# Does this morphism necessarily give rise to a finite extension of residue fields?

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.

• After the quotienting you can ignore the localization (you localize by things that become invertible after you quotient anyway). – Qiaochu Yuan Jun 28 '13 at 3:08
• You are right, the phrasing is confusing. This was corrected this errata. The finiteness is part of the definition. – user18119 Jul 18 '13 at 8:52
• @QiL'8 Thanks for the comment, and thanks for the wonderful book. – Potato Jul 18 '13 at 15:11

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.
• Dear Professor, I actually forgot to say that the field extension should be separable. But now I'm very confused. In Qing Liu's book (page 139), he gives the hypotheses I gave above and requires "the (finite) extension of residue fields $k(y)\rightarrow k(x)$" to be separable. I assumed the parentheses implied that the finite assumption was redundant. You are saying this is not true? Thank you for your help. – Potato Jun 28 '13 at 5:16
• Dear Professor, Here is the exact passage. "Let $f:X\rightarrow Y$ be a morphism of finite type of locally Noetherian schemes. Let $x\in X$ and $y=f(x)$. We say that $f$ is unramified at $x$ if the homomorphism $\mathcal O_{Y,y}\rightarrow \mathcal O_{X,x}$ verifies $\mathfrak m_y \mathcal O_{X,x} = \mathfrak m_x$ (in other words, $\mathcal O_{X,x}/\mathfrak m_y \mathcal O_{X,x} = k(x)$), and if the (finite) extension of residue fields $k(y)\rightarrow k(x)$ is separable." – Potato Jun 28 '13 at 5:25