For the quotient of polynomial rings over complex number field,
its global dimension is finite is equivalent to it is domain.
is this true?
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2 Answers
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No, but Serre proved that for noetherian local rings having finite global dimension is the same as being regular.
So, choose any non-regular local ring which at the same time an integral domain such as the localization of the cuspidal curve at the origin:
$$ k[x,y]_{(x,y)}/(y^2-x^3)k[x,y]_{(x,y)} $$
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$\begingroup$ I am considering the quotient of polynomial. So, I do not think this is a counter example. $\endgroup$– iffCommented Mar 7, 2013 at 13:34
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$\begingroup$ Yes, technically, $k[x,y]/(y^2-x^3)$ is your counter example then. See this: mathoverflow.net/questions/59981/… $\endgroup$ Commented Mar 7, 2013 at 15:57
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It is true. If global dimension of $R$ is finite say $ = r$ then for any maximal ideal $\mathfrak{m}$ we have homological dimension of $R_\mathfrak{m}$ is finite and $\leq r$. By Serre's result $\dim R_\mathfrak{m}$ = global dimension of $R_\mathfrak{m}$. It remains to note that there exists at least one maximal ideal $\mathfrak{m}$ with global dimension of $R_\mathfrak{m}$ equal to $r$. This is so since
$$ r = \max \{ hdim \ R_\mathfrak{m} \ \mathfrak{m} \ \text{a maximal ideal of $R$} \}.$$
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$\begingroup$ Assume $R$ is a domain and prove it has finite global dimension. Let's see how you prove it. (It's false!) $\endgroup$ Commented Mar 5, 2013 at 16:04
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$\begingroup$ I was under the impression that $R$ has finite global dimension $\endgroup$ Commented Mar 6, 2013 at 9:44
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$\begingroup$ Tony: You may be confusing "global dimension" with "finitistic global dimension". Global dimension is the biggest projective dimension of any R-module (and since $R / m$ has infinite projective dimension for some maximal ideal $m$ if $R$ isn't regular, gldim$(R) = \infty$). However, the finitistic global dimension (i.e. the supremum of the projective dimensions of all the $R$-modules that do have finite projective dimension) is, if I remember correctly, the Krull dimension of the ring $R$, assuming $R$ is Noetherian. It still has nothing to do with $R$ being a domain or not, though. $\endgroup$ Commented Mar 6, 2013 at 15:11
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$\begingroup$ TO Tony： what is the meaning of hdim in" r=max hdim R_m" ?thanks a lot $\endgroup$– iffCommented Mar 7, 2013 at 13:39
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$\begingroup$ To Neil Epstein, do you have an example? thanks $\endgroup$– iffCommented Mar 7, 2013 at 13:40