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\title{Algebra III\\December 7, 2004}
\author{Gabriel C. Drummond-Cole}
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So, uh, let me review a couple of things.  The plan for this week is to talk about Dedekind rings, rings of Krull dimension one.  The standard example is 
$$
\xymatrix{
\mathbb{Z}\ar[r]\ar[d]^\subset&O_K\ar[d]_\subset
\\\mathbb{Q}\ar[r] &K\subset \mathbb{C}
}$$
Among other things $O_K$ is an integral extension.    People have been concerned with these because they are natural.  One example is $\mathbb{Z}[i].$  You could have $\mathbb{Z}[\frac{1+\sqrt{-5}}{2}].$  So this is $O_{\mathbb{Q}(\sqrt{-5})}.$  

What happens in a ring like this?  The Gaussian integers are a UFD, but the second ring is not because $3\cdot 3=(2+\sqrt{-5})(2-\sqrt{-5}).$  

You have $z^n=x^n+y^n,$ so the hope is to factor these in a UFD somewhere.  So you go to an algebraic extension.  But unique factorization doesn't hold.  So how far is this from a UFD?  What are its properties?  This gets more complicated with $n$th roots.  But what can we say about this ring?

One consequence of going up and incomparability is that $O_K$ has the same Krull dimension as $\mathbb{Z},$ that is, one.

I had a homework for you, I should have said that.  I'll start talking about such rings in a moment, I'd like to give this:
\begin{exercise}
We talked about what it means for $R\hookrightarrow S$ to be an integral extension, an element satisfies a monic polynomial.  But instead start with $I\subset R$ and look to $\bar{I}=\{x\in R:x$ satisfies a monic polynomial $x^n+a_1x^{n-1}+\cdots+a_n-0$ with $a_j\in I^j$ for all $j.$
\begin{enumerate}
\item We had the lemma with four points, equivalent conditions for integrality.  Use the same methods to show that $\bar{I}$ is an ideal.
\item If $I$ is radical then $I$ is integrally closed in $R.$
\item This has to do with convex geometry.  This is due to Caratheodory.  If $I\subset k[x_1,\cdots,x_n]$ is generated by a set of monomials $\Gamma,$ then $\tilde{\Gamma},$ the collection of exponents of the monomials in $\Gamma.$  This is in $\mathbb{N}^n\subset \mathbb{R}^n_+.$  Take the cone $\Lambda$ these generate (their convex hull), defined as $\tilde{\Gamma}+\mathbb{R}^n_+.$  Then the monomials with generators in $\Lambda$ are the generators of $\bar{I}.$
\end{enumerate}
\end{exercise}
Try to solve this: it is easy and instructive.

Okay, so how do we characterize $O_K?$  What is special about these?  The Krull dimension is one, and these are domains, which is equivalent to the statment that every nonzero prime ideal is maximal.  I will show that every ring with this property has unique factorization of ideals.
\begin{prop}
Let $R$ be a Noetherian domain such that every nonzero prime ideal is maximal.  Then every nonzero ideal can be uniquely expressed as a product of primary ideals whose radicals are all distinct.
\end{prop}
I don't have unique factorization into primes of various powers; instead I get primary ideals for each prime.

Let's prove this.  It's some unique factorization statement.  This is better than nothing.  This is specific to this situation; it wouldn't hold in general.  For almost two centuries people believed that this would prove Fermat.  Kummer got a little progress, but it didn't pan out.

This is the subject of basic algebraic number theory, but also basic arithmetic algebraic geometry.

How do we prove this?  Take $I\subset R,I\ne 0.$  How do I get this decomposition?  What decompositions did we learn, in general for any Noetherian ring?  Primary decomposition, so $I=\cap_{i=1}^n q_i,$ where $q_i$ are primary.  I don't know if it's unique or not.  I would know that $\cap q_i=\prod q_i$ (from the Chinese remainder lemma) if I knew that $q_i+q_j=(1)$ for all $i\ne j.$  So $p_i=\sqrt{q_i}.$  So the radical is bigger than the ideal $p_i\supset q_i\supset I\ne 0$ so $p_i$ is maximal.  So if I take a minimal decomposition, I can assume these are distinct maximal $p_i.$  If you have two maximal ideals then $p_i+p_j=1.$  This is not quite what I want, I want this for $q_i,q_j.$  It is enough to show that $\sqrt{q_i+q_j}=1;$ this is $\sqrt{\sqrt{q_i}+\sqrt{q_j}}=\sqrt{p_i+p_j}=\sqrt{1}=1.$

Now why is this unique?  Is primary decomposition unique?  What was unique in a primary decomposition?  This is not so long ago?  The unique ones were the isolated, minimal ones.  But here all the $p_i$ are minimal associated primes, so all the $q_i$ are unique.

Now let's make a change.  I don't want to assume only this, I want to assume a little more.  Let's make a little, let's start with $R$ a Noetherian domain with Krull dimension one.  What's missing is that every primary ideal is a power of a prime.

Assume that as well.  Then every nonzero ideal is uniquely a product of prime ideals.  This sounds like unique factorization, but it is in ideals.  Maybe I can do algebra with these, form fractions or whatever.  This is called the class group or for geometers the Picard group.

What can you say about such a ring?  How special is it?  I want to localize at a nonzero prime?  If I localize at a maximal ideal the chains are the same; it's not very easy but you can show that if you localize at a maximal ideal you get the same.  Here, though, that's easy.  You get a local ring with a unique nonzero maximal ideal.  If something was a power of a prime to start out with, it's also a power of a prime after localization.  Then every nonzero ideal is a power of the single maximal ideal.  This is quite special.  It's a local ring with every nonzero ideal equal to $(pR_p)^n$ for some $n.$  This is called a discrete valuation ring.

This number $n$ has nice properties; it will behave like a logarithm.
\begin{defi}
A discrete valuation on a field (here the field of fractions) is a function $v:K^*\to\mathbb{Z}$ which is onto with the following property:
\\$v(xy)=v(x)+v(y)$
\\$v(x+y)\ge \min\{v(x),v(y))$ with convention $v(0)=\infty.$\end{defi}  So for each prime you get a different valuation defined by the $n.$  You will sometimes see nondiscrete valuation with values in a group $\mathbb{R}$ or something.

So $R_v=\{x\in K:v(x)\ge 0\}.$  I claim that this is a subring of $K.$  It should contain $0$ and $1.$  I have to show that it's closed under sums, differences, and products, which follow from these properties.

So let me give some examples.

The one we started with we have $\mathbb{Z}\subset \mathbb{Q}$ and I put a valuation which is the unique power of $p$ which can be factored out of a rational.

The big question is, what is $R_{v_p}?$  It is $\mathbb{Z}_p.$ in this case.

How about another one.  Let $K=k(x)$ and $f$ an irreducible element.  Then evaluation is the power of $f$ in your fraction.  Then $R_{v_p}=k[x]_{(f)}.$  If you look to rings of power series $k[[x]]$ in one variable, you can look to the smallest coefficient which is nonzero in the expansion $v(f)=\min\{i:a_i\ne 0\}.$

\begin{defi} An integral domain is called a discrete valuation ring (DVR) if there exists a valuation $v$ on $K$ (the field of fractions) such that the original domain is the evaluation ring of $v.$
\end{defi}
For instance, $\mathbb{Z}_p$ and $k[x]_{(f)}$ are DVRs.

Did I prove that integral closure and localization commute?  Assume I'm in this situation so that $R\subset K=Q(R)$ with $v:K^*\to\mathbb{Z}.$  Let $x\ne 0$ be in $K.$  If $v(x)$ is positive then $x\in R.$  Say $v(x)$ is negative; then $v(x^{-1})$ is $-v(x)$ so $x^{-1}\in R.$  

So now we will show that a discrete valuation ring is a local ring.  The maximal ideal will be $\{x\in K:v(x)>0\}.$  The sum of elements is there because of the one property; the product can only increase the valuation.  So what do I have to show?  It contains the noninvertibles, which is obvious.

It has even more interesting properties.  Now let's prove something about the Krull dimension.
\begin{lemma}
Let $x,y\in R$ with $v(x)=v(y).$  Then $(x)=(y).$  This is equivalent to $x$ and $y$ differing by a unit.  So $v(xy^{-1})=0$ and is thus a unit in $R.$
\end{lemma}
Once you get rid of discrete valuation you are really in general rings.  You have heard of the Hironaka theorem about desingularization.  That's very hard.  I didn't tell you what improved?  The object that keeps track of the improvement is a valuation, but not a discrete one.  This is a baby example because it is discrete.

Let's state what are the ideals of a DVR?
\begin{lemma}
$m_n=\{x:v(x)\ge n\}$ is an ideal for all $n.$

There is a descending sequence $m_1\supset m_2\supset\cdots$  

Every ideal of $I$ is one of these $m_n$ and $m_k=(m_1)^k.$
\end{lemma}
From this lemma, the only chain is this one, so the ring is Noetherian.

If $I\ne 0$ then take $n$ which is minimal for $v(x)$ for $x\in I.$  This should remind you of proving $k[x]$ is a PID.  Then $m_n=\{y\in R:v(y)\ge n\}\subset I.$  If $y\in R$ then $yx^{-1}\in R$ so $y\in (x)\subset I.$  Equality is easy too.  Then $m_n=(x^n)$ for some $x\in m.$

Next time I'll prove some identities and then we'll go back to the nonlocal version.

Thursday we finish with Dedekind rings.
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