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Is k[x^2,x^3] dedekind?

  1. Jul 14, 2010 #1
    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!

  2. jcsd
  3. Jul 14, 2010 #2
    Consider that [itex]k[x^2,x^3] = k[t,u]/(t^2-u^3).[/itex] The polynomial [itex]t^2-u^3[/itex] is irreducible, so it's principle ideal is prime.

    Then, your ring is noetherian and dimension 1.
  4. Jul 15, 2010 #3
    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?
  5. Jul 16, 2010 #4
    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.
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