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Let $R$ be a commutative ring with $1$. For all $R$-modules $V,W$ we have a canonical $R$-linear map $V^{\vee}\otimes W^{\vee}\longrightarrow (V\otimes W)^{\vee}$ from tensor product of dual modules into the dual of tensor product. My question is, why is this map always injective?

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This is false. For instance, let $k$ be a field and let $R=k[x_0,x_1,x_2,\dots]/(x_0,x_1,x_2,\dots)^2$. Let $V=W$ be a free $R$-module with basis $\{e_0,e_1,\dots\}$ and let $\alpha\in V^\vee$ be given by $\alpha(e_i)=x_i$. Note that $\alpha\otimes\alpha$ maps to $0$ in $(V\otimes_R V)^\vee$, since $\alpha(e_i)\alpha(e_j)=x_ix_j=0$ for any $i,j$. However, I claim $\alpha\otimes\alpha$ is nonzero in $V^\vee\otimes_R V^\vee$.

To prove this, note that $R$ is the direct limit of the subrings $R_n=k[x_0,\dots,x_n]/(x_0,\dots,x_n)^2$ and so $V^\vee\otimes_R V^\vee$ is the direct limit of the tensor products $T_n=V^\vee\otimes_{R_n} V^\vee$ over $R_n$. It thus suffices to show that $\alpha\otimes\alpha$ is nonzero in $T_n$ for all $n$.

Note that $V^\vee\cong R^\mathbb{N}$, with the coordinates corresponding to evaluation at the $e_i$. As an $R_n$-module, $R$ splits as a direct sum $R_n\oplus k^{\oplus\mathbb{N}}$ where $k$ is an $R_n$-module by letting the variables act trivially and the second summand is the vector space spanned by the $x_i$ for $i>n$. Thus as an $R_n$-module, $V^\vee\cong R_n^{\mathbb{N}}\oplus (k^{\oplus\mathbb{N}})^{\mathbb{N}}$; let us call these two summands $M$ and $N$. Now notice that $N$ is annihilated by the maximal ideal of $R_n$ and so $N\otimes_{R_n} N$ can be identified with $N\otimes_k N$. But when you take a tensor product over a field, the tensor of any two nonzero vectors is nonzero. In particular, since the projection of $\alpha$ onto $N$ is nonzero, so is the projection of $\alpha\otimes \alpha\in T_n$ onto $N\otimes_{R_n} N$. Thus $\alpha\otimes\alpha$ is nonzero in $T_n$, as desired.

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