# Algebraic geometry notes by Akhil Mathew

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By Akhil Mathew

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Example text

4. (1) The inclusion of an open subset of a ringed space is flat (2) The composite of flat morphisms is flat Proof. Only the second requires a proof; this is because if A → B → C are morphisms of algebras, and B is flat over A and C flat over B, then C is flat over A. Flatness of a module means that tensoring with it is an exact functor, not simply right exact. 5. If f be a flat morphism, then f ∗ is an exact functor from Mod(OY ) → Mod(OX ). Also, if f : X → Y is a morphism of ringed spaces where OY is coherent and such that for each U ⊂ Y and associated map fU : f −1 (U ) → U the induced functor fU∗ : Mod(OY |U ) → Mod(OX |f −1 (U ) is exact, then f is flat.

Proof. If F is of finite type over OX , it is of finite type, then it is of finite type over OX /I, and vice versa. Suppose F ∈ Coh(OX ). Consider an exact sequence for some U OX /I p |U → F|U → 0; we show that the kernel K is of finite type. But this sequence is obtained p by tensoring the sequence of OX -modules OX |U → F|U → 0 with OX /I. Let the kernel here be K . Then K is of finite type. Moreover, we have an exact sequence p 0 → K → OX |U → F|U → 0. Tensoring yields an exact sequence K /IK → OX /I p |U → FU → 0 so that K is a surjective image of K , hence of finite type.

So, first we look at how the ni /ti patch. The quotients ni /ti , nj /tj must become equal each of the localizations Mq for q ∈ D(ti tj ) = SpecAti tj . This means that, by the first part of the proof, the ti nj − tj ni become equal in Ati tj . In particular, there is a power (ti tj )kij such that (ti tj )kij (ti nj − tj ni ) = 0 ∈ M. ), we see that we have reduced our problem to the following: i 2The map M → p Mp is always injective. 2. SHEAVES OF MODULES ON SpecA 57 Suppose given quotients ni /ti ∈ Mti such that for each i, j, we have the gluing condition ti nj = tj ni .