TITLE: A question about a proof of a weak form of Hilbert's Nullstellensatz
QUESTION [8 upvotes]: I'm trying to prove the following (corollary 5.24 page 67 in Atiyah-Macdonald):
Let $k$ be a field and let $B$ be a field that is a finitely generated $k$-algebra, i.e. there is a ring homomorphism $f: k \to B$ and $B = k[b_1, \dots , b_n]$ for $b_i \in B$. Then $B$ is an algebraic (and hence, in this case, finite) extension of $k$. 
There is a proof in Atiyah-Macdonald but it's more like a hint and I'm not sure I understand the details. Can you tell me if this detailed version of the proof is correct? Here goes (thanks!):
We need to show that $b_i$ are algebraic over $k$. Since $k$ is a field we know that $f$ is injective so we may view $k$ as a subfield of $B$ ($f$ is our embedding). Then $k \subset B$ are integral domains and $B$ is finitely generated so we are in the position to apply proposition 5.23 which tells us the following:
If $b$ is a non-zero element in $B$ then we can find a non-zero element $c$ in $k$ such that if $f: k \to \Omega$ is a homomorphism into an algebraically closed field $\Omega$ such that $f(c) \neq 0$ then there exists an extension $g: B \to \Omega$ of $f$ such that $g(b) \neq 0$.
We observe that $1$ is a non-zero element of $B$. The inclusion $i: k \hookrightarrow$ of $k$ into its algebraic closure $\overline{k}$ is a ring homomorphism such that $i(1) \neq 0$. By the previously stated proposition we hence can find a ring homomorphism $g: B \to \overline{k}$ such that $g(1) \neq 0$. Although the fact that $g(1) \neq 0$ doesn't interest us. But since $g$ is a ring homomorphism defined on a field we know that it's injective hence we can view $B$ as a subfield of $\overline{k}$. And now we are done since we have $k \subset B \subset \overline{k}$, hence $B$ is contained in the algebraic closure of $k$ and hence every element of $B$ is algebraic over $k$. In particular, $b_i$.
I wonder why it's called "Nullstellensatz". It doesn't seem to have anything to do with roots of polynomials.
Here is an image of proposition 5.23:

REPLY [6 votes]: First, instead of "a weak form of Hilbert's Nullstellensatz" I prefer the term Zariski's Lemma for the result you're asking about, for instance because (i) it's more informative (it was indeed introduced by Zariski in a 1947 paper) and (ii) it's not so "weak": it is a statement valid over an arbitrary field from which Hilbert's Nullstellensatz -- which holds only over algebraically closed fields -- can be rather easily deduced via the Rabinowitsch Trick.
If I may, I recommend my own commutative algebra notes for a treatment of Zariski's Lemma and the Nullstellensatz.  I first introduce Zariski's Lemma in $\S 11$ and prove it using the Artin-Tate Lemma, which I have come to regard as the simplest, most transparent proof.  But then throughout the notes I come back to give several other proofs of ZL:
$\bullet$ In $\S 12.2$ using Hilbert Rings.
$\bullet$ In $\S 14.5.3$ using Noether's Normalization Lemma (left as an exercise).
$\bullet$ In $\S 17.4$ using valuation rings.
The last of these is the one you're reading about now from Atiyah-Macdonald.  In my opinion it is also the most complicated: you might have better luck with one of the other three...