R/\mathfrak{p}	\]
is injective. The associated primes of the product, however, are just the
minimal primes. So $\ass(R)$ can contain only minimal primes.

That's one direction of the proposition. Let us prove the converse now. Assume
$R$ satisfies the two conditions listed. In other words, $\ass(R)$ consists of
minimal primes, and each $R_{\mathfrak{p}}$ for $\mathfrak{p} \in \ass(R)$ is a
field. We would like to show that $R$ is reduced.
Primary decomposition tells us that there is an injection
\[ R \hookrightarrow \prod_{\mathfrak{p}_i \ \mathrm{minimal}} M_i, \quad M_i
\ \  \mathfrak{p}_i-\mathrm{primary}. \]
In this case, each $M_i$ is primary with respect to a minimal prime. We have a
map
\[ R \hookrightarrow \prod M_i \to \prod (M_i)_{\mathfrak{p}_i},  \]
which is injective, because when you localize a primary module at its
associated prime, you don't kill anything by definition of primariness. Since
we can draw a diagram
\[
\xymatrix{
R \ar[r] \ar[d]  &  \prod M_i \ar[d]  \\
\prod R_{\mathfrak{p}_i} \ar[r] & \prod (M_i)_{\mathfrak{p}_i}
}
\]
and the map $R \to \prod (M_i)_{\mathfrak{p}_i}$ is injective, the downward
arrow on the right injective. Thus $R$ can be embedded in
a product of the fields $\prod R_{\mathfrak{p}_i}$, so is reduced.
\end{proof}

This proof actually shows:
\begin{proposition}[Scholism] A noetherian ring $R$ is reduced iff it injects
into a product of fields. We can take the fields to be the localizations at the
minimal primes.
\end{proposition}

\begin{example}
Let $R = k[X]$ be the coordinate ring of a variety $X$ in
$\mathbb{C}^n$. Assume $X$ is
reduced. Then $\mathrm{MaxSpec} R$ is a union of irreducible components
$X_i$, which
are the closures of the minimal primes of $R$. The fields you get by localizing
at minimal primes depend only on the irreducible components, and in fact are
the rings of meromorphic functions on $X_i$.
Indeed, we have a map
\[ k[X] \to \prod k[X_i] \to \prod k(X_i).  \]

If we don't assume that $R$ is radical, this is \textbf{not} true.
\end{example}

There is a stronger condition than being reduced we could impose. We could say:

\begin{proposition}
If $R$ is a noetherian ring, then $R$ is a domain iff
\begin{enumerate}
\item $R$ is reduced.
\item $R$ has a unique minimal prime.
\end{enumerate}
\end{proposition}
\begin{proof}
One direction is obvious. A domain is reduced and $(0)$ is the minimal prime.

The other direction is proved as follows. Assume 1 and 2. Let $\mathfrak{p}$ be
the unique minimal prime of $R$. Then $\rad (R) = 0 = \mathfrak{p}$ as every
prime ideal contains $\mathfrak{p}$. As $(0)$ is a prime ideal, $R$ is
a domain.
\end{proof}

We close by making some remarks about this embedding of $R$ into a product of
fields.

\begin{definition}
Let $R$ be any ring, not necessarily a domain. Let $K(R)$ be the localized ring
$S^{-1}R$ where $S$ is the multiplicatively closed set of nonzerodivisors in
$R$.  $K(R)$ is called the \textbf{total ring of fractions} of $R$.

When $R$ is a field, this is the quotient field.
\end{definition}

First, to get a feeling for this, we show:

\begin{proposition} Let $R$ be noetherian. The set of nonzerodivisors $S$
can be described by
$S = R- \bigcup_{\mathfrak{p} \in \ass(R)} \mathfrak{p}$.
\end{proposition}
\begin{proof}
If $x \in\mathfrak{p} \in \ass(R)$, then $x$ must kill something in $R$ as it
is in an associated prime.	So $x$ is a zerodivisor.

Conversely, suppose $x$ is a zerodivisor, say $xy = 0$ for some $y \in R -
\left\{0\right\}$. In
particular, $x \in \ann(y)$. We have an injection $R/\ann(y) \hookrightarrow R$
sending 1 to $y$. But $R/\ann(y)$ is nonzero, so it has an associated prime
$\mathfrak{p}$ of $R/\ann(y)$, which contains $\ann(y)$ and thus $x$. But
$\ass(R/\ann(y)) \subset \ass(R)$.
So $x$ is contained in a prime in $\ass(R)$.
\end{proof}

Assume now that $R$ is reduced.  Then $K(R)  = S^{-1}R$ where $S$ is the
complement of the union of the minimal primes.
At least, we can claim:

\begin{proposition} Let $R$ be reduced and noetherian. Then
$K(R) = \prod_{\mathfrak{p}_i \ \mathrm{minimal}} R_{\mathfrak{p}_i}$.
\end{proposition}

So $K(R)$ is the product of fields into which $R$ embeds.

We now continue the discussion begun last time. Let $R$ be noetherian and $M$ a
finitely generated $R$-module. We would like to understand very rough features
of $M$.
We can embed $M$ into a larger $R$-module.
Here are two possible approaches.

\begin{enumerate}
\item  $S^{-1}M$, where $S$ is a large multiplicatively closed subset of $M$.
Let us take $S $ to be the set of all $a \in R$ such that $M
\stackrel{a}{\to}M$ is injective, i.e. $a$ is not a zerodivisor on $M$. Then
the map
\[ M \to S^{-1}M  \]
is an injection. Note that $S$ is the complement of the union of $\ass(R)$.
\item Another approach would be to use a \emph{primary decomposition}
\[ M \hookrightarrow \prod M_i,  \]
where each $M_i$ is $\mathfrak{p}_i$-primary for some prime $\mathfrak{p}_i$
(and these primes range over $\ass(M)$). In this case, it is clear that
anything not in each $\mathfrak{p}_i$ acts injectively. So we can draw a
commutative diagram
\[
\xymatrix{
M \ar[d]  \ar[r] &  \prod M_i \ar[d]  \\
\prod M_{\mathfrak{p}_i} \ar[r] &  \prod (M_i)_{\mathfrak{p}_i}
}.
\]
The map going right and down is injective.
It follows that $M$ injects into the product of its localizations at associated
primes.
\end{enumerate}

The claim is that these constructions agree if $M$ has no embedded primes.
I.e., if there are no nontrivial containments among the associated primes of
$M$, then $S^{-1}M$ (for $S =  R - \bigcup_{\mathfrak{p} \in \ass(M)}
\mathfrak{p}$)
is just $\prod M_{\mathfrak{p}}$.
To see this, note that any element of $S$ must act invertibly on $\prod
M_{\mathfrak{p}}$. We thus see that there is always a map
\[ S^{-1}M \to \prod_{\mathfrak{p} \in \ass(M)} M_{\mathfrak{p}} . \]
\begin{proposition}
This is an isomorphism if $M$ has no embedded primes.
\end{proposition}

\begin{proof}
Let us go through a series of reductions. Let $I = \ann(M) = \left\{a: aM
= 0\right\}$. Without loss of generality, we can replace $R $ by $R/I$. This plays nice with the
associated primes.

The assumption is now that $\ass(M)$ consists of the minimal
primes of $R$.

Without loss of generality, we can next replace $R$ by $S^{-1}R$ and $M$ by
$S^{-1}M$, because that doesn't affect the conclusion; localization plays nice
with associated primes.

Now, however, $R$ is artinian: i.e., all primes of $R$ are minimal (or
maximal). Why is this?
Let $R$ be \emph{any} noetherian ring and $S = R - \bigcup_{\mathfrak{p} \
\mathrm{minimal}} \mathfrak{p}$. Then I claim that $S^{-1}R$ is artinian. We'll
prove this in a moment.

So $R$ is artinian, hence a product $\prod R_i$ where each $R_i$ is local
artinian. Without loss of generality, we can replace $R$ by $R_i$ by taking
products. The condition we are trying to prove is now that
\[ S^{-1}M \to M_{\mathfrak{m}}  \]
for $\mathfrak{m} \subset R$ the maximal ideal. But $S$ is the complement of
the union of the minimal primes, so it is $R - \mathfrak{m}$ as $R$ has one
minimal (and maximal) ideal.  This is obviously an isomorphism: indeed, both
are $M$.
\end{proof}

\add{proof of artianness}
\begin{corollary}
Let $R$ be a noetherian ring with no embedded primes (i.e. $\ass(R)$ consists
of minimal primes).
Then $K(R) = \prod_{\mathfrak{p}_i \ \mathrm{minimal}} R_{\mathfrak{p_i}}$.
\end{corollary}
If $R$ is reduced, we get the statement made last time: there are no
embedded primes, and $K(R)$ is a product of
fields.

\subsection{The image of $M \to S^{-1}M$}
Let's ask now the following question. Let $R$ be a noetherian ring, $M$
a finitely generated
$R$-module, and $S$ the set of nonzerodivisors on $M$, i.e. $R -
\bigcup_{\mathfrak{p} \in \ass(M)} \mathfrak{p}$. We have seen that there is an
imbedding
\[ \phi: M \hookrightarrow S^{-1}M.  \]
What is the image? Given $x \in S^{-1}M$, when does it belong to the imbedding
above.

To answer such a question, it suffices to check locally. In particular:
\begin{proposition}
$x$ belongs to the image of $M $ in $S^{-1}M$ iff for every $\mathfrak{p} \in
\spec R$, the image of $x$ in $(S^{-1}M)_{\mathfrak{p}}$ lies inside
$M_{\mathfrak{p}}$.
\end{proposition}

This isn't all that interesting. However, it turns out that you can check this
at a smaller set of primes.

\begin{proposition}
In fact, it suffices to show that $x$ is in the image of $\phi_{\mathfrak{p}}$
for every $\mathfrak{p} \in \ass(M/sM)$ where $s \in S$.
\end{proposition}
This is a little opaque; soon we'll see what it actually means.
The proof is very simple.

\begin{proof}
Remember that $ x \in S^{-1}M$. In particular, we can write $x = y/s$ where $y
\in M, s \in S$. What we'd like to prove that $x \in M$, or equivalently that
$y \in sM$.\footnote{In general, this would be equivalent to $ty \in tsM$ for
some $t \in S$; but $S$ consists of nonzerodivisors on $M$.}
In particular, we want to know that $y$ maps to zero in $M/sM$. If not, there
exists an associated prime $\mathfrak{p} \in \ass(M/sM)$ such that $y$ does not
get
killed in $(M/sM)_{\mathfrak{p}}$.
We have assumed, however, for every associated prime $\mathfrak{p}\in \ass(M)$,
$x \in ( S^{-1}M)_{\mathfrak{p}}$ lies in the image of $M_{\mathfrak{p}}$. This
states that the image of $y$ in this quotient $(M/sM)_{\mathfrak{p}}$ is zero,
or that $y$ is divisible by $s$ in this localization.
\end{proof}

The case we actually care about is the following:

Take $R$ as a noetherian domain and $M = R$.  Then $S = R - \left\{0\right\}$
and $S^{-1}M $ is just the fraction field $K(R)$.  The goal is to describe $R$
as a subset of $K(R)$. What we have proven is that $R$ is the intersection in
the fraction field
\[ \boxed{ R = \bigcap_{\mathfrak{p} \in \ass(R/s), s \in R - 0}
R_{\mathfrak{p}} . }\]
So to check that something belongs to $R$, we just have to check that in a
\emph{certain set of localizations}.

Let us state this as a result:
\begin{theorem} \label{notreallykrullthm}
If $R$ is a noetherian domain
\[ R = \bigcap_{\mathfrak{p} \in \ass(R/s), s \in R - 0}
R_{\mathfrak{p}}  \]
\end{theorem}

\subsection{Serre's criterion}

We can now state a result.
\begin{theorem}[Serre]
\label{serrecrit1}
Let $R$ be a noetherian domain.	Then $R $ is integrally
closed iff it satisfies
\begin{enumerate}
\item For any $\mathfrak{p} \subset R$ of height one, $R_{\mathfrak{p}}$ is a
DVR.
\item For any $s \neq 0$, $R/s$ has no embedded primes (i.e. all the
associated primes of $R/s$ are height one).
\end{enumerate}
\end{theorem}

Here is the non-preliminary version of the Krull theorem.
\begin{theorem}[Algebraic Hartogs]
Let $R$ be a noetherian integrally closed ring. Then
\[ R = \bigcap_{\mathfrak{p} \ \mathrm{height \ one}} R_{\mathfrak{p}},  \]
where each $R_{\mathfrak{p}}$ is a DVR.
\end{theorem}

\begin{proof}
Now evident from the earlier result \cref{notreallykrullthm} and Serre's criterion.
\end{proof}
Earlier in the class, we proved that a domain was integrally closed if and only
if it could be described as an intersection of valuation rings. We have now
shown that when $R$ is noetherian, we can take \emph{discrete} valuation rings.

\begin{remark}
In algebraic geometry, say $R = \mathbb{C}[x_1, \dots, x_n]/I$.  Its maximal
spectrum is a subset of $\mathbb{C}^n$.  If $I$ is prime, and $R$ a domain,
this variety is
irreducible.  We are trying to describe $R$ inside its field of fractions.

The field of fractions are like the ``meromorphic functions''; $R$ is like the
holomorphic functions. Geometrically, this states to check that a meromorphic
function is holomorphic, you can just check this by computing the ``poleness''
along each codimension one subvariety. If the function doesn't blow up on each
of the codimension one subvarieties,  and $R$ is normal, then you can extend it
globally.

This is an algebraic version of Hartog's theorem: this states that a
holomorphic function on $\mathbb{C}^2 - (0,0)$ extends over the origin, because
this has codimension $>1$.

All the obstructions of extending a function to all of $\spec R$ are in
codimension one.
\end{remark}

Now, we prove Serre's criterion.
\begin{proof}
Let us first prove that $R$ is integrally closed if 1 and 2 occur. We know that
\[ R = \bigcap_{\mathfrak{p} \in \ass(R/x), x \neq 0} R_{\mathfrak{p}} ; \]
by condition 1, each such $\mathfrak{p}$ is of height one, and
$R_{\mathfrak{p}}$ is a DVR. So $R$ is the intersection of DVRs and thus
integrally closed.

The hard part is going in the other direction. Assume $R$ is integrally closed.
We want to prove the two conditions. In $R$, consider the following conditions
on a prime ideal $\mathfrak{p}$:
\begin{enumerate}
\item $\mathfrak{p}$ is an associated prime of $R/x$ for some $x \neq 0$.
\item $\mathfrak{p} $ is height one.
\item $\mathfrak{p}_{\mathfrak{p}}$ is principal in $R_{\mathfrak{p}}$.
\end{enumerate}
First, 3 implies 2 implies 1. 3 implies that $\mathfrak{p}$ contains an element
$x$ which
generates $\mathfrak{p}$ after localizing.
It follows that there can be no prime between $(x)$ and $\mathfrak{p}$ because
that would be preserved under localization.  Similarly, 2 implies 1 is easy. If
$\mathfrak{p}$ is minimal over $(x)$, then $\mathfrak{p} \in \ass R/(x)$ since
the minimal primes in the support are always associated.

We are trying to prove the inverse implications. In that case, the claims
of the theorem will be proved. We have to show that 1 implies 3.
This is an argument we really saw last time, but let's see it again. Say
$\mathfrak{p} \in \ass(R/x)$. We can replace $R$ by $R_{\mathfrak{p}}$ so that
we can assume that $\mathfrak{p}$ is maximal. We want to show that
$\mathfrak{p}$ is generated by one
element.

What does the condition $\mathfrak{p} \in \ass(R/x)$ buy us? It tells us that
there is $\overline{y} \in R/x$ such that $\ann(\overline{y}) = \mathfrak{p}$.
In particular, there is $y \in R$ such that $\mathfrak{p}y \subset (x)$ and $y
\notin (x)$.
We have the element $y/x \in K(R)$ which sends $\mathfrak{p}$ into $R$.  That
is,
\[ (y/x) \mathfrak{p} \subset R.  \]
There are two cases to consider, as in last time:
\begin{enumerate}
\item $(y/x) \mathfrak{p}  = R$. Then $\mathfrak{p} = R (x/y)$ so $\mathfrak{p}
$ is principal.
\item $(y/x) \mathfrak{p} \neq R$. In particular, $(y/x)\mathfrak{p} \subset
\mathfrak{p}$. Then since $\mathfrak{p}$ is finitely generated, we find that
$y/x $ is
integral over $R$, hence in $R$. This is a contradiction as $y \notin (x)$.
\end{enumerate}
Only the first case is now possible. So $\mathfrak{p}$ is in fact principal.
\end{proof}