Gordan's lemma

In convex geometry, Gordan's lemma states that the semigroup of integral points in the dual cone of a rational convex polyhedral cone is finitely generated.^[1] In algebraic geometry, the prime spectrum of the semigroup algebra of such a semigroup is, by definition, an affine toric variety; thus, the lemma says an affine toric variety is indeed an algebraic variety. The lemma is named after the German mathematician Paul Gordan (1837–1912).

Proof

There are topological and algebraic proofs.

Topological proof

Let $\sigma$ be the cone as given in the lemma. Let $u_1, \dots, u_r$ be the integral vectors so that $\sigma = \{ x \mid \langle u_i, x \rangle \ge 0, 1 \le i \le r \}.$ Then the $u_i$ 's generate the dual cone $\sigma^{\vee}$ ; indeed, writing C for the cone generated by $u_i$ 's, we have: $\sigma \subset C^{\vee}$ , which must be the equality. Now, if x is in the semigroup

S_\sigma = \sigma^\vee \cap \mathbb{Z}^d,

then it can be written as

x = \sum_i n_i u_i + \sum_i r_i u_i

where $n_i$ are nonnegative integers and $0 \le r_i \le 1$ . But since x and the first sum on the right-hand side are integral, the second sum is also integral and thus there can only be finitely many possibilities for the second sum (the topological reason). Hence, $S_{\sigma}$ is finitely generated.

Algebraic proof

The proof^[2] is based on a fact that a semigroup S is finitely generated if and only if its semigroup algebra $\mathbb{C}[S]$ is finitely generated algebra over $\mathbb{C}$ . To prove Gordan's lemma, by induction (cf. the proof above), it is enough to prove the statement: for any unital subsemigroup S of $\mathbb{Z}^d$ ,

If S is finitely generated, then

S^+ = S \cap \{ x \mid \langle x, v \rangle \ge 0 \}

, v an integral vector, is finitely generated.

Put $A = \mathbb{C}[S]$ , which has a basis $\chi^a, \, a \in S$ . It has $\mathbb{Z}$ -grading given by

A_n = \operatorname{span} \{ \chi^a \mid a \in S, \langle a, v \rangle = n \}

By assumption, A is finitely generated and thus is Noetherian. It follows from the algebraic lemma below that $\mathbb{C}[S^+] = \oplus_0^\infty A_n$ is a finitely generated algebra over $A_0$ . Now, the semigroup $S_0 = S \cap \{ x \mid \langle x, v \rangle = 0 \}$ is the image of S under a linear projection, thus finitely generated and so $A_0 = \mathbb{C}[S_0]$ is finitely generated. Hence, $S^+$ is finitely generated then.

Lemma: Let A be a $\mathbb{Z}$ -graded ring. If A is a Noetherian ring, then $A^+ = \oplus_0^{\infty} A_n$ is a finitely generated $A_0$ -algebra.

Proof: Let I be the ideal of A generated by all homogeneous elements of A of positive degree. Since A is Noetherian, I is actually generated by finitely many $f_i's$ , homogeneous of positive degree. If f is homogeneous of positive degree, then we can write $f = \sum_i g_i f_i$ with $g_i$ homogeneous. If f has sufficieny large degree, then each $g_i$ has degree positive and strictly less than that of f. Also, each degree piece $A_n$ is a finitely generated $A_0$ -module. (Proof: Let $N_i$ be an increasing chain of finitely generated submodules of $A_n$ with union $A_n$ . Then the chain of the ideals $N_i A$ stabilizes in finite steps; so does the chain $N_i = N_i A \cap A_n.$ ) Thus, by induction on degree, we see $A^+$ is a finitely generated $A_0$ -algebra.