4

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?

Rudy the Reindeer
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  • @DylanMoreland Oh right! Thank you. I'll edit the question. – Rudy the Reindeer Jul 19 '12 at 07:39
  • @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 07:40
  • 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 07:41
  • @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 07:43
  • 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 07:47
  • I should mention that there are much easier ways of proving the Nullstellensatz. Atiyah and MacDonald obtain a lot of interesting information about valuation rings en route to this, but you don't really need any of it if you're only interested in the Nullstellensatz. Zariski's proof (I'll try to dig up a link in the morning) is probably the shortest. – Dylan Moreland Jul 19 '12 at 07:50
  • @DylanMoreland You're right. I was confused. I need to think about whether this resolves my confusion. – Rudy the Reindeer Jul 19 '12 at 07:50
  • @ClarkKent May be helpful: http://math.stackexchange.com/questions/147345/definition-of-a-finitely-generated-k-algebra –  Jul 19 '12 at 09:08
  • @BenjaLim Thanks, I read that thread. I think my confusion was about what I wrote in the comment to the answer, I'm clear about what finite and finitely generated mean though. – Rudy the Reindeer Jul 19 '12 at 09:11
  • @ClarkKent Please see my answer below. –  Jul 19 '12 at 09:23
  • @ClarkKent I think the first paragraph of my answer addresses your comment in Martin's answer. –  Jul 19 '12 at 09:27
  • @BenjaLim Thanks man, yes, I +1-d your answer. It confirms what I wrote in my comment to Martin's answer, right? – Rudy the Reindeer Jul 19 '12 at 09:35
  • @ClarkKent Hmmm I read your comment again I don't really get it; it's kinda phrased as a question can you state it again? –  Jul 19 '12 at 11:04

2 Answers2

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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.

Martin Brandenburg
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  • +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 08:19
5

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.