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Let $k$ be a field and $A=k[X^3,X^5] \subseteq k[X]$.

Prove that:

a. $A$ is a Noetherian domain.

b. $A$ is not integrally closed.

c. $dim(A)=?$ (the Krull dimension).

I suppose that the first follows from $A$ being a subring of $k[X]$, but I don't know about the rest.

Thank you in advance.

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For an easy example of a subring of a Noetherian ring which is not Noetherian: Let $D$ be a non-Noetherian domain. $D$ is a subring of its field of fractions! – rschwieb Jun 5 '12 at 11:18
up vote 4 down vote accepted

a) Not every subring of a noetherian ring is noetherian (there are plenty of counterexamples), so this doesn't work here. Instead, use Hilbert's Basis Theorem.

b) The element $X^2 = \frac{X^5}{X^3}$ is in $\mathrm{Quot}(A)$. Try to show that it is integral over $A$, but not in $A$.

c) The dimension is the transcendence degree of $\mathrm{Quot}(A)$ over $k$. But this field is easy to compute.

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a) You mean thinking $A=k[X^3,X^5]=(k[X^3])[X^5]$ and use Hilbert two times, once for the () and then for A? c) $X^2 \in Quot(A) \Rightarrow X=X^3/X^2 \in Quot(A) \Rightarrow Quot(A)=k[X]$? As for b), it's clear that $X^2 \notin A$, but for integrality, it's one of my weak spots. – AdrianM Jun 5 '12 at 9:58
Well, think about it a while ... – Martin Brandenburg Jun 5 '12 at 10:33
Certainly I will. But was I right about a) and c) in my comment above? :) – AdrianM Jun 5 '12 at 10:34
sorry, I meant: $Quot(A)=k(X)$, of course. – AdrianM Jun 5 '12 at 10:57
I think I can conclude that $X^2\in Quot(A)$ is integral over $A$ by saying that $A[X^2]$ is a f.g. $A$-module. The argument may be a bit to intuitive, though. – AdrianM Jun 5 '12 at 14:57

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