Math 221 : Notes for Oct

Math 221 : Notes for Oct. 9

Alison Miller

1 Second Uniqueness Theorem

I = T q A bit more terminology: if k k is a minimal√ primary decomposition, we say that qk are the primary components of I, and if qk = pk we say that qk is pk-primary. By the First uniqueness theorem from last time we know that every minimal primary decomposition has one primary component for each associated prime. Now we prove the the proposition we stated last time. T Proposition 1.1. Suppose that I = k qk is a minimal primary decomposition, and S is a multiplicatively closed subset of A. Then

\ j∗(I) = j∗(qk) √ qk∩S=∅ is a minimal primary decomposition of j∗(I). and

∗ \ j (j∗(I)) = qk √ qk∩S=∅ ∗ is a minimal primary decomposition of j (j∗(I)).

T −1 Proof. By previous proposition, we have j∗(I) = k j∗(qk), and since j∗(qk) = S A when- ever qk has nonempty intersection with S, we can just drop those terms. This shows it’s a primary decomposition. Likewise, to get the second equation, we just apply j∗ to the whole thing. This is clearly minimal since our original primary decomposition was minimal. Likewise the only bit that’s at all complicated is to show that the ﬁrst decomposition ∗ is minimal. For this, suppose otherwise that j∗(qk) ⊂ ∩j6=kj∗(qj). Then applying j to ∗ this, we get qk = j (j∗(qk)) ⊂ ∩j6=kqj, but we know that’s okay.

1 2 Application to irreducible decomposition

Proposition 2.1. If A is any noetherian ring, and I ⊂ A is a radical ideal, then I can be written T uniquely as k pk where no pk contains any pj for j 6= k, and this is the unique minimal primary decomposition of I.

Proof. Take any minimal primary decomposition I = ∩kqk, and take radicals to get I = ∩kpk where pk = qk. This is still a primary decomposition, and it’s still minimal, so no pk can contain any other pj, hence all the pk are minimal primes. By the second uniqueness theorem, this means that this is the unique minimal primary decomposition. 0 0 Furthermore, if we have any expression of I = ∩ppk such that no pk contains any 0 other pj , this is also a minimal primary decomposition

3 Intersections and Products

Now we are going to move from taking intersections of ideals to multiplying ideals. To do this we need the following lemma.

Lemma 3.1. If I1, I2, ... , In are ideals of A such that Ij + Ik = (1) for j 6= k, then I1 ∩ I2 ∩ · · · ∩ In = I1I2 ··· In.

Proof. HW

We are now going to introduce a property a ring can have that will imply that every ideal is a product of primary ideals.

Deﬁnition. If A is an integral domain, we say that A has (Krull) dimension ≤ 1 if every nonzero prime ideal of A is maximal.

(Comment: in general, we say that the Krull dimension of a ring A is the largest integer n such that there is a sequence of prime ideals p0 ( p1 ( ··· ( pn of A with length n + 1. In the case above, the longest such chain is 0 ( p for any prime ideal of A, assuming that A has any nonzero prime ideals; otherwise A is a field, and has Krull dimension 0.) Example. Z has Krull dimension ≤ 1; so C[t], or k[t] for any field k, in fact we’ll see later that any PID has Krull dimension ≤ 1 (in fact, = 1 if it’s not just a field). As well, for any irreducible polynomial f(x, y), the ring C[x, y]/(f) has Krull dimension ≤ 1 (again, actually = 1) – this will probably be on the problem set.

Proposition 3.2. If A is Noetherian of Krull dimension ≤ 1, then every nonzero ideal I of A can be uniquely expressed as a product of primary ideals with distinct radicals.

2 Proof. We show ﬁrst that for nonzero primary ideals q1, ... , qn with distinct radicals we have ∩ q = q . For this, we use the lemma above; we need to show that q + q = (1) k k k k√ √ √ j k √ for j 6= k. Now, q + q is an ideal containing both ideals q and q . But q √ Q j k j k j q and k are nonzero prime ideals, so√ they are both maximal, and therefore any ideal containing both of them is (1). Hence qj + qk = (1) so qj + qk = 1. Hence it sufﬁces to show that I has a unique primary decomposition. However, the assumption that A has Krull dimension 1 means that none of the associated primes of I can contain any of the others, so they are are minimal, and so the primary decomposition of A is determined by the second uniqueness theorem.

4 Dedekind Rings

We now deﬁne a class of rings in which every ideal has a unique factorization into powers of primes, not just into primary ideals.

Deﬁnition. We say that A is a Dedekind domain if A is Noetherian of dimension ≤ 1 and A is integrally closed in its ﬁeld of fractions.

Example. Again, the rings Z, √C[t] and [˛t] are Dedekind, as are any PID. A non-PID example is Z[ −5]. More generally, we’ll later be able to prove that if K is a ﬁnite extension of Q, the√ integral closure of Z in K is Dedekind; so for instance all of the integral closures in Q[ D] you calculated on the problem set (in the case where D is not square, that is.) Another fact that we’ll see later: the ring C[x, y]/(f) (with f irreducible) is a Dedekind domain if and only if the variety V(f) is smooth. (This means roughly what you expect, ∂f ∂f geometrically; algebraically it means that the polynomials ∂x and ∂y are never both zero at any point of V(f).)