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Let $R$ be any ring. The total quotient ring of $R$, denoted $K(R)$, is $R$ localized at the set of non zero-divisors.

I am trying to solve Exercise 3.15 in Eisenbud's Commutative Algebra:

a) Let $R$ be a Noetherian reduced ring and let $U$ be a multiplicative subset. Then $K(R[U^{-1}])=K(R)[U^{-1}]$.

b) If $R$ is a Noetherian ring, then $K(R[U^{-1}])=K(K(R)[U^{-1}])$.

For a), I was able to show that all prime ideals in $K(R)$ are maximal. Further, there are finitely many of them. Let these ideals be denoted $P_1,\ldots ,P_k$. By the Chinese Remainder Theorem, we have that

$$K(R)=\prod_{i=1}^k K(R/P_i)$$

How should I proceed? Also, can I have a hint to start part b)? Should I use the universal property of the localization somehow?

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  • $\begingroup$ I have changed my question to include that $R$ is Noetherian. I assume Eisenbud meant to include it. $\endgroup$ – Justine Feb 17 '16 at 16:36
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    $\begingroup$ We cannot know what Eisenbud really thought, but the book is called '...with a view towards geometry' and on geometrical practice, one mostly encounters only the noetherian case :) So I think it is pretty satisfying to have a proof in the noetherian case and a counterexample in the nonnoetheriam case, isn't it? $\endgroup$ – MooS Feb 17 '16 at 18:06
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Note that your $P_1, \dotsc, P_k$ are the minimal primes of $R$. Note that they are only finite if $R$ is noetherian, you have to add this assumption as the other answer points out.

For $a)$ you should show the following:

If we localize $\prod_{i=1}^k K(R/P_i)$ at $U$, then the component $K(R/P_i)$ survives if and only if $U \cap P_i = \emptyset$.

Using this, we get:

$K(R)[U^{-1}]$ is the product of all $K(R/P_i)$, where $U \cap P_i=\emptyset$.

Now we start with the reduced ring $R[U^{-1}]$. The minimal primes of this ring are all $P_i[U^{-1}]$ with $P_i \cap U = \emptyset$, i.e. its total quotient ring is the product of all $K(R[U^{-1}]/P_i[U^{-1}])$, where $U \cap P_i=\emptyset$.

Localization commutes with quotients, hence we have

$$K(R[U^{-1}]/P_i[U^{-1}])=K((R/P_i)[U^{-1}])=K(R/P_i),$$ which gives you the result. The latter equality holds, since we have $(R/P_i)[U^{-1}]=R/P_i$ if $U \cap P_i=\emptyset$.

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  • $\begingroup$ How does this give the result: $K(R[U^{-1}])=K(R)[U^{-1}]$? $\endgroup$ – Al Jebr Jun 27 '20 at 6:28
  • $\begingroup$ How do we have $(R/P_i)[U^{-1}]=R/P_i$ if $U \cap P_i=\emptyset$ ? $\endgroup$ – Al Jebr Jun 27 '20 at 6:51
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a) is wrong without the assumption that $R$ is noetherian.

Let $k$ be a field and let $$R=k[X_1, \dots, X_n, \dots]/(X_iX_j:i\ne j,j>2).$$ $R$ is reduced, and its maximal ideal $\mathfrak m=(x_1,\dots,x_n,\dots)$ is the set of the zerodivisors. Hence $K(R)=R_{\mathfrak m}$. Set $\mathfrak p=(x_2, \dots, x_n, ...)$ and $S=R\setminus\mathfrak p$. Then $S^{-1}K(R)=R_{\mathfrak p}$, while $K(S^{-1}R)=K(R_{\mathfrak p})$. (Note that $R_{\mathfrak p}=k[X_1,X_2]_{(X_2)}$ is not a field.)

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  • $\begingroup$ I have changed my question to include that $R$ is Noetherian. I assume Eisenbud meant to include it. $\endgroup$ – Justine Feb 17 '16 at 16:36

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