# 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.

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### 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….