# Dimension of the total ring of fractions of a reduced ring.

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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Your set is never multiplicatively closed (unless it is empty), so the localization doesn't make sense. –  user18119 Feb 4 '13 at 9:40
@QiL: Hmm, if $A=\mathbb{Z}$, $S=\mathbb{Z} \setminus \{0\}$, then $S$ is multiplicatively closed, right? Perhaps I misunderstood you? –  Nils Matthes Feb 4 '13 at 9:56
@NilsMatthes: thanks ! I read "non-zero zerodivisors". Then the answer is clear. –  user18119 Feb 4 '13 at 10:11
In the Noetherian case I think it is $1$, as the zero-divisors are the union of a finite set of minimal prime ideals. –  Alex Becker Feb 4 '13 at 10:13
@AlexBecker You are wrong: math.stackexchange.com/questions/51400/… –  user26857 Feb 4 '13 at 10:39

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.

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However, if $A$ is reduced and its minimal spectrum is finite, then $Q(A)$ is von Neumann regular. –  user26857 Feb 5 '13 at 0:53

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

By modifying a little bit the above example (for instance taking the quotient of $k[X_1, \dots, X_n, \dots]$ by $X_1(X_2, X_3), (X_2, X_3)(X_4, X_5, X_6), ...$,) we find a reduced ring whose zerodivisors consist in the maximal ideal $m=(x_1, x_2, ...)$ and $S^{-1}A=A_m$ has infinite dimension.

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+1 Good example! –  user26857 Feb 4 '13 at 19:10
@YACP: thanks ! –  user18119 Feb 4 '13 at 21:18
Thanks for the first example. Second example needs to be modified as x_1+x_2 is not a zero-divisor. –  manoj Feb 5 '13 at 6:06
For an example of dimension n, we can take quotient of A_m by the ideal generated by x_i.x_j with i,j different and j>n. Then p=(x_{n+1},x_{n+2},...) is a prime ideal and we have a chain of prime ideals p,(p,x_1),(p,x_1,x_2),...,(p,x_1,...,x_n)=m. –  manoj Feb 5 '13 at 6:15
@manoj: you are right. –  user18119 Feb 5 '13 at 20:31