Let $R$ be a PID and $M\cong R^r\!\oplus\bigoplus_{i=1}^s\!R/Ra_i$. Denote the tensor, symmetric, exterior power of $M$ by $T^nM=\bigotimes_{k=1}^nM$ and $S^nM= T^nM/\langle x_{\sigma1}\!\otimes\cdots\!\otimes x_{\sigma n}\!-\!x_1\!\otimes\cdots\!\otimes x_n; x_1,\ldots,x_n\!\in\!M, \sigma\!\in\!S_n\rangle$ and $\Lambda^nM=T^nM/\langle x_1\!\otimes\cdots\!\otimes x_n; x_1,\ldots,x_n\!\in\!M; x_i\!=\!x_j\text{ for some }i\!\neq\!j\rangle$.
By the rules $R^r\!\otimes\!A\cong A^r$, $A\!\otimes\!(B\!\oplus\!C)\cong (A\!\otimes\!B)\!\oplus\!(A\!\otimes\!C)$, $R/I\otimes R/J\cong R/\langle I,J\rangle$, we conclude $$T^n\!M\cong R^{r^n}\!\oplus\bigoplus_{k=1}^n \bigoplus_{1\leq i_1,\ldots,i_k\leq s} (R/R\gcd(a_{i_1},\ldots,a_{i_k}))^{r^{n-k}}.$$ What is the formula for $S^nM$ and $\Lambda^nM$ (free rank and torsion coefficients)?
Use the following:
$S^n(M \oplus N) = \bigoplus_{p+q=n} S^p(M) \otimes S^q(N)$, likewise for $\Lambda^n$.
$S^n(R/I) = T^n(R/I)$
$\Lambda^n(R/I)=0$ for $n>1$