Symmetric and exterior powers of a projective (flat) module are projective (flat)

Question

Assume that $R$ is a commutative ring with unity and $P$ a projective (flat) $R$-module. Why $\mathrm{Sym}^n(P)$ and $\Lambda^n(P)$ are projective (flat) for every $n$?

Answer 1

Here's one approach using the fact that projective modules are direct summands of free modules. Suppose $P\oplus Q\cong R^k$. Then $$R^{k^n}=(P\oplus Q)^{\otimes n}=P^{\otimes n}\oplus\cdots ,$$ so $P^{\otimes n}$ is projective. But then notice that $\Lambda^n(P)$ and $\mathrm{Sym}^n(P)$ are both direct summands of $P^{\otimes n}$.

Edit: As Darij Grinburg points out, and I only just now noticed, this only works for some rings $R$, such as $\mathbb Q$ algebras.