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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 Ris 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

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

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