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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: enter image description here

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I don't have Atiyah-Macdonald at hand, but there is something fishy about the statement of 5.23 you cite (?): it says we can find $c$ such that ..., which is obvious since any $c\in k^\times$ will do, and then concludes something (there exists extension $g$...) in which $c$ plays no part whatsoever. Can't you just cut out the whole part that talks about $c$? – Marc van Leeuwen Jul 19 '12 at 11:52
@Marc, the statement of prop.5.23 from A-M is accurate. I don't understand what you think is "obvious" about $\,c\,$ , but the existence of a homomorphism (in our case, monomorphism as we talk about fields whereas in A-M it's about rings) into an alg. closed field whose kernel doesn't contain $\,c\,$ gurantees the existence of an extensions...etc. – DonAntonio Jul 19 '12 at 13:37
@DonAntonio: Maybe the setting of A-M is different, which would explain something. But as cited here $c$ is a nonzero, so inverible, element of the field $k$, so saying $f(c)\neq0$ is a triviality. An of course a field always contains nonzero elements, so the existence of $c$ isn't adding much either. It would be helpful if the precise assumptions of 5.23 were reproduced here; as it is, I just cannot make up what it is about. – Marc van Leeuwen Jul 19 '12 at 14:40
Dear @MarcvanLeeuwen I added an image of proposition 5.23. Is my statement of the proposition wrong? – Rudy the Reindeer Jul 19 '12 at 15:33
@ClarkKent: thank you. The point is clearly that the proposition only supposes $A$ to be an integral domain, which explains the precaution it must make about $f$, and which makes it a more subtle statement. If $A$ is a field, it can be simplified to "every homomorphism of $A$ into an algebraically closed field $\Omega$ can be extended to $g:B\to\Omega$ with $g(v)\neq0$", the validity of which is quite easy to see. – Marc van Leeuwen Jul 19 '12 at 16:29
up vote 6 down vote accepted

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}


\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.

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Also, another way to prove the statement whose proof you have given is to use Noether normalization. It is a surprisingly useful statement. – Rankeya Jul 19 '12 at 10:56 – user38268 Jul 19 '12 at 11:05
Thanks mate for this awesome answer, I'm just reading it now. Excellent! – Rudy the Reindeer Jul 19 '12 at 14:31
I'm sorry, could you explain how $B/m$ is a field extension of $k$? What happens if $c$ is non-zero in $m \cap k$? It's mapped to zero under the quotient map. But if $B/m$ is an extension of $k$, all elements in $k$ have to remain non-zero in $B/m$. – Rudy the Reindeer Jul 19 '12 at 15:52
@Clark: You have a homomorphism of rings $k \rightarrow B \rightarrow B/m$, right? And $k$ and $B/m$ are fields. But any homomorphism $f$ of rings between fields is injective (i.e., a field extension): the kernel of $f$ is an ideal, which since $f(1) = 1 \neq 0$, does not contain $1$. But in a field, the only ideal which does not contain $1$ is $\{ 0\}$. – Pete L. Clark Jul 19 '12 at 17:46

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...

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Thanks for the link. I always wanted to learn more about commutative algebra and your writing style suits me very well (as in that you don't make it sound more fancy than it actually is for example). – Jonas Teuwen Jul 19 '12 at 14:36
Yes, Zariski's lemma is definitely the better, more descriptive terminology. – Georges Elencwajg Jul 19 '12 at 15:52

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