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I will state my question first:

Suppose $I\subseteq J\subseteq A$ are two ideals in a commutative ring $A$. Furthermore, assume that for every maximal ideal $\mathfrak{m}$ of $A$, the image of $I$ and $J$ under the canonical map $A\to A_{\mathfrak{m}}$ is the same. How can I prove that $I=J$ ?

The aforementioned canonical map $A\to A_{\mathfrak{m}}$ is $a\mapsto a/1$.

My attempts: I think we might need to use the following fact. If $f: M\to N$ is $A$-module homomorphism, then the following statements are equivalent:

1) $f$ is injective.

2) The induced map $f_{\mathfrak{m}}: M_{\mathfrak{m}}\to N_{\mathfrak{m}}$ is injective for every maximal ideal $\mathfrak{m}$ of $A$.

Now, if we let $M=I$ and $N=J$ (as ideals of $A$ are naturally $A$-modules), then the inclusion map $i: I\to J$ is injective. So, the induced maps $i_{\mathfrak{m}}: I_{\mathfrak{m}}\to J_{\mathfrak{m}}$ are also injective, for each maximal ideal $\mathfrak{m}$ of $A$. Now I would like to use the fact that $I$ and $J$ have same extensions in $A_{\mathfrak{m}}$. Can this approach be made to work?

Thanks for your time.

Prism
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  • Prism, I've provided a proof below that applies even if we don't know that $I\subseteq J$ (or vice-versa). However, @YACP's suggestion is great and provides a direct proof in the special case! – Amitesh Datta Aug 10 '13 at 05:51
  • @YACP: Thank you for this direct answer! Feel free to convert it into an answer. – Prism Aug 10 '13 at 06:03

2 Answers2

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Theorem

If $I,J$ are ideals of a commutative ring $A$ such that $I_{\mathfrak{m}}=J_{\mathfrak{m}}$ for all maximal ideals $\mathfrak{m}$ of $A$, then $I=J$.

Proof

Let us assume, for a contradiction, that $I\not\subseteq J$ and let $x\in I\setminus J$. The set $(x:J)=\{a\in A:ax\in J\}$ is an ideal of $A$. If $I_{\mathfrak{m}}=J_{\mathfrak{m}}$ for all maximal ideals $\mathfrak{m}$ of $A$, then $(x:J)$ is not contained in any maximal ideal of $A$. Therefore, $(x:J)=A$, i.e., $x\in J$; a contradiction. Conversely, one can prove that $J\subseteq I$ and conclude that $I=J$. Q.E.D.

An exercise that provides practice with the proof technique above:

Exercise 1: If $A$ is a commutative ring in which every prime ideal is finitely generated, then every ideal is finitely generated, i.e., $A$ is Noetherian. (Hint: use proof by contradiction. Let $S$ be the set of all non-finitely generated ideals of $A$ and note that $S$ is partially ordered with respect to inclusion. If $S\neq \emptyset$, then check that the hypothesis of Zorn's lemma applies, so that $S$ has maximal elements. If $I\in S$ is a maximal element, then prove that $I$ is a prime ideal of $A$; a contradiction. Q.E.D.)

I hope this helps! (The proof is short so you might need to think a little bit in order to digest it.)

Amitesh Datta
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  • Thanks for your answer, Amitesh. By the way, have you interchanged $I$ and $J$? (Especially in your first line). – Prism Aug 10 '13 at 05:51
  • Ohhh so we don't need the hypothesis $I\subseteq J$. Brilliant! – Prism Aug 10 '13 at 05:52
  • You're welcome, @Prism! Yes, it turns out that I didn't notice the hypothesis that $I\subseteq J$ so I just went ahead and provided an answer in general. If I had noticed this hypothesis, then I probably wouldn't have arrived at this argument. I like this argument though and I learnt something from answering your question so thanks for asking it! Just goes to show that in math it's best to question whether you really need all hypothesis. – Amitesh Datta Aug 10 '13 at 05:55
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    I am glad to hear that :) While I am trying to absorb the details of this argument, I am slowly realizing that it "makes sense". Maximal ideals are powerful enough to determine the ideal (regardless of whether one is contained in another)! – Prism Aug 10 '13 at 06:04
  • I've incorporated your nice suggestion @YACP; thanks! – Amitesh Datta Aug 10 '13 at 07:24
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    Yes, absolutely, @Prism in various contexts! For example, here's an example that comes to mind. The Jacobson radical of a commutative ring $A$ is the intersection of all maximal ideals of $A$. Nakayama's lemma states that $J(A)M=M$ for a finitely generated $A$-module $M$ $\implies$ $M=0$. Also, an element of $A$ is a unit if and only if it's not contained in any maximal ideal of $A$. In particular, $a\in J(A)$ $\iff$ $1+ra$ is a unit of $A$ for all $r\in A$! By the way, if you're stuck while unravelling the details of my argument, then please let me know and I'm happy to elaborate! – Amitesh Datta Aug 10 '13 at 07:31
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    Dear Amitesh, thanks for the insight. Nakayama's lemma is my favourite :) As if you read my mind, yes I am somewhat stuck in unravelling your answer :( The key step is that $(x: J)$ is not contained in any maximal ideal. I am trying to understand why. If it were contained in some maximal ideal $\mathfrak{m}$, then my guess is that we could localize at that maximal ideal, and somehow contradict $I_{\mathfrak{m}}=J_{\mathfrak{m}}$. If you could elaborate on this point, that would be fantastic! – Prism Aug 10 '13 at 07:41
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    Dear @Prism, let $\mathfrak{m}$ be a maximal ideal of $A$. (I know this is a strange way to begin a greeting to someone ...) Since $I_{\mathfrak{m}}=J_{\mathfrak{m}}$, it follows that $\frac{x}{1}\in J_{\mathfrak{m}}$, i.e., $\frac{x}{1}=\frac{y}{s}$ for some $y\in J$ and $s\not\in \mathfrak{m}$. However, this is equivalent to the existence of $t\not\in \mathfrak{m}$ such that $(xs-y)t=0$, i.e., $yt=xst$. However, $y\in J$, so $yt\in J$ and so $xst\in J$, i.e., $st\in (x:J)$. We know that $s,t\not\in m$ so $st\not\in m$ (maximal ideals are prime) and $(x:J)$ isn't contained in $\mathfrak{m}$! – Amitesh Datta Aug 10 '13 at 11:14
  • Thanks so much for your corrections @YACP! Third time lucky? – Amitesh Datta Aug 10 '13 at 11:16
  • That is excellent explanation. Thanks a lot :) I finally understand it. Woohoo!!! I have already upvoted you long before :P (Now I see that the technique you are using is also employed in proof of Proposition 3.11 in Atiyah & Macdonald, especially in part (b)). – Prism Aug 10 '13 at 14:45
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    I'm very happy that I helped, @Prism! Yes, this technique is a very useful one to remember. If you're given some information about localizations (such as in this question), then studying the ideals $(I:J)={a\in A:ax\in J\text{ for all }x\in I}$ is quite fruitful. (Of course, $(x:J)=(xA:J)$ in this notation where $xA$ denotes the principal ideal generated by $x$.) I've also added an exercise that uses this technique but not in the context of localisation. If you ever have some time to kill or would like to see further instances of this technique being used, then you might like to have a look! – Amitesh Datta Aug 10 '13 at 22:13
  • Thanks Amitesh. I have done this exercise :) See my previous related question here. – Prism Aug 10 '13 at 23:09
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Use that $M_{\mathfrak m} =0$ for all maximal ideals $\mathfrak m$ iff $M=0$. (Here $M=J/I$, of course.)

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    And if $I \subseteq J$ does not hold, just replace $J$ by $I+J$ and the same argument works. – Martin Brandenburg Aug 10 '13 at 08:28
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    @MartinBrandenburg: Since $(I+J){\mathfrak{m}}=I{\mathfrak{m}}+J_{\mathfrak{m}}=I_{\mathfrak{m}}+I_m = I_{\mathfrak{m}}$ for each maximal ideal $\mathfrak{m}$, and $I\subseteq I+J$, we apply the argument above to get $I+J=I$, which gives $J\subseteq I$, so we are back to the case where one ideal is contained in another. Like this? – Prism Aug 10 '13 at 08:46
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    @Prism I'm not Martin but yes! Alternatively, before you write "... so we are back to the case ...", you could just argue that $I+J=J$ by a symmetrically identical argument, i.e., $I\subseteq J$. – Amitesh Datta Aug 10 '13 at 11:21
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    This is exactly what I meant. – Martin Brandenburg Aug 10 '13 at 12:22
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    @MartinBrandenburg: That's a really slick way of seeing it. Thanks! Will try to keep this kind of argument in mind. – Prism Aug 10 '13 at 14:57