Localization in integral domains

1. Jan 9, 2010

SiddharthM

i'm working through the following text and I think I found an error please let me know if i'm totally wrong.

Janusz, Gerald J. Algebraic Number Fields

and i'm starting with the 3rd exercise on page 3. It is as following:

let R be an integral domain and p a prime ideal of R. Show there is an isomorphism between the fields R/p and R_p/(pR_p).

Note that R_p is the localization at the multiplicative set S=R-p (this is multiplicative b/c p is a prime ideal). Now I believe that pR_p is a maximal ideal: this follows because if it were strictly contained in a maximal ideal of R_p then this would be a prime ideal of R_p different from pR_p but this corresponds to a prime ideal of R distinct from p. This corresponding prime ideal in R is in our multiplicative set at which we localized so we have invertible elements in a maximal ideal, a contradiction (maximality implies proper containment).

I used the fact that there is a bijection between prime ideals of R_p and prime ideals of R that don't intersect S. Also that all maximal ideals are prime.

So I believe that R_p/pR_p is a field (because it's a quotient of a ring by a maximal ideal). But there is nothing to suggest that p is a maximal ideal (in a general ring with a general ideal a quotient R/p is a field if and only if p is maximal) unless we are in a PID (because then prime ideals are maximal). So if this result were true it would mean that in ANY integral domain a prime ideal is maximal i.e. that the quotient of an integral domain by one of it's prime ideals is a field, there is an easy counterexample.

Z[x] is an integral domain and <x> is prime (because Z[x]/<x>=Z is an integral domain) but Z[x]/<x>=Z is most definitely NOT a field.

What do you think?

Cheers

2. Jan 14, 2010

Spartan Math

Hm...I looked at the book and I didn't think that was the exercise verbatim...how odd, you're right, that is definitely not correct. I think what they mean to say is show that there is an isomorphism between the fields Rp/pRp and quotient field of R/P (this is certainly possible since R/P is an integral domain since P is prime). There is certainly an isomorphism between these two fields. Let Q(R/P) denote the quotient field of R/P then you can show this by the map f: Q(R/P) ---> Rp/pRp by f((a+P)/(b+P)) = (a/b) + pRp. Check this is a well-defined map and then show it is an isomorphism (there is of course a slightly simpler way of doing this by using the universality of localization).

You are certainly correct pRp is a maximal ideal, since Rp is a local ring with pRp as its unique maximal ideal. Note that a ring R (commutative with 1) is a local ring if and only if the set of nonunits form an ideal. Thus note in pRp, the nonunits are in pRp and they form an ideal, hence Rp is local with maximal ideal pRp. This is true for any prime ideal of a ring R.

3. Jan 14, 2010

Hurkyl

Staff Emeritus
Or a Dedekind domain -- or any one-dimensional integral domain. Given the subject, the author probably had some of those in mind when he said "integral domain".

4. Jan 16, 2010

Spartan Math

I contemplated this, too and it would certainly make since in light of the fact that algebraic integer rings are Dedekind domains, but I feel it would have been specified and I looked in the textbook and I don't even think the concept of a Dedekind domain had been introduced.