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, let $I \triangleleft k[X_1,\dots,X_n]=S$ be an ideal and fix $f \in S$.

The saturated ideal of $I$ is $I^{sat}=I:f^\infty=\{g \in S \mid \exists m \in \mathbb{N} \ s.t. \ f^mg \in I \}=\displaystyle\bigcup_{i \geq 1} I:f^i$.

Prove that $I^{sat}=I:f^m \Leftrightarrow f^m=f^{m+1}$.

My attempt:

"$\Rightarrow$" Since we have the ascending chain $I:f \subseteq I:f^2 \subseteq \dots$ and $S$ is Noetherian, it follows that the $m$ that we are looking for is exactly the one that stops the chain, i.e. the one from which on all ideals in the chain are equal. From $I^{sat}=\displaystyle\bigcup_{i \geq 1} I:f^i$, we have that $I^{sat}=I:f^m$.

"$\Leftarrow$" We have to show that all of the ideals $I:f^q$ are in $I:f^m$, i.e. the chain stops after $m$ steps. We have to prove $\{g \in S \mid f^mg \in I \} = \{h \in S \mid f^{m+1}h \in I \}$. "$\subseteq$" is clear, from the chain.

What about the reverse inclusion? It seems like going around in circles, so it must be something easy that I don't see.

Thank you.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Perhaps I am misunderstanding your problem, but here goes: You want to show that if $f^m = f^{m+1}$ then this implies that $I^{sat} = (I:f^m)$. Now if $f^m = f^{m+1}$ then by induction it is easily seen that $f^m = f^{m+k}$ for all non-negative integers $k$. Now we claim that

$$\bigcup_{i \geq 1} (I : f^i) = \bigcup_{i=1}^m (I:f^i).$$

One inclusion is obvious, for the other suppose that there is $x \in \bigcup_{i \geq 1} (I : f^i)$ such that $x \notin \bigcup_{i=1}^m (I:f^i)$. Now the former assumption gives that there is a positive integer $k$ such that $f^kx \in I$. The latter assumption gives that $k > m$. However we already proved that $f^{m+1} = f^{m+2} = \ldots = f^k$ so that $f^kx \in I \implies f^mx \in I$, i.e. $x \in (I:f^m)$ which is a contradiction. This establishes the equality above.

Now it is clear that $(I:f^m ) \subseteq I^{sat}$. It now remains to show the other inclusion, namely that $\bigcup_{i=1}^m (I:f^i) \subseteq (I:f^m)$. Now take any $x$ in the left hand side, then $x \in (I : f^n)$ for some $1 \leq n \leq m$. This means that $f^nx \in I$ so that $f^{m-n}(f^nx) \in I$ as well. In other words $x$ is such that $f^mx \in I$, i.e. $x \in (I:f^m)$ and your claim is proven.

share|improve this answer
    
Yes, that seems to be it, thank you. I knew that it was easy, but I don't know how it escaped my mind. –  AdrianM Jun 4 '12 at 8:05
    
@AdrianM Thanks, the problem was not that bad. By the way you may want to look at an answer I posted to your previous question on the jacobson radical of a polynomial ring over an integral domain. –  user38268 Jun 4 '12 at 8:06
    
I did, thanks again. I'm afraid there are some more somehow trivial problems to come, since I am really bad at commutative algebra. Your help is much appreciated. –  AdrianM Jun 4 '12 at 8:09
    
Yes, of course, will edit now. –  AdrianM Jun 4 '12 at 8:13

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.