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Let $A$ be a commutative ring with finitely many minimal prime ideals $\{p_1,\dots,p_n\}$. Let $A_{p_1,\dots,p_n}$ be the localization of $A$ away from the minimal primes, i.e. $S^{-1}A$ where $S = A-\bigcup_{i=1}^np_i$.

Question 1 (Answered): A paper I am reading claims that $A_{p_1,\dots,p_n} \cong \prod_{i=1}^n A_{p_i}$. I have established by hand that the natural map from left to right is injective, but surjectivity seems harder. What's the best way to see this isomorphism?

Also, the paper uses the notation $\text{Frac}(A)$ for $A_{p_1,\dots,p_n}$ (actually, $\text{Frac}(A)$ is defined to be the ring obtained by inverting in $A$ all non-zero-divisors in $A_\text{red} = A/\sqrt{0}$, but it's easy to see that this is the same thing). This seems odd to me, since, as far as I know, the standard meaning of $\text{Frac}(A)$ is the localization of $A$ at all non-zero-divisors in $A$. Note that if $A$ is not reduced, then there could be zero-divisors which are not in any of the minimal primes, and hence are inverted in $A_{p_1,\dots,p_n}$ (example: $y$ in $\mathbb{Q}[x,y]/(x^2,xy)$).

Question 2 (Still open): Is this definition of $\text{Frac}(A)$ common? Is there a good reason for preferring this definition of $\text{Frac}(A)$ over the other in certain situations (when $A$ is non-reduced)?

One more observation: I've noticed that with this definition of $\text{Frac}(A)$, we have that a map of rings $A\to B$ extends to a map $\text{Frac}(A)\to \text{Frac}(B)$ if and only if the inverse image of every minimal prime ideal in $B$ is a minimal prime ideal of $A$. Geometrically, this says that the map $\text{Spec}(B)\to\text{Spec}(A)$ maps generic points of irreducible components to generic points of irreducible components. Maybe this is relevant...

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1) Let $\phi : A \to \prod_{i=1}^n A_{p_i}$ be the natural map (i.e. product of localization maps). Every element of $S$ is sent to a unit under $\phi$, so there is an induced map $\varphi : S^{-1}A \to \prod_{i=1}^n A_{p_i}$. But $\varphi$ is locally an isomorphism (at every maximal ideal $p_i$ of $S^{-1}A$, $\varphi_{p_i} : (S^{-1}A)_{p_i} \cong A_{p_i} \to (\prod_{i=1}^n A_{p_i})_{p_i} \cong A_{p_i}$), so $\varphi$ is globally an isomorphism.

2) For a reduced ring, the set of nonzerodivisors is precisely the complement of the union of the minimal primes (this is easy to see in the Noetherian case, but also holds in general). Thus in this case the two notions of total ring of fractions coincide.

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  • $\begingroup$ (1): Ah, great, I was hoping it was something easy like this. Thanks a lot. (2): Yes, I know they coincide when $A$ is reduced - that's why I pointed out that when $A$ is nonreduced, they might differ. I've added a bit to to the question, and I'll wait to accept your answer to give other people time to respond to (2). $\endgroup$ – Alex Kruckman Sep 22 '14 at 18:17
  • $\begingroup$ @AlexKruckman: Keep in mind that zerodivisors in $A/\sqrt{0}$ are not necessarily the same as zerodivisors in $A$. Which is the one that appears in the paper? $\endgroup$ – zcn Sep 22 '14 at 18:19
  • $\begingroup$ I thought the question was clear about this. The paper defines $\text{Frac}(A)$ to be obtained by inverting in $A$ all non-zero-divisors in $A/\sqrt{0}$. This is the same as inverting all elements which are not in any minimal prime ideal in $A$, but it's different from inverting in $A$ all non-zero-divisors in $A$ (well, at least it can be different when $A$ is non-reduced). That's the whole point of the question. $\endgroup$ – Alex Kruckman Sep 22 '14 at 18:24
  • $\begingroup$ Ah, I see what the question is now (apologies for the confusion - it was clear before). In my opinion this should not be referred to as $\text{Frac}(A)$ in general (as you observe, they are different). However, they can coincide under a milder assumption than reduced: namely Serre's condition $S_1$, which says that every associated prime is minimal (for Noetherian rings) $\endgroup$ – zcn Sep 22 '14 at 18:27
  • $\begingroup$ Interesting, thanks. $\endgroup$ – Alex Kruckman Sep 22 '14 at 18:29

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