Projectivity of exterior powers/algebras?

Question

Suppose that $R$ is a commutative ring and that $P$ is a projective $R$-module. Is it then true that as an $R$-module, $\bigwedge_R(P)$ and/or each exterior power $\bigwedge_R^n(P)$ is projective?

Answer 1

Yes. If $F$ is free, the exterior powers of $F$ are free. It is even well-known how to write down a basis: If $B$ is an ordered basis of $F$, then $\{b_1 \wedge \cdots \wedge b_n : \{b_1 < \cdots < b_n \} \subseteq B\}$ is a basis of $\Lambda^n(F)$.

If $P$ is projective, this means there is a free module $F$ and a split epimorphism $F \twoheadrightarrow P$. Since $\Lambda^n$ is a functor, we get a split epimorphism $\Lambda^n(F) \twoheadrightarrow \Lambda^n(P)$, and $\Lambda^n(F)$ is free, Hence, the exterior powers of $P$ are projectve.

The same is true for the direct sum of all exterior powers. The same proof works. Alternatively, use that free (resp. projective) modules are closed under direct sums.