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By Hopf H.

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Thus, replacing Q by Q/F , we reduce to the case in which Q is sharp and F = 0. We must show in this case that if q ∈ Q belongs to every facet of Q, then q = 0. The complement of a facet F is a prime ideal p of height one, and q ∈ F if and only if νp (q) = 0. Since the set of all such νp generates the cone CQ (H(Q)), it follows that h(q) = 0 for all h ∈ H(Q). 2) implies that q = 0, since Q is sharp. 9 If Q is a fine monoid, the map Q → Qsat induces a homeomorphism Spec(Qsat ) → Spec(Q). 10 Let p be a height one prime ideal in a fine monoid M .

Let C := F0 ⊂ · · · ⊂ Fd = C be a maximal chain of faces of C. Since each Fi is an exact submonoid of C, the inclusions F0gp ⊆ F1gp ⊂ · · · ⊂ Fdgp of linear subspaces of C gp are all strict. Since C gp has dimension d, d ≤ d. We prove gp the opposite inequality by induction on the dimension d of C . If d = 0, C = 0 and the result is trivial. 2) that S is the set of indecomposable elements of C. Our assumptions imply that d ≥ 1, and in particular F1 = 0. Then by (1) it must contain a K-indecomposable element c.

1) that CL∗ f , is a finitely generated group, and since Lt is finitely generated, so is CL∗ . 1) implies that CL is a finitely generated monoid. The finiteness of the saturation of a fine monoid also follows from Gordon’s lemma. 20 Let M be a fine monoid and let C ⊆ K ⊗ M gp be the K-cone it spans. Then M sat = M gp ×C gp C and is finitely generated as a monoid. Proof: The previous result implies that M gp ×C gp C is finitely generated as a monoid and is independent of the choice of K, so we may as well take K = Q.