Exercise 26, § 4

Let $A$ be a Noetherian ring and $M$ a finitely generated $A$-module. For the symmetric algebra $S(M)$ to be a factorial domain, it is necessary and sufficient that $A$ be a factorial domain and that the $S^n(M)$ be reflexive $A$-modules. (To see that the condition is sufficient, observe first that, if $T = A - \{0\}$, $S(M)$ is identified with a subring of $T^{-1}S(M)$; then show that every extremal element $p$ of $A$ is extremal in $S(M)$, by reducing it to proving that, if $p$ divides a product $xy$ of two homogeneous elements in $S(M)$, it divides one of them; finally, apply Proposition 3 of § 3, no. 4.)