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$\def\Rbb{\mathbf{R}}$ Let $F$ be a subfield of the field $K$ and let $V$ be an $n$-dimensional vector space over $F$. Then $K\otimes_FV\cong K^n$.

Considering the case when $F=K=\Rbb$ and $V=\Rbb^1$, I have $$ \Rbb\otimes_\Rbb\Rbb^1=\Rbb^1(=\Rbb)\;. $$

On the other hand, let $V$ and $W$ be two (finite-dimensional) vector spaces over the field $F$. Then $$ V\otimes_F W\cong L(V,W;F)\;, $$ where $L(V,W;F)$ denotes the set of all the bilinear maps from $V\times W$ to $F$.

If I let $V=W=\Rbb^1$, and $F=\Rbb$, then $$ \Rbb^1\otimes_\Rbb\Rbb^1 \cong L(\Rbb^1,\Rbb^1;\Rbb) \cong \Rbb^2 $$

But $\Rbb^1$ cannot be isomorphic to $\Rbb^2$. What is going wrong here?

I guess one cannot "identify" $\Rbb^1$ and $\Rbb$ when talking about tensor products since one is a "vector space" while the other is a "field".

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  • $\begingroup$ Your $L(V,W;F)$ denotes the $F$-linear maps from $V$ to $W$? $\endgroup$
    – peter a g
    Apr 10, 2022 at 15:25
  • $\begingroup$ OK - Then your $L( {\mathbb R}^1 \times {\mathbb R}^1 ; {\mathbb R})$ doesn't "parse" - namely, your $V = {\mathbb R}^1 \times {\mathbb R}^1 $, but what is $W$? $\endgroup$
    – peter a g
    Apr 10, 2022 at 15:34
  • $\begingroup$ on the other hand, the $\mathbb R$-linear maps from ${\mathbb R}^1$ to itself is one-dimensional over the reals. $\endgroup$
    – peter a g
    Apr 10, 2022 at 15:35
  • $\begingroup$ It should be $L(ℝ^1,ℝ^1 ; ℝ^1)$ instead of $L(ℝ^1 × ℝ^1 ; ℝ^1)$. And we rather have $(V ⊗_F W)^* ≅ L(V,W ; F)$. The isomorphism you wrote is valid only if $V$ and $W$ are finite-dimensional, and it's not canonical. $\endgroup$ Apr 10, 2022 at 15:37
  • $\begingroup$ @Dabouliplop typo? you meant $V^*\otimes_F W \simeq L(V,W;F)$... $\endgroup$
    – peter a g
    Apr 10, 2022 at 16:03

1 Answer 1

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$\def\Rb{\mathbf{R}}$ Thanks to the helpful comments from Dabouliplop and peter a g, I identify the mistake is in $L(\Rb^1,\Rb^1;\Rb)\cong \Rb^2$.

This is incorrect. The notation $L(\Rb^1,\Rb^1;\Rb)$ denotes the set of all the bilinear maps from $\Rb^1\times \Rb^1$ to $\Rb$. It is different from the space $L(\Rb^1\times\Rb^1,\Rb)$, which consists of all the linear maps on $\Rb^1\times\Rb^1$.

The dimension of $L(\Rb^1,\Rb^1;\Rb)$ is $1$, while the dimension of $L(\Rb^1\times\Rb^1,\Rb)$ is $2$.

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