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.
This seems to be answered by the OP in comments, but just to get it answered:
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.