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user91132
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The fact about the projectivity of $S$ is also known as the "long Schanuel's Lemma".

In the case you are interested in, the "short Schanuel's Lemma", as explained here

http://en.wikipedia.org/wiki/Schanuel's_lemma

suffices. You have two short exact sequences $0 \to M \to N \to M/N \to 0$$0 \to M \to N \to N/M \to 0$ and $0 \to P_1 \to P_0 \to M/N \to 0$$0 \to P_1 \to P_0 \to N/M \to 0$, where $P_0$ and $P_1$ are projective modules because the projective dimension of $M/N$$N/M$ is $\leq 1$. Therefore $N \oplus P_1$ is isomorphic to $M \oplus P_0$ by the Lemma. So $M$ is projective, being a direct sum of the projective module $N \oplus P_1$.

The fact about the projectivity of $S$ is also known as the "long Schanuel's Lemma".

In the case you are interested in, the "short Schanuel's Lemma", as explained here

http://en.wikipedia.org/wiki/Schanuel's_lemma

suffices. You have two short exact sequences $0 \to M \to N \to M/N \to 0$ and $0 \to P_1 \to P_0 \to M/N \to 0$, where $P_0$ and $P_1$ are projective modules because the projective dimension of $M/N$ is $\leq 1$. Therefore $N \oplus P_1$ is isomorphic to $M \oplus P_0$ by the Lemma. So $M$ is projective, being a direct sum of the projective module $N \oplus P_1$.

The fact about the projectivity of $S$ is also known as the "long Schanuel's Lemma".

In the case you are interested in, the "short Schanuel's Lemma", as explained here

http://en.wikipedia.org/wiki/Schanuel's_lemma

suffices. You have two short exact sequences $0 \to M \to N \to N/M \to 0$ and $0 \to P_1 \to P_0 \to N/M \to 0$, where $P_0$ and $P_1$ are projective modules because the projective dimension of $N/M$ is $\leq 1$. Therefore $N \oplus P_1$ is isomorphic to $M \oplus P_0$ by the Lemma. So $M$ is projective, being a direct sum of the projective module $N \oplus P_1$.

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user91132
  • 3.7k
  • 1
  • 25
  • 26

The fact about the projectivity of $S$ is also known as the "long Schanuel's Lemma".

In the case you are interested in, the "short Schanuel's Lemma", as explained here

http://en.wikipedia.org/wiki/Schanuel's_lemma

suffices. You have two short exact sequences $0 \to M \to N \to M/N \to 0$ and $0 \to P_1 \to P_0 \to M/N \to 0$, where $P_0$ and $P_1$ are projective modules because the projective dimension of $M/N$ is $\leq 1$. Therefore $N \oplus P_1$ is isomorphic to $M \oplus P_0$ by the Lemma. So $M$ is projective, being a direct sum of the projective module $N \oplus P_1$.