# For graded algebras over a field, does finite Krull dimension imply finite generation?

On p. 37 of his book Algorithms in Invariant Theory, Bernd Sturmfels writes,

Let $R$ ... be a graded $\mathbb{C}$-algebra of dimension $n$. This means... that $n$ is the maximal number of elements of $R$ which are algebraically independent over $\mathbb{C}$. This number is the Krull dimension of $R$.... We write $H(R_+)$ for the set of homogeneous elements of positive degree in $R$. A set $\{\theta_1,\dots,\theta_n\}\subset H(R_+)$ is said to be a homogeneous system of parameters (h.s.o.p.) provided $R$ is finitely generated as a module over its subring $\mathbb{C}[\theta_1,\dots,\theta_n]$.... A basic result of commutative algebra, known as Noether's normalization lemma, implies that an h.s.o.p. for $R$ always exists.

Two things about this paragraph are troubling me but they boil down to the same root issue: Does $R$ have to be finitely generated as a $\mathbb{C}$-algebra and if it does, how do we know?

The things that are troubling me are: (1) how can we invoke Noether normalization unless we know $R$ is finitely generated as a $\mathbb{C}$-algebra? and (2) doesn't the identification of Krull dimension with transcendence degree depend on finite generation as well?

Neither finite Krull dimension nor finite transcendence degree imply finite generation over $\mathbb{C}$, as the example of the field $\mathbb{C}(x)$ shows: it's Krull dimension zero and transcendence degree 1, but not finitely generated. So if anything lets us conclude finite generation here, it must involve the grading. Thus my questions are:

• If $R$ is a graded $k$-algebra of transcendence degree $n<\infty$, is it finitely generated as a $k$-algebra? If so, what's the proof?

• If $R$ is a graded $k$-algebra of Krull dimension $n<\infty$, is it finitely generated as a $k$-algebra? If so, what's the proof?

Here $k$ is an arbitrary field (since I strongly suspect Sturmfels doesn't need any special features of $\mathbb{C}$ for what he's doing).

ADDENDUM: I've discovered that the presentation in the relevant section of Sturmfels' book hews pretty closely to the 1979 article Invariants of Finite Groups and Their Applications to Combinatorics by Richard Stanley in the AMS Bulletin. Stanley makes all the same statements except that he has included finite generation over $k$ as part of the definition of a graded $k$-algebra. I think Sturmfels must just have been being careless in not explicitly stating a finite generation requirement.

• The second thing is clearly not true without some additional hypotheses. At the very least you probably want to assume that the degree zero piece is finitely generated, or maybe even that it's just $\mathbb{C}$. Otherwise there are dumb counterexamples, e.g. $R$ could be $\prod_{i=1}^{\infty} \mathbb{C}$ in degree $0$... – Qiaochu Yuan Nov 8 '15 at 8:33
• @QiaochuYuan - Yes I had in mind that it was just $\mathbb{C}$ in dimension zero. Sturmfels includes this in his definition of "graded $\mathbb{C}$-algebra." – Ben Blum-Smith Nov 8 '15 at 15:37
• For both points: consider the graded algebra which coincides with $k[x]$ with its usual grading but such that $x^i\cdot x^k=0$ for all $i,j>0$. – Mariano Suárez-Álvarez Nov 8 '15 at 15:39
• @Mariano: so $x^2= x.x=0$? But then the algebra does not coincide with $k[x]$, does it? – Georges Elencwajg Nov 8 '15 at 19:13
• Which coincides with $k[x]$ as a vector space, I meant to write. – Mariano Suárez-Álvarez Nov 8 '15 at 19:20

$\bullet$The graded $k$ algebra $R=k[x_i|i\in \mathbb N]=k[X_i|i\in \mathbb N]/(X_i^2|i\in \mathbb N)$ has Krull dimension $0$ [since its only prime ideal is $(x_0,x_1,\cdots)$], but it is not finitely generated.
$\bullet$That algebra $R$ is also an example of an algebraic algebra (thus of transcendence degree $0$) over $k$ which is not finitely generated over $k$.