# 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??

• 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.