Lectures on Logarithmic Algebraic Geometry by Arthur Ogus

By Arthur Ogus

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Since h is local, h−1 (0) = F . This proves the first statement. It is clear that S ⊥ is a face of H(Q) if S is any subset of Q and that T ⊥ is a face of Q if T is any subset of H(Q). Furthermore, S2⊥ ⊆ S1⊥ if S1 ⊆ S2 , and S ⊆ (S ⊥ )⊥ . The only nontrivial thing to prove is that F = (F ⊥ )⊥ if F is a face of Q. But this follows immediately from the existence of an h with F = h−1 (0). 5 If Q is fine, then Qsat is again fine. In fact, the action of Q on Qsat defined by the homomorphism Q → Qsat makes Qsat a finitely generated Q-set.

This proves that every exact submonoid of a fine sharp monoid is finitely generated. Slightly more generally, if M is an exact submonoid of any fine monoid N , we can choose a surjection Nr → N , and the inverse image M of M in Nr is an exact submonoid of Nr . It follows that M is finitely generated, and hence so is M . Suppose now that M is an exact submonoid of a saturated monoid N and x ∈ M gp with nx ∈ M for some n ∈ Z+ . Then x ∈ N ∩ M gp = M , so M is also saturated. This proves (2). Let F be a face of an integral monoid M , let x and y be elements of F , and suppose z := x − y ∈ M .

Let P be the submonoid of Q generated by S. Then P is still sharp and the induction hypothesis implies that there exists a local homomorphism h: P → N. Then h induces a homomorphism P gp → Z which we denote again by h. Replacing h by nh for a suitable n ∈ Z+ , we may assume that h extends to a homomorphism Qgp → Z we which still denote by h. If h(t) > 0 there is nothing more to prove. If h(t) = 0, choose any h : Qgp → Z such that h (t) > 0. Then if n is a sufficiently large natural number, nh(s) + h (s) > 0 for all s ∈ S and h (t) > 0, so nh + h ∈ H(Q) and is local.

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