Lemma 30.14.3. Let $A$ be a graded ring such that $A_0$ is Noetherian and $A$ is generated by finitely many elements of $A_1$ over $A_0$. Let $M$ be a finite graded $A$-module. Set $X = \text{Proj}(A)$ and let $\widetilde{M}$ be the quasi-coherent $\mathcal{O}_ X$-module on $X$ associated to $M$. The maps

$M_ n \longrightarrow \Gamma (X, \widetilde{M}(n))$

from Constructions, Lemma 27.10.3 are isomorphisms for all sufficiently large $n$.

Proof. Because $M$ is a finite $A$-module we see that $\widetilde{M}$ is a finite type $\mathcal{O}_ X$-module, i.e., a coherent $\mathcal{O}_ X$-module. Set $N = \bigoplus _{n \geq 0} \Gamma (X, \widetilde{M}(n))$. We have to show that the map $M \to N$ of graded $A$-modules is an isomorphism in all sufficiently large degrees. By Properties, Lemma 28.28.5 we have a canonical isomorphism $\widetilde{N} \to \widetilde{M}$ such that the induced maps $N_ n \to N_ n = \Gamma (X, \widetilde{M}(n))$ are the identity maps. Thus we have maps $\widetilde{M} \to \widetilde{N} \to \widetilde{M}$ such that for all $n$ the diagram

$\xymatrix{ M_ n \ar[d] \ar[r] & N_ n \ar[d] \ar@{=}[rd] \\ \Gamma (X, \widetilde{M}(n)) \ar[r] & \Gamma (X, \widetilde{N}(n)) \ar[r]^{\cong } & \Gamma (X, \widetilde{M}(n)) }$

is commutative. This means that the composition

$M_ n \to \Gamma (X, \widetilde{M}(n)) \to \Gamma (X, \widetilde{N}(n)) \to \Gamma (X, \widetilde{M}(n)$

is equal to the canonical map $M_ n \to \Gamma (X, \widetilde{M}(n)$. Clearly this implies that the composition $\widetilde{M} \to \widetilde{N} \to \widetilde{M}$ is the identity. Hence $\widetilde{M} \to \widetilde{N}$ is an isomorphism. Let $K = \mathop{\mathrm{Ker}}(M \to N)$ and $Q = \mathop{\mathrm{Coker}}(M \to N)$. Recall that the functor $M \mapsto \widetilde{M}$ is exact, see Constructions, Lemma 27.8.4. Hence we see that $\widetilde{K} = 0$ and $\widetilde{Q} = 0$. Recall that $A$ is a Noetherian ring, $M$ is a finitely generated $A$-module, and $N$ is a graded $A$-module such that $N' = \bigoplus _{n \geq 0} N$ is finitely generated by the last part of Lemma 30.14.2. Hence $K' = \bigoplus _{n \geq 0} N_ n$ and $Q' = \bigoplus _{n \geq 0} Q_ n$ are finite $A$-modules. Observe that $\widetilde{Q} = \widetilde{Q'}$ and $\widetilde{K} = \widetilde{K'}$. Thus to finish the proof it suffices to show that a finite $A$-module $K$ with $\widetilde{K} = 0$ has only finitely many nonzero homogeneous parts $K_ d$ with $d \geq 0$. To do this, let $x_1, \ldots , x_ r \in K$ be homogeneous generators say sitting in degrees $d_1, \ldots , d_ r$. Let $f_1, \ldots , f_ n \in A_1$ be elements generating $A$ over $A_0$. For each $i$ and $j$ there exists an $n_{ij} \geq 0$ such that $f_ i^{n_{ij}} x_ j = 0$ in $K_{d_ j + n_{ij}}$: if not then $x_ i/f_ i^{d_ i} \in K_{(f_ i)}$ would not be zero, i.e., $\widetilde{K}$ would not be zero. Then we see that $K_ d$ is zero for $d > \max _ j(d_ j + \sum _ i n_{ij})$ as every element of $K_ d$ is a sum of terms where each term is a monomials in the $f_ i$ times one of the $x_ j$ of total degree $d$. $\square$

Comment #6604 by Yuto Masamura on

I have a problem reading "we have a canonical isomorphism $\widetilde N\to\widetilde M$ such that $M_n\to N_n=\Gamma(X,\widetilde M(n))$ is the canonical map." How the map $M_n\to N_n$ is obtained from the isomorphism $\widetilde N\to\widetilde M$? (I think we want to say that the induced map $\widetilde M\to\widetilde N$ by the map $M\to N$ is equal to the inverse of the isomorphism $\widetilde N\to \widetilde M$, but...)

Comment #6605 by Yuto Masamura on

@6604 Sorry, my typo: "How the map ... is obtained" should be "How is the map ... obtained".

Comment #6668 by WhatJiaranEatsTonight on

I have some confusion about the summand. By lemma 28.28.5, the direct sum of $N$ is over all the integers while in the proof here we only take nonnegative parts.

But obviously it is harmful because $A$ is generated by the degree $1$ part.

The problem is whether it holds for $n<<0$? Obviously, $M_n=0$ for $n\ll 0$. But does $\Gamma(X,\tilde{M}(n))$ vanish for $n$ small enough?

Comment #6669 by WhatJiaranEatsTonight on

I mean it is harmless. Sorry for my typo.

Comment #6850 by on

OK, I fixed the problem with the arrow pointing in the wrong direction, see here. I fixed the problem with the stuff in negative degrees here. Sometimes I wish that Proj had never been invented. I think it is much better to think about proper varieties endowed with ample invertible modules...

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