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What property should $A$ satisfy so that $A[x_1, \ldots, x_n]$ satisfies the dimension formula, $$\mathrm{dim}(A[x_1, \ldots, x_n]) = \mathrm{dim}(A[x_1, \ldots, x_n]/\mathfrak{p}) + \mathrm{ht}(\mathfrak{p}),$$ for any prime ideal $\mathfrak{p}$ in $A[x_1, \ldots, x_n]$?

For instance, this property holds when $A$ is a field. Is there a general property that ensures this formula is satisfied?

Zoey
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    Even $A$ regular isn't enough as I pointed out in this answer. It certainly holds for $A$ a field. – user26857 Jul 22 '15 at 07:12
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    looks like my question was answered already ? -- http://math.stackexchange.com/questions/49136/operatornameheight-mathfrakp-dim-a-mathfrakp-dim-a/49285#49285 – Zoey Aug 03 '15 at 10:42

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