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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!2012-06-05

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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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    Thank you. As for integrality of $X^2$ over $A$, I finally found 2 basic examples $p,q \in A[Y], \ p(Y)=Y^3-X^3\cdot X^3$ and $q(Y)=Y^4-X^3\cdot X^5$ for which $p(X^2)=q(X^2)=0$, hence the conclusion. I now call the problem solved. Thanks to all the comentators!2012-06-06