The Stacks project

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 $N = \bigoplus _{n \in \mathbf{Z}} \Gamma (X, \widetilde{M}(n))$. 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 $N' = \bigoplus _{n \geq 0} N_ n$ is finitely generated by the last part of Lemma 30.14.2. Hence $K' = \bigoplus _{n \geq 0} K_ n$ 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 $d$. $\square$

Comments (8)

Comment #6604 by Yuto Masamura on

I have a problem reading "we have a canonical isomorphism such that is the canonical map." How the map is obtained from the isomorphism ? (I think we want to say that the induced map by the map is equal to the inverse of the isomorphism , 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 is over all the integers while in the proof here we only take nonnegative parts.

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

The problem is whether it holds for ? Obviously, for . But does vanish for 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...

Comment #7920 by y1k on

There are two in the sentence containing the non-diagram equation.

Comment #7921 by y2k on

Shall we make the following changes: 1. , the definition of in the 2nd sentence of the proof, changes to ; 2. changes to ; 3. changes to .

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