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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)?

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1 Answer 1

up vote 2 down vote accepted

Use the following:

  1. $S^n(M \oplus N) = \bigoplus_{p+q=n} S^p(M) \otimes S^q(N)$, likewise for $\Lambda^n$.

  2. $S^n(R/I) = T^n(R/I)$

  3. $\Lambda^n(R/I)=0$ for $n>1$

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Thus $S^n(M_1\oplus\ldots\oplus M_k)\cong \bigoplus_{i_1+\ldots+i_k=n} S^{i_1}(M_1)\otimes\ldots\otimes S^{i_k}(M_k)$ and $\Lambda^n(M_1\oplus\ldots\oplus M_k)\cong \bigoplus_{i_1+\ldots+i_k=n} \Lambda^{i_1}(M_1)\otimes\ldots\otimes\Lambda^{i_k}(M_k)$, so (confirm) $$S^nM\cong R^{\binom{r+n-1}{n}}\!\oplus\bigoplus_{k=1}^n \bigoplus_{1\leq i_1\leq\ldots\leq i_k\leq s} (R/R\gcd(a_{i_1},\ldots,a_{i_k}))^{\binom{r+n-k-1}{n-k}}\text{ and }$$ $$\Lambda^nM\cong R^{\binom{r}{n}}\!\oplus\bigoplus_{k=1}^n \bigoplus_{1\leq i_1<\ldots<i_k\leq s} (R/R\gcd(a_{i_1},\ldots,a_{i_k}))^{\binom{r}{n-k}}?$$ –  Leon Mar 3 '14 at 4:49
I haven't checked the formulas (and also won't do it). I don't think that they will be useful. It is really more important to know the general techniques how to compute symmetric and exterior powers. –  Martin Brandenburg Mar 3 '14 at 11:57
In general $S^2((R/I)^3) \cong (R/I)^6$. There are $6$ partitions $2=a+b+c$, and each summand gives you $R/I$. –  Martin Brandenburg Mar 3 '14 at 12:56
Explicitly, $R^6 \cong S^2(R^3)$ corresponds to the basis $\{x_1^2,x_2^2,x_3^2,x_1 x_2,x_1 x_3,x_2 x_3\}$ of $S^2(R^3)$ when $\{x_1,x_2,x_3\}$ is a basis of $R^3$. –  Martin Brandenburg Mar 3 '14 at 13:16
Yes, that's right! –  Martin Brandenburg Mar 4 '14 at 18:09

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