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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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  • $\begingroup$ In the Noetherian case I think it is $1$, as the zero-divisors are the union of a finite set of minimal prime ideals. $\endgroup$ Feb 4, 2013 at 10:13
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    $\begingroup$ @AlexBecker You are wrong: math.stackexchange.com/questions/51400/… $\endgroup$
    – user26857
    Feb 4, 2013 at 10:39
  • $\begingroup$ @YACP I meant to say $0$. I believe that is the case. $\endgroup$ Feb 4, 2013 at 19:48

2 Answers 2

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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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  • $\begingroup$ 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. $\endgroup$
    – manoj
    Feb 5, 2013 at 6:15
  • $\begingroup$ @manoj I guess your example is in fact the quotient of $k[X_1,\dots,X_n,\dots]$ by the ideal $(X_iX_j:i\neq j \text{ and } j>n)$. However, the existence of a chain of prime ideals of length $n$ can prove only that the height of $m$ is at least $n$, not necessarily equal to $n$. (However, in this case it's true that the height of $m$ is $n$.) $\endgroup$
    – user26857
    Feb 5, 2013 at 21:47
  • $\begingroup$ We saw that the dimension of total quotient ring Q(A) of a reduced ring need not be zero. I was trying to prove that it is zero (I was wrong) since I want to know whether finitely generated projective modules of constant rank over Q(A) are free. Over a zero dimensional ring, we know that all finitely generated projective modules of constant rank are free. So my original question over Q(A) remains to be answered. Any answer! $\endgroup$
    – manoj
    Feb 6, 2013 at 7:00
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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.

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  • $\begingroup$ However, if $A$ is reduced and its minimal spectrum is finite, then $Q(A)$ is von Neumann regular. $\endgroup$
    – user26857
    Feb 5, 2013 at 0:53

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