
Lemma 10.70.1. Let $R$ be a ring. Let $M$ be an $R$-module.

1. There exists an exact complex

$\ldots \to F_2 \to F_1 \to F_0 \to M \to 0.$

with $F_ i$ free $R$-modules.

2. If $R$ is Noetherian and $M$ finite over $R$, then we can choose the complex such that $F_ i$ is finite free. In other words, we can find an exact complex

$\ldots \to R^{\oplus n_2} \to R^{\oplus n_1} \to R^{\oplus n_0} \to M \to 0.$

Proof. Let us explain only the Noetherian case. As a first step choose a surjection $R^{n_0} \to M$. Then having constructed an exact complex of length $e$ we simply choose a surjection $R^{n_{e + 1}} \to \mathop{\mathrm{Ker}}(R^{n_ e} \to R^{n_{e-1}})$ which is possible because $R$ is Noetherian. $\square$

Comment #139 by on

In 2, replace "finite $R$" by "finite over $R$". Moreover, 2 seems to have too much hypotheses: If $R$ is coherent, then $M$ is finitely presented if and only if there is such an exact complex such that each $F_i$ is finite free. The proof is the same as above, with "because $R$ is Noetherian" replaced by "by 05CW and 05CX". (Unfortunalty, this needs the notion of and results about coherence appearing only in 05CU. It should probably be put after 05CX. The noetherian statement follows then from 05CY.)

Comment #141 by on

Maybe the material about coherent modules over coherent rings should be put in the chapter More on Algebra? Somewhere in the section on pseudo-coherent modules? In that language it says that a coherent module over a coherent ring is pseudo-coherent. In any case I think we leave this lemma like it is now and we add an additional one to cover the coherent case. OK?

Comment #142 by on

Oh, I have not seen that chapter. So, the coherent case might be incorporated in 066E.

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