Ok, I feel a little bit ashamed by my question.

This afternoon in the train, I looked for a counter-example:
— $k$ a field
— $A$ a finitely generated $k$-algebra
— $B$ a $k$-subalgebra of $A$ that is not finitely generated

Finally, I have found this:
— $k$ any field
— $A=k[x,y]$
— $B=k[xy, xy^2, xy^3, \dots]$

(proof : exercise)

My questions are:

1) What is your usual counter-example ?
2) Under which conditions can we conclude that $B$ is f.g. ?
3) How would you interpret geometrically this counter-example ?

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    $\begingroup$ 2) For example, if $B$ is the invariant ring of $A$ under the action of a group $G$, and $A$ is a completely reducible $G$-module, then $B$ is f.g.. This is Hilbert's theorem. It is far from being an if-and-only-if, however, and it seems hard to construct non-f.g. invariant rings even without complete reducibility. $\endgroup$ – darij grinberg Dec 9 '10 at 18:44
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    $\begingroup$ @nicojo, I see no need to start your question by "Ok, I feel a little bit ashamed by my question", there are a lot worse MO questions out there.. $\endgroup$ – J.C. Ottem Dec 9 '10 at 19:21
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    $\begingroup$ Although I'm sure this is no surprise, it might be worth adding that subalgebras of an algebra with a single generator are finitely generated (ams.org/journals/proc/1957-008-05/S0002-9939-1957-0091273-0/…). So the case you have with 2 generators is best-possible in this sense. $\endgroup$ – George Lowther Dec 9 '10 at 22:50

It is easy to make examples of such subrings. For example, take $A=k[x,y]$ and consider the subring

$$ B=k[x^a y^b : 0\le \frac{b}{a}<\sqrt{2}]. $$Geometrically, $B$ is spanned by monomials whose exponent vectors lie below the line $y=\sqrt{2}x$.

I think your question is quite interesting in the setting where $B=A^G\subset A$ is the invariant ring of some group action on $A$ (or equivalently, on the space $X=\mbox{Spec }A$). In many cases this subalgebra is finitely generated, which allows one can define a quotient space $X/G$ by $Y=\mbox{Spec }A^G$ with many good properties. This happens for example if $G$ is finite or reductive. However, as shown by Nagata's famous counterexample to Hilbert's 14th problem, $A^G$ may be infinitely generated, so the problem of defining such quotients in general is subtle. (Nagata's construction is indeed very geometrical, but a bit too complicated to restate here).


Dear Nicojo, since you now have many counter-examples, let me give you a situation where $B$ is finitely generated, in line with your question 2). I am going to adopt your notations with the important caveat that $k$ is a ring which needn't be a field .

Theorem of Artin-Tate Consider the inclusions of rings $k \subset B \subset A$ . Suppose that $k$ is Noetherian, that $A$ is a finitely generated algebra over $k$ and that $A$ is a finitely generated module over $B$. Then $B$ is a finitely generated algebra over $k$.

You might interpret this as saying that when $B$ is sufficiently close to $A$, finite generation is preserved.

You can find the proof in Atiyah-Macdonald, Proposition 7.8, page 81. From this theorem you can then prove Zariski's result that an extension of fields that is finitely generated as an algebra is actually a finite-dimensional extension (Proposition 7.9 page 82 loc.cit.) and then Hilbert's Nullstellensatz is literally an exercise: exercise 14, page 85 . So this result of Artin-Tate is really basic in commutative algebra and algebraic geometry, not surprisingly if you consider the authors (the Artin here is Emil, Mike's father.)


With regards to your questions.

1) Here's another example. $k[y, xy, y/x, y/x^2, y/x^3, \dots]$. The localization of this at the origin is a valuation ring (and this idea can be used to construct many other examples).

2+3) If you are constructing examples of this type, many are constructed by gluing. In other words, as pushouts of diagrams of affine schemes $$ \{ X \leftarrow Z \rightarrow W \}. $$ where $Z \to X$ is a closed immersion and $Z \rightarrow W$ is arbitrary. The condition you then want in (2) is for $Z \rightarrow W$ to be a finite map. Some relevant references include Ferrand, "Conducteur, descente et pincement", MR2044495 (2005a:13016) and Artin, "Algebraization of Formal Moduli II: Existence of Modifications", MR0260747 (41 #5370)

For example, the ring $k[x, xy, xy^2, \dots]$ is the pushout of $$ \{ \mathbb{A}^2 \leftarrow \text{coordinate-axis} \rightarrow \text{point} \}.$$ This gives a nice geometric interpretation, you just contracted a coordinate axis to a point, you can contract other schemes and get new examples. Note the $Z \to W$ in this example is not finite.

My example in 1) is the pushout of

$$ \{ \mathbb{A}^2 \setminus{V(x)} \leftarrow \text{Spec } k[x,y,x^{-1}]/(y) \rightarrow \text{Spec } k[x] \}.$$

Where the maps are the obvious ones. The $Z \rightarrow W$ map is not finite in this example either.

  • $\begingroup$ I think that your description corresponds actually to B=k[x, xy, x.y^2, x.y^3, ...]. $\endgroup$ – user2330 Dec 9 '10 at 19:53
  • $\begingroup$ nicojo, you are right, I misread it. I'll fix it now. $\endgroup$ – Karl Schwede Dec 9 '10 at 21:26
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    $\begingroup$ One might also want to look at Karl's paper on glueing: ams.org/mathscinet/search/… $\endgroup$ – Sándor Kovács Dec 9 '10 at 21:44

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