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:
Best Answer
Your proof looks okay.
Here is a reason why it is called the Nullstellensatz:
Assume that your ground field $k$ is algebraically closed. Let $B = k[x_1,...,x_n]$ where the $x_i$ are indeterminates. Let $m \subset B$ be a maximal ideal. Then $B/m$ is a field extension of $k$ which is also clearly a finitely generated $k$ algebra. Then by what you proved above, $k \hookrightarrow B/m$ is a finite ring map. However, as $k$ is algebraically closed, this means that the ring map must be an isomorphism. Let $a_i \in k$ be those elements that are mapped to $x_i + m$ in $B/m$. Then it follows that $x_i - a_i \in m$, which implies that $(x_1 - a_1, ..., x_n - a_n) \subset m$. But, $(x_1 - a_1, ..., x_n - a_n)$ is a maximal ideal, so we must have $(x_1 - a_1, ..., x_n - a_n) = m$. What we have just proved is usually referred to as Hilbert's Weak Nullstellensatz (or so I think).
You can use the Weak Nullstellensatz to prove the Strong Nullstellensatz (a reference would be Mumford's Red Book), which says that following:
For an algebraically closed field $k$, given an ideal $\alpha \subset k[x_1,...,x_n]$, we have $I(V(\alpha)) = \sqrt{\alpha}$, where \begin{equation} V(\alpha) = \{\vec{a} \in k^n: \forall f \in \alpha, f(\vec{a}) = 0 \} \end{equation}
and
\begin{equation} I(Z) = \{f \in k[x_1,...,x_n]:\forall \vec{a} \in Z, f(\vec{a}) = 0\} \end{equation} for $Z \subset k^n$.
You can now see what this has to do with the zeroes of polynomials. As a side note, the Strong Nullstellensatz also has an equivalent algebraic formulation that amounts to proving that $k[x_1,...,x_n]$ is a Jacobson ring, that is a ring where any radical ideal is the intersection of the maximal ideals containing it. Hope this helps, and let me know if you are confused by anything.
$\textbf{Later edit:}$ I should mention that the Weak Nullstellensatz is also a very geometric statement. It clearly implies that the elements of any proper ideal of $k[x_1,...,x_n]$ have a common zero.
Also, while I like the proof of the statement that I refer to as the Strong Nullstellensatz using Rabinowitsch's Trick, I find it difficult to remember the trick. Here is an alternate proof to the algebraic version of the Strong Nullstellensatz using techniques that I am sure you are familiar with, since you are reading chapter $5$ of A-M.
$\textbf{The Nullstellensatz:}$ Let $k$ be any field, not necessarily algebraically closed, and let $A$ be a nonzero finitely generated $k$ algebra. Then we have the following:
(a) $\sqrt{(0)} \subset A$ is the intersection of all maximal ideals containing it.
(b) Given an ideal $I \subset A$, $\sqrt{I}$ is the intersection of all maximal ideals containing it.
$\textbf{Proof:}$ (a) I will let $M$ denote the set of maximal ideals of $k[x_1,...,x_n]$. Since $\sqrt{(0)}$ is the intersection of all prime ideals containing it, it follows that $\sqrt{(0)} \subset \cap_{m \in M} m$. Let $f$ be an element contained in every maximal ideal, and suppose that $f \notin \sqrt{(0)}$. Then $A_f$ is a nonzero ring, hence contains a maximal ideal $m$. Note that $A_f$ is also a finitely generated $k$ algebra (why?), and as a result, so is $A_f/m$. By Zariski's Lemma, this means that $A_f/m$ is a finite extension of $k$. Let $p: A \rightarrow A_f/m$ be the standard map. Then we have the inclusions $k \subset A/p^{-1}(m) \subset A_f/m$. Since $A_f/m$ is integral over $k$, it follows that $A/p^{-1}(m)$ is also integral over $k$. But $A/p^{-1}(m)$ is a domain, so it must be a field. It follows that $p^{-1}(m)$ is a maximal ideal of $A$, and by construction $f \notin p^{-1}(m)$, which contradicts our choice of $f$. This proves (a).
(b) follows from (a) by passing to the quotient $A/I$ (do you see why?). $\blacksquare$
As an exercise try to show that $I(V(\alpha)) = \sqrt{\alpha}$ over an algebraically closed field $k$, using the above theorem.