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Let $E$ be a free module, we define the $r$-th divided power as the dual of the symmetric power $D_r(E):=(S_r(E^*))^*$. For every $u \in E$ we can define its $r$-th divided power $u^{(r)} \in D_r$ by the formula: $$ \left(\sum_{i=1}^nu_ie_i \right)^{(r)}= \sum_{p_1+ \cdots p_n=r} u_1^{p_1} \cdots u_n^{p_n} e_1^{(p_1)}\cdots e_n^{(p_n)} .$$ Now I read that the divided powers have the following property: $u^{(p)}u^{(q)} $= $p+q \choose p$ $u^{(p+q)}$. I tried but this formula doesn't work just for $p=q=1$. Infact $u^{(1)}u^{(1)}=u_1^2e_1^2+u_2^2e_2^2+2u_1u_2e_1e_2 $ but $u^{(2)}= u_1^2e_1^2+u_2^2e_2^2+u_1u_2e_1e_2 $ and $2 \choose 1 $ $=2$... How have I to read this formula? Thank you!

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  • $\begingroup$ $e_1^2 = 2 e_1^{(2)}$. These parentheses aren't just there to keep terms together. $\endgroup$ – darij grinberg Jun 12 '15 at 11:45
  • $\begingroup$ ... i think $e_1^2=\frac{1}{2}e_1^{(2)}$ $\endgroup$ – user137921 Jun 12 '15 at 12:59
  • $\begingroup$ No. It's "divided powers", not "multiplied powers". $\endgroup$ – darij grinberg Jun 12 '15 at 13:19

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