# Ideals in the localization $R_p$

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.

• By $IR_p$ do you mean the ideal in $R_p$ generated by $I$? Commented Apr 6, 2021 at 21:22
• @marlasca23, yes, exactly. Thank you. Commented Apr 6, 2021 at 21:23
• math.stackexchange.com/questions/1430835/… Commented Apr 6, 2021 at 21:35
• @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 :) Commented Apr 6, 2021 at 22:06
• Oh, I see now that we can take $p=(x,y), I=(x(x-1),y)$...\ Commented Apr 6, 2021 at 22:17

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 $$I_\mathfrak{p}\supseteq\mathfrak{p}_{\mathfrak{p}}=\left\langle x\big/1\right\rangle$$.
• 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 $p^n\mathbb{Z}_{(p)}$ Commented Apr 6, 2021 at 22:05
• Thank you very much both of you! Please, are there interesting special cases where $I=p$ after all? Commented Apr 6, 2021 at 22:07
• @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! Commented Apr 6, 2021 at 22:09
• 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 $I=\langle 6\rangle\subset\langle 2\rangle=\mathfrak{p}$.) Commented Apr 6, 2021 at 22:14