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Let $A$ be a commutative reduced ring (need not be noetherian). Let $S$ be the set of all non-zerodivisors of $A$. What is the Krull dimension of $S^{-1}A$ ?

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This is just a remark.

In the noetherian case one knows that $\dim Q(A)=0$. (Look here for a proof.) But in general this is not true. Assume that $\dim Q(A)=0$. Since $Q(A)$ is reduced, it follows that $Q(A)$ is von Neumann regular. But there are examples of (non-noetherian) reduced rings (with compact minimal spectrum) such that their total ring of fractions is not von Neumann regular.

A concrete example. Let $k$ be a field and let

$$A=k[X_1, \dots, X_n, \dots]/(X_iX_j)_{i\ne j}=k[x_1, \dots, x_n, \dots].$$

It is reduced, its maximal ideal $\mathfrak m$ is the set of the zerodivisors. Hence $S=A\setminus\mathfrak m$ and $S^{-1}A=A_{\mathfrak m}$. But $A_{\mathfrak m}$ has dimension $\ge 1$ (in fact equal to $1$) because $(x_2, \dots, x_n, …)$ generate a prime ideal in $A$ strictly contained in $\mathfrak m$.

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