Fun Comm. Alg. Result

 

The representability of \mathbb{P}E was finished around the end of october.  Then Ogus talked a little bit about some generalities of \mbox{Proj} G for a graded ring G.  For example, there is not an equivalence of categories between Graded modules and qcoh sheaves of modules on \mbox{Proj} G, but before that Ogus proved the following result.

Thm: A a ring, M an A module and S = \mbox{Spec} A.

  1.  M is finitely generated iff S has an affine open cover \{U_i\} s.t. \widetilde{M}(U_i) is finitely generated over \mathcal{O}_S(U_i).
  2. Same as 1 replacing ‘finitely generated’ with ‘finitely presented.’
  3. M is finitely generated and projective iff \widetilde{M} is locally free of finite rank.

——————————————————


Proof of 1

One direction is clear: if M is finitely generated, then M \otimes_A A_a is finitely generated over A_a, for a \in A.  For the converse, WLOG, replace the covering with a finite covering of distinguished affines: \{ \mbox{Spec} A_{a_i}\}.  That is we have 

M_{a_i} finitely generated (say by) m_{i1}, \dotsc, m_{in}

Multiplying by powers of a_i we can assume m_{ij} \in M.  Then since we work with finitely many affines and finitely many generators, there is a finite set S and a map 

A^S \to M

Such that the map become a surjection when localized at the a_i, i.e. its locally surjective, and in particular surjective when localized at primes, so 

\widetilde{A^S} \to \widetilde{M}

is a surjective map of sheaves.  By the exactness of \sim on A modules, we get A^S \to M is surjective.  QED.

Proof of 2

Again, the same direction as before is clear.  For the converse the following lemma is useful

Lemma: Let N be a finitely presented A module.  And S and finite set of generators.  Then \ker (A^S \to N) is finitely generated.

pf: Say k(f) \to A^T \xrightarrow{f} \to N is a finite presentation for N.  As A^S is free, and A^T \to N is surjective, this map lifts to factor through A^S \to N, also using the universal property of kernel I get the following commutative diagram:

\begin{array}{ccccccccc} 0 & \to & k(f) & \to & A^T & \to & M & \to & 0 \\ \mbox{} & \mbox{} & \downarrow & \mbox{} & \downarrow & \mbox{} & || & \mbox{} & \mbox{} \\ 0 & \to & k(g) & \to & A^S & \to & M & \to & 0 \\ \mbox{} & \mbox{} & \downarrow & \mbox{} & \downarrow & \mbox{} & \downarrow & \mbox{} & \mbox{} \\ \mbox{} & \mbox{} & \mbox{Cok1} & \to & \mbox{Cok2} & \to & 0 & \mbox{} & \mbox{} \end{array}  

The snake lemma says there is an exact sequence 0 = \ker(M \xrightarrow{id} M) \to \mbox{Cok1} \to \mbox{Cok2} \to \mbox{Coker}(M \xrightarrow{id} M) = 0.  In particular, \mbox{Cok1} \cong \mbox{Cok2} and \mbox{Cok2} if finitely generated as there is a surjection A^S \to \mbox{Cok2}.  Therefore, the first and last terms of k(f) \to k(g) \to \mbox{Cok1} are finitely generated, hence so is the middle term. QED.

Now to prove 2 the other part of 2, say M is locally finitely presented, note that by 1 M is finitely generated,  hence there is an exact sequence 0 \to K = A^{(I)} \to A^S \to M \to 0.  It remains to show K is finitely generated.  Localizing the exact sequence I get (because localization is an exact functor)

 0 \to K_{a_i} \to (A^S)_{a_i} \to M_{a_i} \to 0

So by 1, it suffices to show K_{a_i} is finitely generated.  And this follows from the lemma as M_{a_i} is assumed to be finitely presented. QED.

Proof of 3

Say first that \widetilde{M} is locally free of finite rank.  The required statement is that if 0 \to E' \to E \to E'' \to 0 is exact then 0\to \hom(M,E') \to \hom(M,E) \to \hom(M,E'') \to 0 is exact.  By lemma one in this post, it suffices to show \mbox{{\it Hom}}(M,E) is qcoh, and for this it suffices to show the natural map

\hom_A(M,E)_a \cong \hom_{A_a}(M_a,E_a)

is an isomorphism.  This holds if M is free (because localization commutes with direct sums).  Using the 5-lemma and the previous observation, it can be shown that the required isomorphism holds also if M is finitely presented.  Now by assumption, M_{a_i} is locally free, hence finitely presented, hence by 2, M is finitely presented.  So this direction is done.

For the converse, it is enough to prove that M_P is locally free of finite rank, as M_P will be finitely generated and also projective.  So we can reduce this case to proving the following claim: if A is a local ring and M is finitely generated and projective, then M is free

pf: for some finite S we have an exact sequence 0 \to K \to A^S \to M \to 0.  Now K\otimes_A A/m = 0, that is, M/mM is a finite vector space.  Subclaim: a basis for M/mM lifts to a generating set for M (pf: let \{m_i \} be arbitrary lifts of some basis of M/mM, let N \subset M be the submodule they generate.  Then M = N + mM so by Nakayama’s lemma says M = N).  Now, as M is projective, there is a lift M \to A^s, so the short exact sequence above splits, hence

A^S \cong K \oplus M

which shows K is finitely generated (since there is a surjection A^S \to K), also K/mK = 0 so applying Nakayama again gives K = 0. QED.

 

The next big topic Ogus talked about was fibers a morphism and the valuation criterion for separatedness and properness.  The treatment was not radically different (although it was more general) than what is found in Hartshorne, so I don’t plan on putting that stuff here.  Next big topic: differentials…. 

Advertisements

About this entry