Are $\mathbb{C} \otimes _\mathbb{R} \mathbb{C}$ and $\mathbb{C} \otimes _\mathbb{C} \mathbb{C}$ isomorphic as $\mathbb{R}$-vector spaces?

I am having a very hard time at digesting tensor products and I do not know how to "compare" tensor products over different rings. My hunch is that they are not isomorphic. It is easy to see that $\mathbb{C} \otimes _\mathbb{R} \mathbb{C}$ is an $\mathbb{R}$-vector space of dimension $4$. I suspect that $\mathbb{C} \otimes _\mathbb{C} \mathbb{C}$ is also an $\mathbb{R}$-vector space but of lower dimension, but I have no idea how to show this or if indeed this intuition is correct.

Thank you.

  • 1
    $\begingroup$ $\mathbb{C} \otimes _\mathbb{C} \mathbb{C} \simeq \mathbb{C}$ $\endgroup$ – Jeff Jun 28 '13 at 14:18
  • $\begingroup$ For the first tensor product, see math.stackexchange.com/questions/263192 $\endgroup$ – Watson Nov 24 '18 at 16:48

No, they are not. Your intuition is correct.

Given a ground field $k$, and two finite-dimensional $k$-vector spaces $V,W$, then $\dim_k(V \otimes_k W) = (\dim_k V)(\dim_k W)$. So:

  • $\dim_{\mathbb C}(\mathbb C \otimes_{\mathbb C} \mathbb C) = (\dim_{\mathbb C} \mathbb C)^2= 1$, therefore $\dim_{\mathbb R} (\mathbb C \otimes_{\mathbb C} \mathbb C) = 2$.
  • $\dim_{\mathbb R}(\mathbb C \otimes_{\mathbb R} \mathbb C) = (\dim_{\mathbb R} \mathbb C)^2= 2^2 = 4$.

So the two are not isomorphic. In fact, $\mathbb C \otimes_{\mathbb C} \mathbb C \simeq \mathbb C$ canonically (and more generally $V \otimes_k k \simeq V$), while $\mathbb C \otimes_{\mathbb R} \mathbb C \simeq \mathbb C \oplus \mathbb C$.

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    $\begingroup$ Moreover, as hinted at by your first bullet, $C\otimes_{\mathbb{C}}\mathbb{C}\cong\mathbb{C}$, as $z_1\otimes z_2=z_1z_2\otimes 1$ for any $z_1,z_2\in\mathbb{C}$. $\endgroup$ – mdp Jun 28 '13 at 14:17
  • $\begingroup$ @MattPressland My problem was indeed justifying the first bullet, thank you for your illuminating remark. $\endgroup$ – user39280 Jun 28 '13 at 14:20
  • $\begingroup$ @dado: Can you prove the general fact about vector spaces over $k$? If so, try to apply it to $\mathbb C \otimes_{\mathbb C} \mathbb C$. $\endgroup$ – Najib Idrissi Jun 28 '13 at 14:22
  • $\begingroup$ @MattPressland I think I can prove it using the theorem (a sort of distributive law) relating tensor product and direct sum and using that $R \otimes _R N \cong N$ for any ring R and left $R$-module N $\endgroup$ – user39280 Jun 28 '13 at 14:30
  • $\begingroup$ @dado: You can do it that way, indeed, but since we're talking about finite dimensional vector spaces, there's a much easier proof. Try with a basis for each vector space. $\endgroup$ – Najib Idrissi Jun 28 '13 at 14:57

Your intuition is indeed correct.

$\mathbb C\otimes_\mathbb R \mathbb C$: $\mathbb C$ as a real vector space is nothing by $\mathbb R^2$. $\mathbb R^2\otimes_{\mathbb R}\mathbb R^2\cong \mathbb R^4$ (you can write down a basis if you like).

On the other hand, very generally $A\otimes_A A\cong A$ always. Why? Well, the $\mathbb C$ in $\otimes_{\mathbb C}$ tells you that you can move scalars across. So you can make all of these tensors look like $v\otimes 1$. So again you can write down a basis. So this module is just $\mathbb C\cong \mathbb R^2$.


In this case the first one is saying that you can move any part of $\mathbb R$ from "side to side" - e.g. $2(1+i)\otimes 3i = (1+i)\otimes 6i$. However, it would not be true that $i\otimes 1 = 1 \otimes i$.

In the second case, $\mathbb C \otimes \mathbb C$, you can move elements of $\mathbb C$ (and hence also $\mathbb R$) from "side to side". e.g. $1\otimes i = i \otimes 1$.

I find this a relatively concrete way to think of it, even if not the best to generalise.


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