Lemma 10.109.3. Let $R$ be a ring. Suppose that $M$ is an $R$-module of projective dimension $d$. Suppose that $F_ e \to F_{e-1} \to \ldots \to F_0 \to M \to 0$ is exact with $F_ i$ projective and $e \geq d - 1$. Then the kernel of $F_ e \to F_{e-1}$ is projective (or the kernel of $F_0 \to M$ is projective in case $e = 0$).

**Proof.**
We prove this by induction on $d$. If $d = 0$, then $M$ is projective. In this case there is a splitting $F_0 = \mathop{\mathrm{Ker}}(F_0 \to M) \oplus M$, and hence $\mathop{\mathrm{Ker}}(F_0 \to M)$ is projective. This finishes the proof if $e = 0$, and if $e > 0$, then replacing $M$ by $\mathop{\mathrm{Ker}}(F_0 \to M)$ we decrease $e$.

Next assume $d > 0$. Let $0 \to P_ d \to P_{d-1} \to \ldots \to P_0 \to M \to 0$ be a minimal length finite resolution with $P_ i$ projective. According to Schanuel's Lemma 10.109.1 we have $P_0 \oplus \mathop{\mathrm{Ker}}(F_0 \to M) \cong F_0 \oplus \mathop{\mathrm{Ker}}(P_0 \to M)$. This proves the case $d = 1$, $e = 0$, because then the right hand side is $F_0 \oplus P_1$ which is projective. Hence now we may assume $e > 0$. The module $F_0 \oplus \mathop{\mathrm{Ker}}(P_0 \to M)$ has the finite projective resolution

of length $d - 1$. By induction applied to the exact sequence

of length $e - 1$ we conclude $\mathop{\mathrm{Ker}}(F_ e \to F_{e - 1})$ is projective (if $e \geq 2$) or that $\mathop{\mathrm{Ker}}(F_1 \oplus P_0 \to F_0 \oplus P_0)$ is projective. This implies the lemma. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: