Let $X=Spec\;A$ be a scheme of finite type over a field $k$ and $x\in X$. Let $q\subset A$ be an ideal corresponding to $x$.
How one can show that $\dim \overline{\lbrace x \rbrace} = \text{tr.deg} (A_q/qA_q)$? Maybe it is some corollary of the Nagata altitude formula?
Note that $A_q/qA_q=k(x)$, the residue field of $x$. As the height of $q$ is equal to the codimension of $\overline{\{x\}}$ in $X$, and $\dim X=\dim_\text{Krull}A$, then as the Krull dimension of $A$ is equal to $\mathrm{tr.deg}\,k(x)$, we are done. Hartshorne 1.8A refers to some references with proofs.