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This question is from a lecture notes from which I am studying commutative algebra and this question was left as an exercise for students.

For two ideals I , J in A, prove that $I \subset J$ iff $I_M \subset J_M$ in $A_M$ for all maximal ideal M.

Attempt : Let $I_M \subset J_M$ for all maximal ideals means that {i/m }$\subset ${j/m} for all maximal ideals. But how to proceed now?

Let on the other hand $I\subset J$ is true, how should I proceed?

I don't have any intuition regarding this problem. Can you please give some hints? I want to complete it by myself.

Thanks!

  • If $I\subseteq J$ then trivially $S^{-1}I\subseteq S^{-1}J$ for any multiplicative set $S$, and in particular for localizations. For the other direction, we assume $I_M\subseteq J_M$ for all maximal ideals $M$ and want to prove $I\subseteq J$. Hint: let $x\in I$, and define the ideal $\mathfrak{a}={r\in A: xr\in J}$. Now try to show that $1\in\mathfrak{a}$. – Mark Dec 12 '21 at 15:25
  • @Mark Can you please prove your hint? –  Dec 16 '21 at 11:18
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    If we show that $1\in\mathfrak{a}$ then we are done. So suppose otherwise. Then by Zorn's lemma $\mathfrak{a}$ is contained in some maximal ideal $M$. Now, by assumption $I_M\subseteq J_M$, and so $\frac{x}{1}\in J_M$, i.e there are $j\in J, s\notin M$ such that $\frac{x}{1}=\frac{j}{s}$. By definition this means there is some $u\notin M$ such that $(xs-j)u=0$, and so we have $xsu=ju\in J$. But this means $su\in\mathfrak{a}\subseteq M$. Since $M$ is a prime ideal it follows that $s\in M$ or $u\in M$, a contradiction. – Mark Dec 16 '21 at 18:30
  • @Mark Doesn't in 3rd line of your comment it should be $s\in M$ instead of $s\nin M$? –  Jan 09 '22 at 06:41
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    No, because here we use the multiplicatively closed set $S=A\setminus M$. The elements of $S$ are exactly the elements which are not in $M$. – Mark Jan 09 '22 at 12:29
  • @Mark In 4th line of your comment,how does $xsu =ju \in J$ means that $su\in a$? –  Jan 15 '22 at 06:53
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    Pretty sure you can answer this question yourself. Just take a look at the definition of $\mathfrak{a}$. – Mark Jan 15 '22 at 10:50
  • @Mark thanks you very much –  Jan 19 '22 at 08:18

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