I am lacking the knowledge of multi-linear algebra while I come up with the following problems.

Suppose $k$ is a field of characteristic $p>0$. Let $E$ and $F$ be two $k$-vector spaces of dimension $r$ and $s$ respectively.

  1. How to describe $\bigwedge^n(E \otimes F)$? Direct sum? Filtration?

  2. I would like to have explicit answers to the previous question for $n =2, 3$ or even $n = 4$. If $\dim F = 1$, then $\bigwedge^n(E \otimes F) \simeq \bigwedge^n E \otimes \operatorname{T}^n F \simeq \bigwedge^n E \otimes \operatorname{Sym}^n F$. How about $\dim F =2$?

  3. Do we have natural map $\bigwedge^n(E \otimes F) \to \bigwedge^n(E) \otimes \operatorname{Sym}^n(F)$?

  4. I don't like to confuse the symmetric tensors with symmetric algebra, and alternating tensor with exterior powers. So in my definition, $\operatorname{Sym}^n(E)$ and $\bigwedge^n(E)$ are all quotients of $\operatorname{T}^n(E)$, while the space of symmetric tensors and alternating tensors are subspaces of $\operatorname{T}^n(E)$.

  5. It seems that $\operatorname{Sym}(E^\vee)$ is naturally isomorphic to $\Gamma(E)^\vee$, where $\Gamma(E)$ is the divided power algebra. What is correct definition of $\Gamma(E)$? quotient or sub of something? What is the corresponding sub or quotient?

  6. How does taking dual behave under these constructions?

  7. What is the reference for such problems?

Problem 3 solved. Consider the natural map $\operatorname{T}^n(E \otimes F) \xrightarrow{\sim} (\operatorname{T}^n E) \otimes (\operatorname{T}^n F) \to (\bigwedge^n E) \otimes (\operatorname{Sym}^n F)$. Then we can easily see that $\ker\big(\operatorname{T}^n (E \otimes F)\to \bigwedge^n (E \otimes F)\big)$ goes to zero under this natural map, hence it factors through $\bigwedge^n (E \otimes F)$.

3'. Is $(\bigwedge^n E) \otimes (\operatorname{Sym}^n F)$ a sub/quotient/subquotient of $\wedge^n (E\otimes F)$.

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    $\begingroup$ 3. Yes: it maps $\left( e_1 \otimes f_1 \right) \wedge \left( e_2 \otimes f_2 \right) \wedge \cdots \left( e_n \otimes f_n \right)$ to $\left( e_1 \wedge e_2 \wedge \cdots e_n \right) \otimes \left( f_1 f_2 \cdots f_n\right) $. The proof of well-definedness is not too hard: First check that it is well-defined as a map from the tensor power; then show that it factors through the exterior power because if two adjacent tensorands are equal, the output is zero (caveat: the tensorands might not be pure tensors). $\endgroup$ – darij grinberg Jun 28 '17 at 10:45
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    $\begingroup$ 6. As you probably know, for finitely-generated free $k$-modules $V$, there is a canonical isomorphism $\left(\wedge^n V\right)^\ast \to \wedge^n \left(V^\ast\right)$ that comes from the non-degenerate bilinear form $\left(\wedge^n V\right) \times \left(\wedge^n \left(V^\ast\right)\right) \to k, \ \left(v_1 \wedge v_2 \wedge \cdots \wedge v_n, f_1 \wedge f_2 \wedge \cdots \wedge f_n\right) \mapsto \det \left( \left( f_i\left(v_j\right) \right)_{1\leq i\leq n,\ 1\leq j\leq n}\right)$. In the same situation, there is also a canonical ... $\endgroup$ – darij grinberg Jun 28 '17 at 14:40
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    $\begingroup$ ... isomorphism $\left(D^n V\right)^\ast \to \operatorname{Sym}^n \left(V^\ast\right)$, where $D^n V$ stands for the $n$-th divided-powers module of $V$ (which is isomorphic to the $n$-th symmetric-tensors module of $V$, i.e., the $S_n$-invariant subspace of $V^{\otimes n}$). $\endgroup$ – darij grinberg Jun 28 '17 at 14:41
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    $\begingroup$ 4. If $V$ is a finitely-generated free $k$-module, then the $n$-th "alternating-tensors" module of $V$ is canonically isomorphic to the $n$-th exterior power $\wedge^n V$. In fact, the isomorphism from $\wedge^n V$ to the $n$-th "alternating-tensors" module simply sends each $v_1 \wedge v_2 \wedge \cdots \wedge v_n$ to $\sum\limits_{\sigma \in S_n} \left(-1\right)^\sigma v_{\sigma\left(1\right)} \otimes v_{\sigma\left(2\right)} \otimes \cdots \otimes v_{\sigma\left(n\right)}$. But of course, no such isomorphism exists between "symmetric-tensors" and the symmetric power. $\endgroup$ – darij grinberg Jun 28 '17 at 14:43
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    $\begingroup$ @darijgrinberg -- That's a lot of comment! I think it should be an answer. $\endgroup$ – mr_e_man Dec 15 '18 at 6:17

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