# A question about a weak form of Hilbert's Nullstellensatz

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?

• @DylanMoreland Oh right! Thank you. I'll edit the question. – Rudy the Reindeer Jul 19 '12 at 7:39
• @DylanMoreland I edited it. If an extension is finite then it's algebraic. So if an algebra is finitely generated and it is a field then it's a finite extension. – Rudy the Reindeer Jul 19 '12 at 7:40
• When they say "finite" they mean "finitely generated as a module". This is a stronger condition. You seem to be taking it to mean "finitely generated as an algebra" or "finite type" in Grothendieck's terminology. – Dylan Moreland Jul 19 '12 at 7:41
• @DylanMoreland Yes, I know, I had to look up the definitions on page 30. But the corollary is stated as I quote it in the question, so they don't use finite in the corollary. Or did I misunderstand what you were saying? – Rudy the Reindeer Jul 19 '12 at 7:43
• I'm afraid that I don't understand: I'm reading the words "finite algebraic extension" both here and in my copy of the book. If you can prove that $B$ has finite dimension as a vector space over $k$ then you are done, but I don't see that in the above. – Dylan Moreland Jul 19 '12 at 7:47

## 2 Answers

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.

• +1 for this fine answer, thank you. I think you helped me resolve my confusion, can you tell me if that's it: If $B$ is a finitely generated $k$-algebra, we have $B = k[b_1, \dots, b_n]$. But this doesn't imply that $B$ is an algebraic extension over $k$ since $b_i$ are not necessarily algebraic over $k$. – Rudy the Reindeer Jul 19 '12 at 8:19

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.

• I'll read that proof later, right now I'm working on a proof using proposition 5.23. – Rudy the Reindeer Jul 19 '12 at 9:35