Krull dimension on localization 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 $A[x_1, \ldots, x_n]$?
 A: 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? :-)
