# Finite fiber- unramified morphisms

I'm in trouble understandig the proof of Proposition 3.2 Chapter 1 of Milne's Book "Étale Cohomology". Let $f:Y\rightarrow X$ be locally of finite-type. The following are equivalent.

$(a)$ f is unramified.

$(b)$ for all $x\in X$ the fiber $Y_x\rightarrow spec(k(x))$ over $x$ is unramified.

$(d)$ for all $x\in X$, $Y_x$ has an open covering by spectra of finite separable $k(x)$-algebras.

A morphism is unramified if for all $y \in Y$ $k(y)$ is a finite separable extension of $k(x)$ and $m_y=m_x\mathcal{O}_{Y,y}$.

For $(b) \Rightarrow (d)$

Let $U$ an open afine subset of $Y_x$ and $\mathfrak{q}$ a prime ideal in $B=\Gamma(U,\mathcal{O}_{Y_x})$, acording to $(b)$ $B_\mathfrak{q}=k(\mathfrak{q})$ is a finite separable field extension of $k(x)$. Also

$$k(x)\subset B/\mathfrak{q}\subset B_\mathfrak{q}/\mathfrak{q}B_\mathfrak{q}=B_\mathfrak{q}$$

then $B/\mathfrak{q}$ is also a field.

I don't understand why $B/\mathfrak{q}$ is a field. I understand that the result follows from that if $B/\mathfrak{q}$ is a field then $\mathfrak{q}$ is maximal and then $B$ is an Artin ring. Is that because the field extension is finite and separable??

Thanks a lot in advance for your help.

• Can you fill in the details of what exactly it is you are asking or, at the very least, explain the notation. It is usually bad form to ask questions which sre unreadable without having a specific book open in a specific page. – Mariano Suárez-Álvarez Jan 12 '14 at 5:11
• I'm sorry I edited it, and also I realized why it happened. It is because the extension is finite and then $B_\mathfrak{q}$ is integer over the domain $B/\mathfrak{q}$. Due to the first is field then $B/\mathfrak{q}$ also is a field. – user120731 Jan 12 '14 at 8:11

We have the inclusion $k(x) \subset B/\mathfrak q \subset B_{\mathfrak q}$, where $k(x)$ is a field and $B_{\mathfrak q}$ is a finite field extension of $k(x)$. Any intermediate ring is then automatically a field, hence $B/\mathfrak q$ is a field.