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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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|cite|improve this question
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|cite|improve this answer
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
Ok, so $R=K[t]$ and $M\cong R^2\oplus (R/Rt)^3$ should imply $S^2M\cong R^3\!\oplus\!(R/Rt)^{12}$ and $\lambda^2M\cong R\!\oplus\!(R/Rt)^9$, right? – Leon Mar 4 '14 at 17:55
Yes, that's right! – Martin Brandenburg Mar 4 '14 at 18:09

Your Answer


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.