algebraic geometry - Equivalent condition for being a Jacobson ring

(Atiyah-Macdonald, Ex. 5.25)

Let $A$ be a ring. Show that the following are equivalent:
i) $A$ is a Jacobson ring;
ii) Every finitely generated $A$-algebra $B$ which is a field is finite over $A$.

I'm trying to solve $i)\Rightarrow ii)$. Almost done, but I need some help. My trial:

Let $f:A\to B$, $B$ a finitely generated $A$-algebra which is a field. Since $f(A)$ is Jacobson and $A$-algebra/module actually means that $f(A)$-algebra/module, we can assume that $A \subseteq B$. Pick $s\in A$ as in Ex. 5.21 (see below). Because $A$ is Jacobson, $J(A)=0$ so we can find a maximal ideal $m$ not containing $s$. Let $k:=A/m$ and $f:A \twoheadrightarrow k \hookrightarrow \bar{k}$, then $f(s) \neq 0$, so $f$ extends to $g:B\to \bar{k}$. Because $B$ is a field, $g$ is either injective or trivial. But $g(s)=f(s) \neq 0$ so $g$ is injective and $B \simeq g(B)$. Suppose $B=A[z_1,\cdots,z_m]$, then $g(B)$ $=$ $g(A)[g(z_1),\cdots,g(z_m)]$ $=$ $k[g(z_1),\cdots,g(z_m)]$. Each $g(z_i)$ is in $\bar{k}$ hence integral over $k$, so $g(B)$ is finitely generated $k$-module.

Now the question:
1. How can I deduce that $B$ is a finitely generated $A$-module?
2. $g(A)=f(A)=k$ but $g$ is injective. How can it be possible? I think that $A$ strictly(?) contains $k$.

(Ex. 5.21) Let $A$ be a subring of an integral domain $B$ such that $B$ is finitely generated over $A$. Show that there exists $s \neq 0$ in $A$ such that, if $\Omega$ is an algebraically closed field and $f:A\to \Omega$ is a homomorphism for which $f(s) \neq 0$, then $f$ can be extended to a homomorphism $B\to \Omega$.

Answer

Since $A$ is assumed to be a subring of $B$, we have $A\hookrightarrow B \hookrightarrow \bar{k}$, so the composition is injective. But the composition is nothing but $A\twoheadrightarrow k$. It follows that $A=k$ and the maximal ideal $m$ is $0$. Now $B=k(z_1, \ldots, z_m)$ with all $z_i$ algebraic over $k$, hence a finite extension of $k=A$.