ALGEBRA III DECEMBER 7, 2004

GABRIEL C. DRUMMOND-COLE

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{ℤ \ar[r] \ar[d]^{⊂} & O_{K} \ar[d]_{⊂} \\ ℚ \ar[r] & K ⊂ ℂ}

Among other things O_{K} is an integral extension. People have been concerned with these because they are natural. One example is ℤ[i]. You could have ℤ[\frac{1+\sqrt{−5}}{2}]. So this is O_{ℚ(\sqrt{−5})}.

What happens in a ring like this? The Gaussian integers are a UFD, but the second ring is not because 3 · 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 nth 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 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:

Exercise 1. We talked about what it means for R ↪ S to be an integral extension, an element satisfies a monic polynomial. But instead start with I ⊂ R and look to Ī = {x ∈ R : x satisfies a monic polynomial x^{n} + a_{1}x^{n−1} + ⋯ + a_{n} − 0 with a_{j} ∈ I^{j} for all j.

(1) We had the lemma with four points, equivalent conditions for integrality. Use the same methods to show that Ī is an ideal.

(2) If I is radical then I is integrally closed in R.

(3) This has to do with convex geometry. This is due to Caratheodory. If I ⊂ k[x_{1}, ⋯, x_{n}] is generated by a set of monomials Γ, then Γ̃, the collection of exponents of the monomials in Γ. This is in ℕ^{n} ⊂ ℝ_{+}^{n}. Take the cone Λ these generate (their convex hull), defined as Γ̃ + ℝ_{+}^{n}. Then the monomials with generators in Λ are the generators of Ī.

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.

Proposition 1. 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.

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 ⊂ R, I ≠ 0. How do I get this decomposition? What decompositions did we learn, in general for any Noetherian ring? Primary decomposition, so I = ∩_{i=1}^{n}q_{i}, where q_{i} are primary. I don’t know if it’s unique or not. I would know that ∩q_{i} = ∏ q_{i} (from the Chinese remainder lemma) if I knew that q_{i} + q_{j} = (1) for all i ≠ j. So p_{i} = \sqrt{q_{i}}. So the radical is bigger than the ideal p_{i} ⊃ q_{i} ⊃ I ≠ 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.

Definition 1. A discrete valuation on a field (here the field of fractions) is a function v : K^{∗} → ℤ which is onto with the following property: v(xy) = v(x) + v(y) v(x + y) ≥ min{v(x), v(y)) with convention v(0) = ∞.

So for each prime you get a different valuation defined by the n. You will sometimes see nondiscrete valuation with values in a group ℝ or something.

So R_{v} = {x ∈ K : v(x) ≥ 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 ℤ ⊂ 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 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} ≠ 0}.

Definition 2. 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.

For instance, 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 ⊂ K = Q(R) with v : K^{∗} → ℤ. Let x ≠ 0 be in K. If v(x) is positive then x ∈ R. Say v(x) is negative; then v(x^{−1}) is −v(x) so x^{−1} ∈ R.

So now we will show that a discrete valuation ring is a local ring. The maximal ideal will be {x ∈ 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.

Lemma 1. Let x, y ∈ 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.

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?

Lemma 2. m_{n} = {x : v(x) ≥ n} is an ideal for all n.

There is a descending sequence m_{1} ⊃ m_{2} ⊃ ⋯

Every ideal of I is one of these m_{n} and m_{k} = (m_{1})^{k}.

From this lemma, the only chain is this one, so the ring is Noetherian.

If I ≠ 0 then take n which is minimal for v(x) for x ∈ I. This should remind you of proving k[x] is a PID. Then m_{n} = {y ∈ R : v(y) ≥ n} ⊂ I. If y ∈ R then yx^{−1} ∈ R so y ∈ (x) ⊂ I. Equality is easy too. Then m_{n} = (x^{n}) for some x ∈ m.

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

Thursday we finish with Dedekind rings.