Given $A$ an integral domain and $A[x_1, \ldots, x_n]$ a polynomial ring over $A$. Let $s = \prod_{1\leq i \leq n}{x_i}^{\alpha_i}$. What is the Krull dimension of $A[x_1, \ldots, x_n]_s$? Will it be equal to Krull dimension of $A[x_1, \ldots, x_n]$?


1 Answer 1


Under the assumption that $A$ is noetherian, yes, $\dim A[x_1, \ldots, x_n]_s =\dim A[x_1, \ldots, x_n]$.

It is enough to prove that for a noetherian ring $A$ of finite dimension $n$ the ring $B:=A[x]_x$ has dimension $\dim B=n+1$, since by noetherianity $\dim A[x]=n+1$.
Indeed let $\mathfrak p_0\mathfrak \subsetneq \cdots \subsetneq \mathfrak p_n$ be a chain of prime ideals of length $n$ in $A$.
Then $\mathfrak p_0B\mathfrak \subsetneq \cdots \subsetneq \mathfrak p_n B \subsetneq \mathfrak p_n B+(x-1)B$ is a chain of prime ideals of length $n+1$ in $B$, proving that $$\dim B\geq n+1$$ Since the inequality $\dim B=\dim A[x]_x\leq \dim A[x]$ is obvious and since $\dim A[x]=n+1$ (as already observed) we obtain $$\dim B\leq n+1$$ so that finally we get what we wanted:$$\dim B=n+1 $$ Fair Trade?
Notice that I have added the hypothesis that $A$ is noetherian but subtracted the hypothesis that $A$ be a domain: in the answer above $A$ is an arbitrary noetherian ring.
Do you think this is an honest swap? :-)

  • $\begingroup$ So using above how can I show that $\dim A[x,y]_{xy}= \dim A[x,y] ,$ where $xy$ is a non-nilpotent element of $A[x,y]$ and $A$ is a Noetherian ring ? $\endgroup$
    – user185640
    May 12, 2016 at 20:31

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