Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)?

share|improve this question

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 $n>1$

share|improve this answer
    
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 Lampret Mar 3 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 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 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 at 13:16
1  
Yes, that's right! –  Martin Brandenburg Mar 4 at 18:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.