Question

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 sub-k-algebra of A that is not finitely generated

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

(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 ?

Answer 1

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.)

Answer 2

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.

Answer 3

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).