Let $k$ be a field, $R$ a finitely generated $k$-algebra, $f \in R$ nonzero. Assume $R$ is a domain. I understand how to show that $\dim(R_f) = \dim(R)$, but I began exploring the following argument, and there is a mistake somewhere since it gives a different answer. Dimension in this problem means Krull dimension.
Now $R_f$ is a domain, and since $R_f \simeq R[X]/\langle 1 - fX \rangle$, $\langle 1 - fX\rangle$ is prime. Since $R[X]$ is a finitely generated $k$-algebra and a domain, we have \begin{equation} \dim(R[X]_{\langle 1 - fX \rangle}) + \dim(R[X]/\langle1 - fX\rangle) = \dim(R[X]). \end{equation} We have the tower of fields $k \subset \text{Frac}(R) \subset R(X)$, and \begin{align} \dim(R[X]) & = \text{tr.deg.}_k(R(X)) \\ &= \text{tr.deg.}_k(\text{Frac}(R)) + \text{tr.deg.}_{\text{Frac}(R)}(R(X)) \\ &= \dim(R) + \text{tr.deg.}_{\text{Frac}(R)}(R(X)). \end{align} But $R(X) = \text{Frac}(R)(X)$, so $\text{tr.deg.}_{\text{Frac}(R)}(R(X)) = 1$. Since again $R_f \simeq R[X]/\langle 1- fX \rangle$, we thus have \begin{equation} \dim(R[X]_{\langle 1 - fX \rangle}) + \dim(R_f) = \dim(R) + 1. \end{equation} Now $\langle 1 - fX \rangle R[X]_{\langle 1- fX \rangle}$ is a prime ideal (in fact, maximal) in $R[X]_{\langle 1 - fX \rangle}$. But $R[X]_{\langle 1 - fX\rangle}$ is a domain, so we have the strictly increasing chain of prime ideals $0 \subsetneq \langle 1 - fX \rangle R[X]_{\langle 1- fX \rangle}$ in $R[X]_{\langle 1 - fX \rangle}$. Thus $\dim(R[X]_{\langle 1 - fX \rangle}) \geq 2$, and then $\dim(R_f)\leq \dim(R) - 1$. Help please thanks!