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Let $A$ be an affine $K$-algebra and $f$ be a non-zero divisor of $A$ then can one say that $\dim A=\dim A_f$ ?

What I proved that if $A$ is an affine domain and $f$ is a non-zero element in $A$ then $\dim A=\dim A_f.$The sketch of the proof goes as following: Since $A_f \cong A[x]/(fx-1)$ and $A[x]$ is an affine domain with $(fx-1)$ prime ideal in $A[x]$ one can use the result $\dim A[x]=\dim A[x]/(fx-1) + ht(fx-1).$ And the final conclusion is given by krull's principal ideal theorem that $ht(fx-1)=1 .$

But here in this problem i can only show that $(fx-1)$ is a non-zero divisor of $A[x] $ because if $(fx-1)$ was a zero-divisor in $A[x] $ there exist a non-zero element $a\in A$ such that $a(fx-1)=0$ in $A[x]$ which gives that $af=0$ in $A$, giving $f$ zero-divisor. I cannot proceed further . I am not also sure if it can be proved or not. Can any one help me in proving or disproving this ?

Thank You.

user26857
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    Can you show $\dim A_f\leq \dim A$? For the reverse inequality, use the fact that $\dim A$ is the maximum of $\dim A/P$, as $P$ varies over the minimal primes of $A$ and thus reduce to the domain case, since $f\not\in P$ for any minimal prime $P$. – Mohan May 13 '16 at 21:09

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$\dim A=\sup\{\dim A/{\mathfrak p}:\mathfrak p\text{ minimal prime}\}$, and $\dim A_f=\sup\{\dim A_f/{\mathfrak p_f}:\mathfrak p\text{ minimal prime}\}$. (Note that $f\notin\mathfrak p$ for any minimal prime.)

But $A_f/{\mathfrak p_f}\simeq(A/\mathfrak p)_{\bar f}$, where $\bar f$ is the residue class of $f$ modulo $\mathfrak p$. We have $\bar f\ne\bar 0$ in $A/\mathfrak p$ for every minimal prime $\mathfrak p$, so $\dim A_f/{\mathfrak p_f}=\dim A/\mathfrak p$.

user26857
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