ALL TITLE:Is k[x^2,x^3] a Dedekind domain?

[sidm]

I need help.

So the obvious answer is no, because it's not integrally closed (incidentally it's integral closure is in fact k[x]. Here obviously k is a field. But I want to show that it is both noetherian and dimension 1 (nonzero prime ideals are maximal), here is my idea for a proof: (i immediately apologize for the fact that I've forgotten how to tex on this forum)

1) Noetherian: k[x^2]<R<k[x] and k[x] is a finitely generated k[x^2] module (i.e. 1 and x generate k[x] as a k[x^2]-module) thus it is a noetherian k[x^2]-module since it is therefore the quotient of a free k[x^2]-module of finite rank. So if we take I<R an ideal of R (i.e. a R-submodule of R) then it is also a k[x^2]-submodule of R and therefore of k[x], thus it must be finitely generated as k[x^2]-module and so a (k[x^2,x^3]=)R -module as well.

2) B=k[x^3]<k[x^2,x^3]=A, then let P be a prime in A then P' the contraction in B. Note that contractions of primes are always prime and since B is a PID (it is in fact euclidean*) as long as we can prove the contraction is nonzero maximality follows.

It is nonzero since A is integral over B: x^2 satisfies Z^3-(x^3)^2 a monic polynomial with coefficients in B. Then let c \in P \neq 0 we know that c satisfies a monic polynomial over B and so
c^n+b_{n-1}c^{n-1}+...+a_0=0, we can assume a_0\neq 0 for otherwise factor out by the largest power of c that divides this equation and use the fact that we're in an integral domain. Then c^n+...+ca_1=-a_0 \in P \cap B.

Then B/P' is a field, K, living inside the integral domain A/P. Clearly A/P is algebraic over K: let y^2 denote the image of x^2 then it satisfies the polynomial Z^3-(y^3)^2 with coefficients in K (note y^3 is the image of x^3 in K). Now to finish we prove an unrelated result:

3) Every integral domain (R) containing a field (K) over which it itself is algebraic is a field: Let b \in R be nonzero, then R is a n-dimensional K-space. So then 1,b,..,b^n are linearly dependent over K: k_0+k_1b...+k_nb^n=0 so that b(b^{n-1}k_n+...+k_1)=k_0, if k_0 is zero then the polynomial in in brackets is zero but the 1,...,b^{n-1} are linearly independent so the k_i's are all zero but we chose them not to be. So k_0 is nonzero and in K and thus b has an inverse in B.

If someone could check my argument I would appreciate it, a simpler one would be nice too. A side note: I'm currently teaching myself algebraic number theory (using James Milne's notes) and if anyone would like to work with me let me know!

SM

 

[tmccullough]

Consider that $k[x^2,x^3] = k[t,u]/(t^2-u^3).$ The polynomial $t^2-u^3$ is irreducible, so it's principle ideal is prime.

Then, your ring is noetherian and dimension 1.

 

[sidm]

thanks! I had that exact isomorphism in mind though I abandoned it because for some reason or another I thought it was false! Now however it's clear that k[x^2,x^3]=R is a k-algebra with 2 generators and thus there is an induced map from k[t,u] onto R. The kernel most certainly includes this prime ideal (t^2-u^3) and since k[t,u] is dim 2 it is either this prime or a maximal ideal that is the kernel (bc R is a domain) but R isn't a field so the isomorphism follows.

it seems that dimension theory a powerful tool! i am aware of the theorem that k[x_1,..,x_n] has dimension n over a field k though I have never seen a proof! I will investigate further. Another question is this, if a ring, polynomial ring in n variables or otherwise, has dimension n does every prime appear in a chain of length n? Obviously that t^2-u^3 does is clear since (t,u) is maximal and contains it but is it true in general?

 

[tmccullough]

Dimension theory (krull dimension anyways) is very much a fundamental part of algebraic geometry, and that's where I first learned it.

Polynomial rings (finite number of variables) over a field are very "homogeneous", so every maximal tower of ideals has the same length. This will then hold true for every finitely generated algebra over a field, just by pulling back to the polynomial ring (as in my very brief previous comment). It's been a really long time since I studied this stuff, so I don't know to what degree this holds for algebras that are not finitely generated. Maybe there are towers of ideals of different lengths in some strange algebra?

It's a great reference book, and pretty cheap, so you may want to look into the book Commutative Algebra: with a View Toward Algebraic Geometry by David Eisenbud. You can't really use it as a textbook, commutative algebra isn't really that kind of subject, but you can learn a ton.

 

[blue_raver22]

ALL TITLE:

Your argument seems sound and well thought out. It is always beneficial to have someone else review your work to ensure accuracy. As for your interest in algebraic number theory, it is a fascinating subject and I encourage you to continue exploring it. Collaboration and discussion with others can greatly enhance one's understanding and learning. Best of luck in your studies!