# Serre Duality and Tor Stuff

This post contains mostly statements of theorems, and essentially no proofs.  I guess I’m getting lazy.

Thm(Serre): Let $X/k$ be smooth and proper of dim n (smooth = geometrically regular and flat).  Then if $E$ is locally free sheaf of finite rank on $X$, then there exists natural perfect parings

$H^i(X,E) \times H^{n-1}(X, E^\vee\otimes \Omega^n_{X/k}) \to k$

Ogus proved this in the case $X/k$ is projective.  In fact Ogus went on to discuss duality in the derived category.  Along the way this definition came up: a complex of R-modules is strictly perfect if its bounded and all its terms are projective and finitely generated.  Then perfect is defined to be quasi isomrophic to a strictly perfect complex.

Ogus talked about a dualzing sheaf and a trace map much like Harthshrone and proved the following thm called the general duality thm:

Thm: Let $X/S$ ($S = \mbox{Spec} R$, R Noeth)  be projective, then there exists a canonical pair $(\omega^\bullet_{X/S}, \mbox{tr}_{X/S})$ where the first term is in the derived category $D^+(X/S)$ of quasicoherent complexes with coherent cohomology, and the trace map is from $R\Gamma(X, \omega^\bullet_{X/S}) \to R$.  This pair has the property that for all $E \in D^b(X)$ the composition

$R\hom(E, \omega^\bullet) \to R\hom(R\Gamma(E), R\Gamma(\omega^\bullet)) \to R\hom(R\Gamma(E), R)$

is an isomoprhism.

Later, Ogus proved that if $X/S$ is smooth of dim d then $\omega_{X/S} \cong \Omega^d_{X/S}$.

### Tor Stuff

The notation is A is a ring, M,N are A-modules and for $P^\bullet \to N$ a projective resolution $Tor_i(M, N ) = Tor^{-i} (M, N ) = h^{-1}(M \otimes P^\bullet)$.  A handy fact is that flat modules N are $Tor(M, - )$ acyclic.  And Tor can be computed by taking a projective resolution of either term.  Here’s a fun fact:

Prop: $R$ is a Noeth. local ring and $M$ is fin. gen. R -module.  TFAE

1.  M is free
2. M is projective
3. M is flat
4. $Tor_1(M, R/m = k) = 0$

proof: 1 implies 2 implies 3 implies 4 is straightforward.  For the last step, choose a basis to get an isomorhpism

$k^d \cong M \otimes k = M/mM$

Nakayama’s lemma says that lifting the basis in M/mM to M gives a generating set, so there is a short exact sequence

$0 \to K \to R^d \to M \to 0$

Tensor with the residue field to get a long exact sequence

$\to 0 = Tor_1(M, k) \to K\otimes k \to k^d \cong M\otimes k \to 0$

It follows that $K \otimes k = 0$ and Nakayama’s lemma says $K = 0$.QED

Here’s another result

Thm: If $X/R$ is projective and $E \in Coh(X)$ then $R\Gamma(X,E) \in D^b(R)$ is a perfect complex and E is flat over R.

This result is used to prove basic results usually termed under cohomology with base change.  A nice treatment is done in Mumford’s book ‘Abelian Varieties.’  I think at some point I’ll put up a post about it.