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.

  • $\begingroup$ I think you wrote the statement incorrect. A counterexample: $I=\langle y\rangle\subset k[x,y]$, $f=x\in k[x,y]$. Now there is no $m\geq 2$ such that $x^m=x^{m+1}$ but $I\colon f^\infty=I:f^m$ for every $m\in\mathbb{N}$. You may say $m=1$ solves the above example, but then consider the following example: $I=\langle x^3\rangle\subset k[x]$, $f=x\in k[x]$. Now there is no $m\geq 2$ such that $x^m=x^{m+1}$ but $I\colon f^\infty=I:f^m$ for every $m\geq 3$ So not $m=1$.. By looking at your $\Leftarrow$ part, I think you meant if and only if $I\colon f^m=I\colon f^{m+1}$, am I right? $\endgroup$ – AmirHosein SadeghiManesh Nov 27 '17 at 10:45

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.

  • $\begingroup$ Yes, that seems to be it, thank you. I knew that it was easy, but I don't know how it escaped my mind. $\endgroup$ – Adrian Manea Jun 4 '12 at 8:05
  • $\begingroup$ @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. $\endgroup$ – user38268 Jun 4 '12 at 8:06
  • $\begingroup$ 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. $\endgroup$ – Adrian Manea Jun 4 '12 at 8:09
  • $\begingroup$ Yes, of course, will edit now. $\endgroup$ – Adrian Manea Jun 4 '12 at 8:13

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