Fractional Ideals, Part 2

First, note to any reader who doesn’t understand these posts: feel free to ask questions – that’s what I’m here for.

Second, continuing from my previous post about ideal theory, I’m going to show that given any fractional ideal I of a Dedekind domain, there exists another fractional ideal J such that IJ is precisely the domain.

Recall that a Dedekind domain R (with fraction field K) is defined by three properties:

  1. It is integrally closed. If c in K satisfies a polynomial x^n + a(n-1)x^(n-1) + … + a(1)x + a(0) = 0, then it must be in R.
  2. It is Noetherian. If I1, I2, I3… is a chain of ideals such that I(n+1) contains I(n) for all n, then for some m, we have I(m) = I(m+1) = I(m+2) = … Equivalently, if I = (a1, a2, a3…), then for some m we have I = (a1, a2… a(m)).
  3. Every prime ideal, except (0), is maximal.

I’m going to proceed in four steps. In step 1, I show that every ideal A contains a product of prime ideals. In step 2, I show that for every prime ideal P, the inverse of P defined in my previous post, call it S, satisfies PS = R. In step 3, I show that every integral ideal can be uniquely factored into prime ideals. In step 4, I show that every fractional ideal can be factored into prime ideals and their inverses. In step 5, I show that the theorem in step 2 applies to all fractional ideals.

Step 1: if an ideal I1 does not contain a product of prime ideals, it’s clearly not prime, in which case it just contains itself. So we can find ideals A and B such that AB is contained in I1, but neither A nor B is contained in I1. Now, (I1 + A)(I1 + B) = (I1^2 + AI1 + BI1 + AB) which is contained in I1. So at least one of I1 + A and I1 + B doesn’t contain a product of prime ideals; call that one I2, and note that since A and B are not contained in I1, I2 is strictly bigger than I1. We can then similarly define I3, I4, I5… which gives us an ascending chain of ideals that doesn’t terminate. But R is Noetherian, so we have a contradiction.

Step 2: if P is a prime ideal, then it’s a maximal ideal. Define Q = {q in K: qP is contained in R}. Clearly, QP is contained in R. Every element of R is in Q, since rP is in P by the definition of an ideal. Therefore, QP contains RP = P; this means that QP is P or R, since P is a maximal ideal. We show that QP = R is two steps: first we show that Q is bigger than R, and only then we show QP = R.

Let a be any nonzero member of P. The ideal (a) must contain a product of prime ideals, say P1P2…P(n); we can choose n to be the smallest number for which this holds. The product is then contained in P; since P is a prime ideal, at least one of the ideals, say P1, is contained in P. But P1 is a prime ideal, so it’s maximal. This means it’s equal to P, or else P is between P1 and R, contradicting the definition of maximality.

Now, let’s look at P2P3…P(n). By our choice of n, it is not contained in (a), so it contains an element not in (a), call it b. If b were divisible by a, it would be of the form ac, which is in (a); therefore, b is not divisible by a, so the element b/a is in K but not in R. But now bP = bP1 is in P1P2…P(n), which is in (a), so (b/a)P is in (a/a) = R, and (b/a) is in Q.

If QP = P, then QQP = QP = P, so if c and d are in Q, then so is cd. Let b/a be in Q but not in R as above; then Q contains (1, b/a, b^2/a^2, b^3/a^3…). As Q is a fractional ideal, multiplying by a suitable r in R gives an integral ideal of R. As R is Noetherian, the integral ideal has only finitely many generators, i.e. is of the form (r, rb/a, rb^2/a^2… rb^m/a^m). We then get that the fractional ideal is of the form (1, b/a, b^2/a^2… b^m/a^m). We write (b/a)^(m+1) as the sum of the generators of the ideal; by moving terms around, we get (b/a)^(m+1) + c(m)(b/a)^m + … + c(1)(b/a) + c(0) = 0. As R is integrally closed, it means b/a is in R, which is a contradiction. Therefore, we must have QP = R.

Step 3: first, we show that each ideal can be expressed as a product of prime ideals; then we show uniqueness. If I1 can’t be factored into prime ideals of R, then I1 at least contains the product P1P2…P(n); we can choose n to be the smallest integer for which this holds. Given P1, define Q1 as in step 2; then I1Q1 contains P2P3…P(n). If I1Q1 can be expressed as a product of prime ideals, then so can I1 = I1Q1P1. If I1Q1 = I1, then I1 contains P2P3…P(n), contradicting the choice of n. So I2 = I1Q1 is a bigger ideal than I1 that isn’t a product of prime ideals. We then get a chain I1, I2, I3… which is impossible since R is Noetherian.

Now, if I = P1P2…P(n) = P(n+1)P(n+2)…P(n+m), then P1 contains the product P(n+1)…P(n+m); hence it contains one of these ideals, say P(n+1); since P(n+1) is prime and hence maximal, P1 = P(n+1). We multiply I by Q1 = Q(n+1) and continue this process until all ideals in the two factorizations have been paired off, which means the two factorizations are the same.

Step 4: let I be a fractional ideal of R, and let r be an element of R such that rI is an integral ideal of R. Both (r) and rI are integral ideals, so we can write them as products of prime ideals, P1P2…P(n) for (r) and P(n+1)…P(n+m) for rI. Then I = rI/r = Q1Q2…Q(n)P(n+1)…P(n+m).

Now, if I = sI/s = Q(n+m+1)…Q(n+m+k)P(n+m+k+1)…P(n+m+k+l), then look at rsI = P(n+1)…P(n+m+k) = P1…P(n)P(n+m+k+1)…P(n+m+k+l). By uniqueness of ideal factorization, we can pair off these two sets of prime ideals. This allows us to pair off the factorizations of rI/r and sI/s, possibly after killing off pairwise inverse Q(i)’s and P(i)’s (for example, if Q1 is the inverse of P(n+1)).

More generally, if we have two equal factorizations into P’s and Q’s, we multiply each side by the P’s corresponding to the other’s Q’s. We can do this, because if Q is the inverse of P, then P is the inverse of Q, i.e. P = P* where P* = {k in K: kQ is contained in R}. To see why, remember that Q strictly contains R, so if k is not in R, then k*1 = k is not in R, which means P* is an integral ideal of R; 1Q = Q which is not contained in R, so P* is proper; and PQ = R, so P* contains P; then P* = P since P is a maximal ideal.

Step 5: if I =  Q1Q2…Q(n)P(n+1)…P(n+m), then its unique inverse is clearly P1P2…P(n)Q(n+1)…Q(n+m).
I’m going to explore fractional ideals more later, and explain a little bit about why they’re useful. But I’m going to introduce a theorem I’m not going to prove – while its proof is fairly easy, it requires a ton of theory behind it, most of which I’ve omitted so far on the grounds that it’s confusing and for now unnecessary.

About these ads

One Response to Fractional Ideals, Part 2

  1. Outlets and switches, occurrences of the?Popularity websites like, This is completely.Storyville (red light), find any women.Self-esteem booster Geile Treffen mit Sex vor der Chatcam, example is an your whole family.Or move while, horrible your mornings.,

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

%d bloggers like this: