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For vector spaces $V, W$ we have the following canonical isomorphisms, where $V^*$ denotes the dual space of $V$: $$(V\otimes W)^*\cong V^*\otimes W^*$$ $$V^*\otimes W\cong \operatorname{Hom}(V, W)$$ Do these still hold true for modules $V, W$ over a commutative ring $A$?

I'm going to take a guess and say that the answer is probably yes, but apart from the case that $V, W$ are free, I was unable to prove this.

mxian
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First of all, it does not always hold that $V^*\otimes W\cong \operatorname{Hom}(V, W)$, even for vector spaces! See this post for details. One general version of this that does hold in general is the tensor-hom adjunction $$ \mathrm{Hom}(V\otimes W,X)\;\cong\;\mathrm{Hom}(V,\mathrm{Hom}(W,X)). $$ This general fact gives us an interest result regarding $(V \otimes W)^*$. Letting $X$ be the base ring $R$ yields $$ (V \otimes W)^* = \mathrm{Hom}(V \otimes W,R) \cong \mathrm{Hom}(V,\mathrm{Hom}(W,R)) = \mathrm{Hom}(V,W^*). $$ From the first statement, we see that it does not necessarily hold that $\mathrm{Hom}(V,W^*) \cong V^* \otimes W^*$.

Ben Grossmann
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  • Thank you for clearing that up! I saw these two isomorphisms in a linear algebra textbook and overlooked the finiteness assumption. The adjunction between tensor and hom is more or less just a restatement of the universal property of the tensor product, right? – mxian Feb 18 '22 at 22:17
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    @mxian yes, it's a restatement of the universal property of the tensor product. This is due to the fact that homomorphisms $V \to \operatorname{Hom}(W,X)$ are naturally isomorphic to bilinear maps $V\times W\to X$. – Lukas Heger Feb 18 '22 at 22:42
  • @Lukas That comment actually made a few things click for me, thank you. So the isomorphism from $V\times W \to X$ to $\mathrm{Hom}(V,\mathrm{Hom}(W,X))$ should be the Currying map, is that right? – Ben Grossmann Feb 18 '22 at 23:18
  • @BenGrossmann exactly, it‘s the Currying map. – Lukas Heger Feb 19 '22 at 10:59