HARTSHORNE III, 6.10

The Five Lemma: Below is the five lemma as stated in Matsumura’s Commutative Theory (p 281). Suppose we have the following . A / B / C / D / E

f1 f2 f3 f4 f5      A0 / B0 / C0 / D0 / E0. Then (1) If f1 is surjective and f2 and f4 are injective then f3 is injective. (2) If f5 is injective and f2 and f4 are surjective then f3 is surjective.

6.10 Duality for a Finite Flat Morphism. (a) Let f : X → Y be a finite morphism of Noetherian schemes. For any quasi-coherent OY - G , H omY (f∗OX , G ) is a quasi-coherent f∗OX -module, hence corresponds to a ! quasi-coherent OX -module, which we call f G (II, Ex 5.17e).

Proof. There really isn’t much to do here. One could show that H omY (f∗OX , G ) is really a quasi-coherent f∗OX -module. But it certainly is locally which is all that matters.  (b) Show that for any coherent F on X and any quasi-coherent G on Y , there is a natural ! φ : f∗H omX (F , f G ) → H omY (f∗F , G ). Proof. Let us first show that φ exists. We will define it on an open affine set Spec A = U ⊂ Y (note that V = f −1(U) = Spec B is affine as well). So we must show there is a map ! Γ(U, f∗H omX (F , f G )) → Γ(U, H omY (f∗F , G )). But since V is affine we have ! ∼ ! ∼ ! Γ(U, f∗H omX (F , f G )) = Γ(V, H omX (F , f G )) = HomB(Γ(V, F ), Γ(V, f G )) ! . But we form f G by finding the module corresponding to H omY (f∗OX , G ) via f and ! ∼ ∼ (II, Ex 5.17). The way that we do this ensures that Γ(V, f G ) = Γ(U, H omY (f∗OX , G )) = HomA(B, Γ(U, G )) in a natural way. So we then have ! ∼ Γ(U, f∗H omX (F , f G )) = HomB(Γ(V, F ), HomA(B, Γ(U, G ))). On the other hand ∼ ∼ Γ(U, H omY (f∗F , G )) = HomA(Γ(U, f∗F ), Γ(U, G )) = HomA(Γ(V, F ), Γ(U, G ))

Letting M = Γ(U, G ) and N = Γ(V, F ) we merely need to find a map from HomB(N, HomA(B, M)) ∼ ∼ to HomA(N, M). But HomB(N, HomA(B, M)) = HomA(B ⊗B N, M) = HomA(N, M). If f ∈ HomB(N, HomA(B, M)), this map just sends f to the map that takes n ∈ N to f(n)(1) ∈ M. Clearly this map will commute with localization as will all identifications used to get to this point. Thus these will glue and we have a global iso- ! morphism. Alternately we could note that we have global map(s) f∗H omX (F , f G ) → 1 2 HARTSHORNE III, 6.10

! H om f∗F , f∗f G → H om f∗F , G → H om H om f∗F , G f∗OX ( ) Y ( ) Y ( Y ( ) in the style of the hint to the proof of (c) and then we would need to note that this map restricts down to the identification we have exhibited above.  (c) For each i ≥ 0, there is a natural map i F !G i F G φi : ExtX ( , f ) → ExtY (f∗ , ) i F !G i F !G Proof. First we will construct a map ExtX ( , f ) → ExtY (f∗ , f∗f ) as the hint sug- i F gests. Let us first note that ExtX ( , ·) is a universal delta functor. Now note that i F ExtY (f∗ , f∗·) is itself a delta functor since f∗ is exact (because f is finite). Finally note that we have a natural map HomX (F , ·) → HomY (f∗F , f∗·) (you can even factor it through f∗F , f∗· Homf∗OX ,Y ( )). These maps will commute with the long . Thus we we have the desired map by the defining property of a universal delta functor. ! The hint also suggests composing with a map from f∗f G to G . However, such a map always exists since we can define it (locally if desired) as is just H omY (f∗OX , G ) → G by evaluating our at 1f∗OX , also, this is the same as the global sections map of part i F !G i F G (b). We then compose to get a map ExtX ( , f ) → ExtY (f∗ , ) as desired.  (d) Now assume that X and Y are separated, Coh(X) has enough locally frees, and assume that f∗OX is locally free on Y (this is equivalent to saying f flat). Show that φi is an isomorphism for all i, all F coherent on X, and all G quasi-coherent on Y . ! ∼ Proof. First let us do i = 0. So we need to show that HomX (F , f G ) = HomY (f∗F , G ). But this is just global sections of the isomorphism from part (b). Next let us consider F i !G ∼ i G the case when = OX . So we need to show that ExtX (OX , f ) = ExtY (f∗OX , ) i !G ∼ i !G via φi. We have ExtX (OX , f ) = H (X, f ) by (III. 6.3c) and by (III, ex 4.1) we i !G ∼ i !G ∼ i !G have H (X, f ) = H (Y, f∗f ) = ExtY (OY , f∗f ). But, since f∗OX is locally free we i G ∼ i ∨ G ∼ i H G ∼ also have ExtY (f∗OX , ) = ExtY (OY , (f∗OX ) ⊗ ) = ExtY (OY , omY (f∗OX , )) = i !G !G H G ExtY (OY , f∗f ) since f∗f = omY (f∗OX , ) by (b). Both this map and the φi are induced by f∗ and (b). Carefully checking these maps will show that they are the same. F i F !G ∼ For the more general locally free we can do the same thing. Again we have ExtX ( , f ) = i ∨ ! ∼ i ∨ ! ∼ i ! H (X, (F ) ⊗ (f G )) = H (Y, f∗((F ) ⊗ (f G ))) = H (Y, f∗(H omX (F , f G )). Then we i F G ∼ i F ∨ G ∼ i H F G also have ExtY (f∗ , ) = H (Y, (f∗ ) ⊗ ) = H (Y, omY (f∗ , )). By part (b) these two are isomorphic naturally (since the sheaves are). This is the φi map for the same reason as above. In general we can write 0 → K → L → F → 0 with L locally free. We proceed by induction, so suppose it is true for all n <= i and all sheaves. We then have the following exact diagram i L !G / i K !G / i+1 F !G / i+1 L !G / i+1 K !G ExtX ( , f ) ExtX ( , f ) ExtX ( , f ) ExtX ( , f ) Ext ( , f )

φi φi φi+1 φi+1 φi+1      i L G / i K G / i+1 F G / i+1 L G / i+1 K G ExtY (f∗ , ) ExtY (f∗ , ) ExtY (f∗ , ) ExtY (f∗ , ) Ext (f∗ , ) where the second downward arrow is an isomorphism by induction and the first and fourth are isomorphisms since L is locally free. The five lemma guarantees that the middle (third) downward arrow is injective. But we are doing this for a general sheaf F which implies that HARTSHORNE III, 6.10 3

the fifth arrow is injective as well. Then the five lemma also tells us that the middle arrow is surjective and thus an isomorphism as desired.