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Let $R$ be a commutative Noetherian ring and let $p$ be a maximal ideal of $R$. The localization of $R$ at $p$, $R_p$, is a local ring with unique maximal ideal $pR_p$.

Now let $I$ be an arbitrary (= not necessarily prime) ideal of $R$ such that $I \subseteq p$.

Assume that $IR_p=pR_P$.

Question: Is it true that $I=p$?

Of course, if we knew that $I$ is a prime ideal of $R$, then by the known result concerning the one-one correspondence between prime ideals of $R_p$ and prime ideals of $R$ contained in $p$, we would have obtained that $I=M$. However, here $I$ is not known to be a prime ideal of $R$.

Relevant questions: 1, 2, 3; the second answer in reference 3 quotes Theore 5.32 from "Steps in commutative agebra" by Sharp, and it seems that I need some version of Theorem 5.30, just without the primality assumption (still with contraction and extension of ideals).

Remark: $R=\mathbb{Z}$, $p=0$ is not a counterexample.

Edit: What if $R$ is an integral domain?

Thank you very much!

Edit 2: Now asked this more general question.

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  • $\begingroup$ By $IR_p$ do you mean the ideal in $R_p$ generated by $I$? $\endgroup$ Apr 6, 2021 at 21:22
  • $\begingroup$ @marlasca23, yes, exactly. Thank you. $\endgroup$
    – user237522
    Apr 6, 2021 at 21:23
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    $\begingroup$ math.stackexchange.com/questions/1430835/… $\endgroup$ Apr 6, 2021 at 21:35
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    $\begingroup$ @user237522 apologies, I deleted my answer briefly when I saw that you were asking for $\mathfrak{p}$ to be a maximal ideal. in my initial example $\mathfrak{p}$ was only prime, and not maximal. but this is easy to correct, and you can see a corrected example in my answer below :) $\endgroup$ Apr 6, 2021 at 22:06
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    $\begingroup$ Oh, I see now that we can take $p=(x,y), I=(x(x-1),y)$...\ $\endgroup$
    – user237522
    Apr 6, 2021 at 22:17

1 Answer 1

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No, this is not true, even if $R$ is an integral domain. Consider the case when $R$ is the polynomial ring $F[x]$ over your favorite field $F$, and let $\mathfrak{p}=\langle x\rangle$ and $I=\langle x(x-1)\rangle$. Then $\mathfrak{p}$ is a maximal ideal and $I$ is a strict subset of $\mathfrak{p}$, but, in the localized ring $R_\mathfrak{p}$, the element $(x-1)\big/1$ is a unit, and we hence have $$I_{\mathfrak{p}}=\left\langle x(x-1)\big/1\right\rangle\ni \left(x(x-1)\big/1\right)\left(1\big/(x-1)\right)=x\big/1,$$ whence $I_\mathfrak{p}\supseteq\mathfrak{p}_{\mathfrak{p}}=\left\langle x\big/1\right\rangle$.

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    $\begingroup$ Thanks for pointing this out. After I saw this it occurred to me that $\mathbb{Z}_{(p)}$ also provides a ton of counterexamples since its only nonzero ideals are $p^n\mathbb{Z}_{(p)}$ $\endgroup$ Apr 6, 2021 at 22:05
  • $\begingroup$ Thank you very much both of you! Please, are there interesting special cases where $I=p$ after all? $\endgroup$
    – user237522
    Apr 6, 2021 at 22:07
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    $\begingroup$ @BrianMoehring indeed! :) I'm very geometrically-minded so I like polynomial ring examples when possible, but perhaps an example from $\mathbb{Z}$ is even simpler! $\endgroup$ Apr 6, 2021 at 22:09
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    $\begingroup$ I'm not sure if this is the kind of thing you are looking for, but, for example, if $I=\langle \lambda a\rangle$ and $\mathfrak{p}=\langle a\rangle$ are both principal ideals, then we will have $I_{\mathfrak{p}}=\mathfrak{p}_{\mathfrak{p}}$ if and only if $\lambda\notin\mathfrak{p}$. (so, eg, the example above works because $\lambda=x-1\notin\mathfrak{p}$, and, as @BrianMoehring alludes to, you can whip up similar examples in $\mathbb{Z}$ like $I=\langle 6\rangle\subset\langle 2\rangle=\mathfrak{p}$.) $\endgroup$ Apr 6, 2021 at 22:14
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    $\begingroup$ @AtticusStonestrom, interesting, thank you again. $\endgroup$
    – user237522
    Apr 6, 2021 at 22:19

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