Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.