## Dedekind Domains, Prime Ideals are Maximal

### All Nonzero Prime Ideals are Maximal

This is a rather complicated proof. If you have a simpler one, please send it along.

I will show one prime ideal cannot contain another (discounting 0); hence every nonzero prime ideal is maximal. The proof is broken up into several steps.

I use the axiom of choice all over the place. I wonder if a dedekind domain, with chains of prime ideals, might exist, if we set aside the axiom of choice?

### Principal Ideal Contains no Primes

Let the nonunit g generate a principal ideal that contains the prime ideal Q. Suppose g lies outside of Q. Given x in Q, x = gy for some y. With g outside of Q, y lies in Q. Therefore Q lies in gQ. Of course gQ lies in Q, hence Q = gQ. Think of {g} as the ideal generated by g, and Q = {g}Q contradicts unique factorization.

If Q lives properly inside a maximal ideal M, M cannot be principal.

### Minimal Prime

If U is an ideal inside a prime ideal M, there is a minimal prime ideal P containing U and contained in M. Now factor U in the dedekind domain. Thus U becomes a product of primes that happens to lie in P, and since P is prime, one of these prime factors lies in P. Each prime factor of U contains U, so this prime factor is trapped between U and P. Since P is minimal over U, this prime factor is indeed P. In other words, P is a factor of U. A minimal prime over an ideal is always one of its factors.

A minimal ideal over a generator g is a factor of the ideal {g}.

Assume a prime ideal P properly contains a prime ideal Q. Let x be an element in P-Q. Pull P down to a minimal prime ideal containing Q and x. In the same way, raise Q up to a maximal prime ideal that lies in P and misses x. This can be done because the union of a chain of prime ideals is prime, and the union lies in P, and the union misses x, so Zorn's lemma applies.

Suppose another prime ideal lies properly between P and Q. It either contains x or it doesn't. Either way we have a contradiction, since P is minimal over x, and Q is maximal without x. Therefore P and Q are adjacent prime ideals with nothing in between.

### Localization

Localize R about P, and let f be the map from the ideals of R onto the ideals of RP. We only care about the ideals contained in P, since these become proper ideals in RP. Remember that f(P) is the only maximal ideal in the local ring RP. Also, prime ideals correspond, and multiplication and localization commute. That is, f(A)*f(B) = f(A*B).

Let J be an ideal in RP, and pull back to an ideal H in R. Factor H in R, and push this product forward into RP. Prime ideals remain prime, and only the primes in P persist. Thus J can be factored into prime ideals in the local ring.

Apply this to P when P is a minimal nonzero prime ideal. By correspondence, there is only one nonzero prime ideal in RP, namely f(P). By factorization, every ideal in RP is a product of prime ideals, hence a power of f(P). These ideals are linearly ordered by containment. This means the local ring is a valuation ring. Its valuation group extends forever, hence the powers of the maximal ideal f(P) form a strictly descending chain of ideals.

Let J be the intersection of f(P)n, and suppose J is nonzero. Now J is a power of f(P). Yet the powers of f(P) are strictly descending. This is a contradiction, hence the powers of f(P) intersect in 0.

The powers of P intersect in an ideal H, with f(H) contained in 0. A nonzero ideal embeds in RP, and then extends, thus only 0 has f(0) = 0. Therefore H = 0, and the powers of P intersect in 0.

An ideal such as P7 becomes f(P)7 in the local ring, and then pulls back to P7, or perhaps something larger. But it can't pull back to include P6. If f(P6) = f(P7) then f(P8) = f(P)f(P7) = f(P)f(P6) = f(P7). All images beyond P6 stabilize, and their intersection is f(P6), which is definitely nonzero. This is a contradiction, hence the saturation of each power of P is distinct.

To be more specific, suppose the saturation of P7, call it H, is larger than P7. If the factorization of H includes P7, then H is too small. Thus H is P6 times some other primes outside of P. Push this forward and get f(P)6, not f(P)7. Therefore the saturation of Pn = Pn.

### Quotient

Let Q be a prime ideal and mod out by Q, giving an integral domain. Pull J in the quotient ring back to H, and factor H. If a factor of H does not contain Q then neither does the product H, thus all factors contain Q. Push this forward into the quotient ring, and remember that prime ideals correspond. Thus J is the product of prime ideals. Every ideal factors into primes.

Apply this to the adjacent primes P and Q. Now P becomes a minimal nonzero Prime, and as described above, the powers of P close in on 0. Pull back to R, and the powers of P close in on Q.

What if we localize first, and then mod out by f(Q)? Since Q is saturated, one can mod out by Q, and localize about P, in either order, and get the same ring. So the ideals in RP that contain f(Q) are the ideals of RP mod f(Q), are the ideals in the localization of R/Q, are the powers of f(P). Since they pull back to the powers of P in R/Q, they pull back to the powers of P in R. In other words, the saturation of Pn in R is Pn, whether we mod out by Q or not.

### No New Primes over g

Let g be anything in P-Q. Let G be the ideal generated by g. Since G is principal, it cannot equal P, else it would contain the smaller prime ideal Q. Thus G is properly contained in P.

Suppose there is a prime ideal containing G and properly contained in P. Pull these two prime ideals together, so that they are adjacent. Call these adjacent prime ideals S and T, remembering that S might equal P. As shown above, the powers of S intersect in T. This is contained in the powers of P, which intersect in Q. Therefore T is contained in Q. However, T contains g, which is not in Q. This is a contradiction, hence there are no primes between G and P. In other words, P is minimal over G, and is a factor of G.

If P contains another prime ideal T, that is not contained in Q, select g in T-Q, and argue as above to show that T cannot exist. All the prime ideals properly inside P lie inside Q.

### The Powers of a Prime Intersect in a Prime

Let P be prime and assume the powers of P intersect in a nonzero ideal H. Let Q be a minimal prime ideal between H and P, whence Q is a factor of H. Suppose Q is properly between H and P. Select adjacent primes S and T between P and Q. The powers of S intersect in T, and the powers of P intersect in something larger. Yet T contains Q properly contains H, whence the powers of P intersect in something smaller than T. This is a contradiction, thus Q = H or Q = P.

Suppose P is the least prime over H. Thus P to some power, times some primes that are not contained in P, but still contain H, equals H. Suppose P contains some other prime Q that is not inside H. We know that Q does not contain H. Select adjacent primes S and T between P and Q. By the above, T does not contain H. The powers of S intersect in T, which includes some x not in H, yet the powers of P cannot contain x outside of H. This is a contradiction, hence P has no other primes inside it, unless those primes live in H.

Let Q be the largest prime in H. (Remember that Q could be 0). Since P and Q are adjacent, the powers of P intersect in Q, and Q = H.

The powers of any prime intersect in another prime.

Let the powers of P intersect in Q, and suppose some other prime ideal lies properly between. Find adjacent primes S and T and build the usual contradiction: the powers of S intersect in T, but the powers of P intersect in something smaller. Thus there is nothing in between.

This completes the circle. Adjacent primes P and Q have Q as the intersection of Pn, and the intersection of Pn is a prime Q, where P and Q are adjacent. (Q might be 0.)

### Chain of Prime Ideals

If P contains Q, think of P as P0, and let P1 be the intersection of the powers of P0. This is an adjacent prime, and any other primes in P0, such as Q, are in P1. Let P2 be the intersection of the powers of P1, and so on until you run into Q. If the process goes on forever, switch from integers to ordinals, and let Pω be the intersection of Pn. This is the intersection of prime ideals, and it is prime. Since Q is in each Pn it is in Pω. Assuming we have not reached Q, let Pω+1 be the intersection of the powers of Pω. By transfinite induction, we eventually run into Q. Thus P and Q determine a linearly ordered chain of prime ideals, such that any other primes in P are in Q.

Set Q = 0, and find a unique chain of Prime ideals from P down to 0. This can be done for each maximal ideal.

### Chain of 2

If every chain from maximal ideal down has length 1, then every prime is maximal, and we're done. So suppose there is a chain of length at least 2, starting with P and Q. If a third prime is present, mod out by the third prime, and then localize about P. Every ideal factors into prime ideals in this local ring; but we want to prove the factorization is unique.

First, show the local ring is a valuation ring, by showing ideals are linearly ordered. This is the case if the preimages are linearly ordered. The powers of P are linearly ordered, and so are the powers of Q, with Q below all the powers of P. The trick is to show that Q3 times the powers of P all contain Q4. A copy of the chain based at P fits neatly between successive powers of Q.

Let x = yz, where y is a product of 3 elements drawn from Q, and z is another element drawn from Q. Note that yz is in each Q3Pn, and in their intersection. Thus the intersection, which is an ideal, contains Q4. The ideals based on P and Q are well ordered by the exponent on Q followed lexicographically by the exponent on P. A well ordering is a linear ordering, so we're ok. Push this forward and all ideals in the local ring are linearly ordered, hence it is a valuation ring.

The valuation group is an ordered group, and looks the same under any translation. Shift to Q3, and Q3 times the powers of P must increase in valuation, just as Pn increases in valuation. The images in the local ring are distinct, and their preimages are distinct, and factorizations based on P and Q in R correspond uniquely to factorizations in the local ring. Therefore our local ring, with two primes P and Q, is dedekind.

Now for the long sought contradiction. Let g be anything in P-P2. Let g generate G, and factor G. The only prime containing G is P, and P2 misses g, hence G = P. P is principal, and contains Q, which is a contradiction.

Our original hypothesis, one prime ideal containing another, has failed. Every nonzero prime ideal in a dedekind domain is maximal.

### A UFD that is not Dedekind

Let R be the ufd K[x,y], for some field K. The ideal generated by x is prime, since the quotient ring K[y] is an integral domain. Yet this ideal is not maximal, being contained in the ideal generated by x and y. hence K[x,y] is a ufd that is not dedekind.

If you want a more direct proof, an ideal that will not factor, let H be generated by x2 and y2. If P is a prime factor of H it contains H. Since x times x lies in H, P contains x. Similarly, P contains y, hence P is the maximal ideal K adjoin x and y. Some power of P equals H, and it isn't P1. Since Pn contains xyn-1, which is not in H, H does not factor in R.