Corollary 5.24 on page 67 in Atiyah-Macdonald reads as follows:
Let $k$ be a field and $B$ a finitely generated $k$-algebra. If $B$ is a field then it is a finite algebraic extension of $k$.
We know a field extension $E$ over $F$ is algebraic if it’s finite, that is, $E = F[e_1, \dots, e_n]$. By definition, a finitely generated $k$-algebra is of the form $k[b_1, \dots , b_n]$. So the corollary above seems to directly follow from these two facts.
I hope I misunderstand something fundamental because I worked through the propositions and proofs this corollary is using and it was rather lengthy and not very enjoyable. What am I missing?
Perhaps the statement will become more clear in the following language:
A ring homomorphism $R \to S$ is finite if $S$ is finitely generated as a module over $R$. A ring homomorphism $R \to S$ is called of finite type if $S$ is finitely generated as an algebra over $R$. Clearly, finite implies of finite type. The converse is not true, in general. We have that finite <=> integral and of finite type.
But for fields, the converse is true: Every field extension which is of finite type, is already finite (and therefore algebraic). This is an easy consequence of Noether’s normalization lemma. It is not a consequence of the definitions, because it is not clear a priori that our algebra generators are algebraic.
Your confusion is coming from the fact that you are assuming $B$ finitely generated as a $k$ – algebra implies that it is a finite (and hence algebraic) extension of $k$. Consider the polynomial ring $k[x]$ in one indeterminate – this is finitely generated as a $k$ – algebra by definition. But this is definitely not a finite extension of $k$ because $x$ is an indeterminate and hence is transcendental over $k$!
Here is the proof of the weak Nullstellensatz using Noether Normalisation that Martin mentioned below. Recall that Noether Normalisation states that if $k$ is a field and $B$ a finitely generated $k$ – algebra, then there exists an integer $d$ and algebraically independent elements $x_1,\ldots,x_d$ such that $B$ is finitely generated as a module over $k[x_1,\ldots, x_n]$. The following visualisation may be helpful:
$$k \longrightarrow k[x_1,\ldots,x_d]\longrightarrow B.$$
Recall that because now $B$ is a field, in particular it is an integral domain and so is $k[x_1,\ldots,x_d]$. Since $B$ is finitely generated over this guy, we have by Proposition 5.7 of Atiyah – Macdonald that $k[x_1,\ldots,x_d]$ is a field. This is a field if and only if $d = 0$ so that $A$ is finitely generated as a module over $k$, proving the weak Nullstellensatz.