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I am studying the localization of a ring R at a submonoid S. I am really confused of the form of the ideals in the localization.

In the case of the quotient ring A by an ideal I, it is defined a bijection between the ideals of A containg I and the ideals of A/I.

Is this rule true anymore in the localization of a ring at a submonid?

Can we say that there exists a bijection between the ideals of the ring R that does not meet S and the ideals of the localization of the ring R at the submonid S? Can we say that every ideal of the localization of the ring R at a submonoid S is of the form Is, where I is an ideal of R that does not meet S?

May you help me, please? Thank you in advance.

user404634
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    This answer might be helpful: https://math.stackexchange.com/a/2061480/133781 – Xam Jun 04 '17 at 22:38

2 Answers2

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There is such a bijection between the prime ideals of $S^{-1}R$ and the prime ideals of $R$ which do not meet $S$. For not necessarily prime ideals, every ideal in $S^{-1}R$ has the form $S^{-1}\mathfrak a$ for some ideal $\mathfrak a\subset R$, but the correspondence is not necessarily injective.

Bernard
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  • Thank you very much for your answer! So, you mean that if J is an ideal in S^{-1}R, then J has the form S^{-1}\mathfrak I, for some ideal I of R, but not for all the ideals I of R. Is this correspondence always surjective? I think yes, since the correspondence between R and S^{-1}R such that a goes to a/s is surjective, but I do not know if I am right. – user404634 Jun 04 '17 at 23:26
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    I mean that there may be several ideals in $R$ extending to the same ideal in $S^{-1}R$. So yes it is surjective. A reference (in the more genral context of submodules) is Bourbaki, Commutative Algebra, Ch. II, Rings and Modules of Fractions, §2, n°4, prop. 10. – Bernard Jun 04 '17 at 23:40
  • Ok, I will look for them when I'll pass to modules. Thank you! – user404634 Jun 04 '17 at 23:43
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    F.y.i., if $\varphi:R\longrightarrow S^{-1}R$ is the canonical map, the ideal in $R$ which extends to $J$ in $S^{-1}R$ is simply $\mathfrak I=\varphi^{-1}(J)$. – Bernard Jun 04 '17 at 23:56
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For completion, let me add a somewhat generic counterexample that I've just run into. Let $R$ be a commutative ring, $\mathfrak{a}$ a proper ideal that is not prime, and $\mathfrak{p} \supset \mathfrak{a}$ a prime ideal that is minimal among the primes that contain $\mathfrak{a}$ (if $\mathfrak a$ has a primary decomposition, then $\mathfrak p$ is just an isolated prime of $\mathfrak p$).

Now, localization at $S = R - \mathfrak p$ cuts off all primes except those contained in $\mathfrak p$. Any prime in $S^{-1} R$ that contains the image $S^{-1} \mathfrak a$ must be the image of a prime in $R$ between $\mathfrak a$ and $\mathfrak p$. By minimality, the only such prime is $\mathfrak p$ itself. So $S^{-1} \mathfrak p$ is the only prime that contains $S^{-1} \mathfrak a$, meaning that $\sqrt{S^{-1} \mathfrak a} = S^{-1} \mathfrak p$.

Thus, since localization commutes with radicals, we have $S^{-1} \sqrt{\mathfrak a} = S^{-1} \mathfrak p$. But in general $\sqrt{\mathfrak a}$ and $\mathfrak p$ may be distinct; not all radical ideals are prime. (In fact, in a Noetherian ring, a radical ideal is either prime or the intersection of two strictly bigger radical ideals.)

csaltachin
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