Let $f:X\rightarrow Y$ be a morphism of finite type of locally Notherian schemes. Let $x\in X$ and $y=f(x)$. Recall that $f$ is said to be unramified if the map of stalks $g:\mathcal O_{Y,y} \rightarrow \mathcal O_{X,x}$ satisfies $g(m_y)=m_x$, where $m_x$ denotes the maximal ideal of the quotient ring.
The condition on the maximal ideal shows that this map descends to a map $k(y)\rightarrow k(x)$. Is this necessarily a finite extension? I believe so, because taking appropriate affine neighborhoods, we can reduce to the case of a map of affine schemes with associated ring map $A\rightarrow A[t_1,\dots,t_n]$ for some elements $t_i$ of the ring of global sections of $X$. It seems to me after the localization and quotienting happens to get the map of residue fields, the second ring (now a field) should still be finitely generated over the first. Unfortunately, I don't know enough commutative algebra to do this rigorously. Is this correct? If so, what is the proof?
My motivation comes from studying étale morphisms, where the finiteness of the extension above is usually presupposed. However, in Qing Liu's book, on page 139, he seems to imply that the finiteness condition is redundant.
Consider the map $X =$ Spec $k[x] \to Y = $ Spec $k$, where $k$ is a field. The generic point of $X$ maps to the unique point of $Y$, and the map on local rings is just the inclusion of fields $k \subset k(x)$, so the map is unramified at $x$ according to your definition. But the map on residue fields is evidently not finite.
This is why the definition of unramified in EGA is not the one you gave (see e.g. here).