The Stacks project

Lemma 10.109.8. Let $R$ be a ring. Let $M$ be an $R$-module. Let $n \geq 0$. The following are equivalent

  1. $M$ has projective dimension $\leq n$,

  2. $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N) = 0$ for all $R$-modules $N$ and all $i \geq n + 1$, and

  3. $\mathop{\mathrm{Ext}}\nolimits ^{n + 1}_ R(M, N) = 0$ for all $R$-modules $N$.

Proof. Assume (1). Choose a free resolution $F_\bullet \to M$ of $M$. Denote $d_ e : F_ e \to F_{e - 1}$. By Lemma 10.109.3 we see that $P_ e = \mathop{\mathrm{Ker}}(d_ e)$ is projective for $e \geq n - 1$. This implies that $F_ e \cong P_ e \oplus P_{e - 1}$ for $e \geq n$ where $d_ e$ maps the summand $P_{e - 1}$ isomorphically to $P_{e - 1}$ in $F_{e - 1}$. Hence, for any $R$-module $N$ the complex $\mathop{\mathrm{Hom}}\nolimits _ R(F_\bullet , N)$ is split exact in degrees $\geq n + 1$. Whence (2) holds. The implication (2) $\Rightarrow $ (3) is trivial.

Assume (3) holds. If $n = 0$ then $M$ is projective by Lemma 10.77.2 and we see that (1) holds. If $n > 0$ choose a free $R$-module $F$ and a surjection $F \to M$ with kernel $K$. By Lemma 10.71.7 and the vanishing of $\mathop{\mathrm{Ext}}\nolimits _ R^ i(F, N)$ for all $i > 0$ by part (1) we see that $\mathop{\mathrm{Ext}}\nolimits _ R^ n(K, N) = 0$ for all $R$-modules $N$. Hence by induction we see that $K$ has projective dimension $\leq n - 1$. Then $M$ has projective dimension $\leq n$ as any finite projective resolution of $K$ gives a projective resolution of length one more for $M$ by adding $F$ to the front. $\square$

Comments (1)

Comment #690 by Keenan Kidwell on

In the second paragraph of the proof, there are two instances where "" should be "."

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