Gordan's lemma

Let ${\displaystyle \sigma }$ be the dual cone of the given rational polyhedral cone. Let ${\displaystyle u_{1},\dots ,u_{r}}$ be integral vectors so that ${\displaystyle \sigma =\{x\mid \langle u_{i},x\rangle \geq 0,1\leq i\leq r\}.}$ Then the ${\displaystyle u_{i}}$'s generate the dual cone ${\displaystyle \sigma ^{\vee }}$; indeed, writing C for the cone generated by ${\displaystyle u_{i}}$'s, we have: ${\displaystyle \sigma \subset C^{\vee }}$, which must be the equality. Now, if x is in the semigroup

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

then it can be written as

${\displaystyle x=\sum _{i}n_{i}u_{i}+\sum _{i}r_{i}u_{i},}$

where ${\displaystyle n_{i}}$ are nonnegative integers and ${\displaystyle 0\leq r_{i}\leq 1}$. But since x and the first sum on the right-hand side are integral, the second sum is a lattice point in a bounded region, and so there are only finitely many possibilities for the second sum (the topological reason). Hence, ${\displaystyle S_{\sigma }}$ is finitely generated.

The proof[3] is based on a fact that a semigroup S is finitely generated if and only if its semigroup algebra ${\displaystyle \mathbb {C} [S]}$ is finitely generated algebra over ${\displaystyle \mathbb {C} }$. To prove Gordan's lemma, by induction (cf. the proof above), it is enough to prove the statement: for any unital subsemigroup S of ${\displaystyle \mathbb {Z} ^{d}}$,

If S is finitely generated, then ${\displaystyle S^{+}=S\cap \{x\mid \langle x,v\rangle \geq 0\}}$, v an integral vector, is finitely generated.

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

${\displaystyle A_{n}=\operatorname {span} \{\chi ^{a}\mid a\in S,\langle a,v\rangle =n\}}$.

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

Lemma: Let A be a ${\displaystyle \mathbb {Z} }$-graded ring. If A is a Noetherian ring, then ${\displaystyle A^{+}=\oplus _{0}^{\infty }A_{n}}$ is a finitely generated ${\displaystyle A_{0}}$-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 ${\displaystyle f_{i}'s}$, homogeneous of positive degree. If f is homogeneous of positive degree, then we can write ${\displaystyle f=\sum _{i}g_{i}f_{i}}$ with ${\displaystyle g_{i}}$ homogeneous. If f has sufficiently large degree, then each ${\displaystyle g_{i}}$ has degree positive and strictly less than that of f. Also, each degree piece ${\displaystyle A_{n}}$ is a finitely generated ${\displaystyle A_{0}}$-module. (Proof: Let ${\displaystyle N_{i}}$ be an increasing chain of finitely generated submodules of ${\displaystyle A_{n}}$ with union ${\displaystyle A_{n}}$. Then the chain of the ideals ${\displaystyle N_{i}A}$ stabilizes in finite steps; so does the chain ${\displaystyle N_{i}=N_{i}A\cap A_{n}.}$) Thus, by induction on degree, we see ${\displaystyle A^{+}}$ is a finitely generated ${\displaystyle A_{0}}$-algebra.