TITLE: Dimension of the total ring of fractions of a reduced ring.
QUESTION [9 upvotes]: 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 $S^{-1}A$ ?

REPLY [7 votes]: 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 $\mathfrak m$.