Dedekind Domains, Localization Produces a DVR

Localization Produces a DVR

Dedekind → Noetherian and Local DVRs

Let R be dedekind and let P be one of its prime/maximal ideals. Let S be the localization of R about P. In other words, S consists of fractions with numerators in R, and denominators not contained in P. Let Q be the localization of P in S. These are fractions with numerators in P and denominators not in P.

Remember that all prime ideals are maximal. Take any prime ideal other than P and find an element x not in P. The localization of this prime ideal in S includes x/1. Since S also contains 1/x, the localization of this prime ideal, i.e. a prime ideal other than P, brings in all of S. Localization kills off all prime ideals except P → Q.

Let J be an ideal in S, and pull back to H, an ideal in R. Write H as a product of prime ideals in R. Since product commutes with localization, find the corresponding ideals in S, then take their product to find J. All the prime ideals other than P map to all of S. Only the powers of P remain. If H includes P^k in its factorization, then J = Q^k.

Every ideal in S is a power of Q. Ideals are linearly ordered by containment, and that makes S a valuation ring. Furthermore, the chain of ideals corresponds to the positive integers, the exponent on Q. This gives S a valuation group of Z, hence S is a dvr, and a pid.

In summary, a dedekind domain is noetherian, and localization about every prime ideal produces a valuation ring, which happens to be a dvr and a pid.

Noetherian and Local DVRs → Invertible Ideals

Since R is noetherian, the localization of R about P is also noetherian. (An ascending chain of ideals in the localization would pull back to an ascending chain of ideals in R.) And a noetherian valuation ring is a pid. Thus every localization about a prime ideal produces a pid.

Let H be a nonzero proper ideal of R, and let H′ be the elements of F that drive H into R. Recall that H′ is an R submodule of F. If H is invertible, then H′ is indeed the inverse of H.

Suppose H is not invertible. Thus H*H′ is a proper ideal of R. Embed HH′ in a maximal ideal P. Of course a maximal ideal is prime, which is why I'm calling it P.

Now localize about P. In other words, bring in denominators from S, where S = R-P. Now R/S is a dvr, and a pid. Its maximal ideal is P/S, which I will call Q.

Remember that HH′ lies in P, and since H′ contains 1, H also lies in P. Since H lies in P, H/S lies in Q, and is principal. Let x generate H/S, where x is an element of H.

At this point we need the fact that R is noetherian. Let y₁ y₂ y₃ … y_n generate H in R. Bring in a denominator of 1, and these generators embed in Q. For each generator y_i, write the following.

y_i/1 = x * a_i/s_i (x generates the ideal)

y_i/x = a_i/s_i

Let t be the product over s_i, an element in R-P. Within the fraction field F, consider t/x times y_i. This is the same as t times y_i/x. Since y_i/x is the same as a_i/s_i, multiplying by t gives something in R. Thus t/x times y_i is in R.

This holds for each y_i, so t/x drives H into R, and t/x is a member of H′.

Multiply t/x times x. This is part of HH′. Therefore t lies in P. Yet we know t is in R-P. This is a contradiction, hence HH′ is all of R, and H is invertible.

Every ideal of R is invertible.

Invertible Ideals → Invertible Prime Ideals

Invertible Prime Ideals → Invertible Maximal Ideals and Noetherian

An invertible ideal is finitely generated, and if the prime ideals are finitely generated, then R is noetherian.

Invertible Maximal Ideals and Noetherian → Dedekind

Let H be any proper, nonzero ideal of R. Pull out the factors of H, one by one, where each factor is a maximal ideal of R. Multiply H by M′, since each maximal ideal M is invertible. That's how we "pull out" the factor of M. Repeat this process until you are left with a maximal ideal. Thus H is a product of maximal ideals.

Before we explore this process in detail, let's prove the factorization is unique.

Let H be the product of maximal ideals in two different ways. If the maximal ideal M is common to both factorizations, multiply through by M′. This effectively cancels M. Keep doing this, until the ideals in the first factorization are completely different from the ideals in the second.

Suppose the first factorization has fewer ideals, and let M be an ideal in this factorization. Now the product of the ideals on the right gives an ideal that is contained in M. Since M is maximal it is prime, which means one of the ideals on the right is contained in M. Those ideals are maximal too. If one maximal ideal contains another they are equal. Yet the primes on the right are suppose to be different from the primes on the left. This is a contradiction, hence the factorization of H into maximal ideals is unique. We only need show that a factorization exists.

It is enough to find one factor M of H, and prove that HM′ is larger than H. Divide a number by one of its prime factors and the result is smaller. Pull a maximal ideal out of H and the resulting ideal is larger. At least it's suppose to be (the proof is not trivial). If this can always be done, then pull out one factor after another, and multiply H by M inverse at each step. This builds an ascending chain of ideals. Since R is noetherian the chain cannot continue forever. At some point H itself becomes a maximal ideal, and the process terminates. This establishes a factorization for H. The hard part is showing HM′ is strictly larger than H. Here we go.

Let H be a nonzero ideal of R that is not maximal. Embed H in a maximal ideal M, and let M′ be the inverse of M. Note that M′ includes all of R, plus some fractions not found in R.

Let J = M′H. Since H is contained in M, M′H is contained in R. This makes J an ideal.

If J = R, write H = HR = HM′M = JM = RM = M. We already said H is properly contained in M, so this is a contradiction. Therefore J is a proper ideal of R.

Since M′ contains 1, M′H contains H, and J is a proper ideal that contains H. If J is larger than H we are done. That's all we need to finish the proof, and R becomes a dedekind domain. Suppose J = H, and derive a contradiction.

Since M′H = H, multiply through by M, giving H = MH. Multiplying H by the maximal ideal that contains it doesn't change H at all.

We are about to localize about M, and that means R must have a well defined fraction field. Well - if R has invertible ideals, I guess it has a fraction field eh? I mean, this kinda goes without saying, but I felt the need to say it.

Localize R about M, and recall that the localization of an invertible ideal remains invertible. This means M_M is an invertible ideal in the local ring R_M. Call this ideal W. Thus W is both invertible and maximal in R_M.

If W = W² then multiply through by W′, and W equals the entire ring R_M. This is a contradiction, hence W² is properly contained in W.

Let x be an element of W that is not int W².

Let V be the principal ideal in R_M that is generated by x.

Consider the product VW′. This is contained in WW′, and is therefore an ideal in R_M.

If VW′ is contained in W, then VWW′ is contained in W². This means V is in W², and x is in W², which is a contradiction. Therefore VW′ includes elements outside of W.

Remember that R_M is a local ring. If VW′ is a proper ideal it is contained in a maximal ideal - the maximal ideal, which is W. Since VW′ is not contained in W it is not a proper ideal. Therefore VW′ is the entire ring, V is the inverse of W′, V = W, and W is principal, generated by x.

Remember our working assumption, H = HM. Localize about M, and let G = H_M (for notational convenience). Since product and localization commute, we can write GW = G. Now everything in W is a multiple of x, so GW is actually G scaled by x. You can do this again and again, as often as you like. Thus G = Gxⁿ for all powers of x.

Suppose G includes an element c that is in Wⁿ, but not in Wⁿ⁺¹. Then xG does not include c, and xG cannot equal G. Therefore G lies in all the ideals generated by the powers of x. To be precise, G lies in the intersection of the descending chain of ideals generated by the powers of x. If this intersection is empty, then G cannot exist, H cannot equal J, and we are done. Let's prove the intersection is empty.

For each n, let c = xⁿf_n. Consider the successive principal ideals generated by f_n and f_n+1. Note that f_n = xf_n+1. Since f_n+1 generates f_n, The second ideal contains the first. If the ideals are equal then x is a unit, which is impossible. Therefore the second ideal properly contains the first. The factors f_n generate an infinite ascending sequence of ideals, which is impossible in a noetherian ring. The powers of W have no intersection, and that completes the proof.

Summary

These are all equivalent characterizations of a dedekind domain.