$\Bbb{C} \otimes_{\Bbb{R}} \Bbb{C}$ and $\Bbb{C} \otimes_{\Bbb{C}} \Bbb{C}$ are not isomorphic as $\Bbb{R}$-vector spaces.

Clearly each tensor product is both a left and right $R$-module. But how do I show that they're not isomorphic?

I'm trying to argue as in other examples, so in $M = \Bbb{C} \otimes_{\Bbb{C}} \Bbb{C}$, there's the simple tensor $(-1)\otimes i = (i^2) \otimes i = i \otimes(-1)$. But those two simple tensors are not equal in $N = \Bbb{C} \otimes_{\Bbb{R}} \Bbb{C}$. I'm not sure how to show that. Let $\phi : M \to N$ be an isomorphism. Then $\phi((-1)\otimes i)$ must equal $\phi (i \otimes (-1))$.


marked as duplicate by Watson, Namaste abstract-algebra Nov 24 '18 at 19:11

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    $\begingroup$ Look at them as $\;\Bbb R - $ vector spaces... $\endgroup$ – DonAntonio Nov 24 '13 at 17:25
  • $\begingroup$ @DonAntonio: Yanked the words right out of my mouth! $\endgroup$ – Asaf Karagila Nov 24 '13 at 17:26
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    $\begingroup$ If $V$ and $W$ are finite dimensional vector spaces over a field $K$, then $$\dim_K (V\otimes_KW)=(\dim_KV)(\dim_KW).$$ $\endgroup$ – Jyrki Lahtonen Nov 24 '13 at 17:26
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    $\begingroup$ Oh, dear! אני נורא מצטער, אסף...I shall put those words back at the first opportunity...:) $\endgroup$ – DonAntonio Nov 24 '13 at 17:27
  • $\begingroup$ @DonAntonio: If you're not a dentist, please keep your mitts out of my mouth. :-) $\endgroup$ – Asaf Karagila Nov 24 '13 at 18:10

Let $z \otimes w$ be a simple tensor in $\mathbb{C} \otimes_\mathbb{C} \mathbb{C}$.

Certainly we have $z \otimes w = 1 \otimes zw$. Every simple tensor may be written in this form, so that certainly every element of $\mathbb{C} \otimes_\mathbb{C} \mathbb{C}$ has the form $1 \otimes (a+bi)$. Note that $1 \otimes (a+bi) = 1 \otimes a + 1 \otimes bi = a \otimes 1 + b \otimes i = a \cdot (1 \otimes 1) + b \cdot (1 \otimes i)$. In particular, every element of $\mathbb{C} \otimes_\mathbb{C} \mathbb{C}$ may be written as an $\mathbb{R}$-linear combination of $1 \otimes 1$ and $1 \otimes i$.

We claim that $1 \otimes 1$ and $1 \otimes i$ form a basis for $\mathbb{C} \otimes_\mathbb{C} \mathbb{C}$ over $\mathbb{R}$, so that this module is free with rank 2 over $\mathbb{R}$.

by the universal property of free modules, the mapping $(1,0) \mapsto 1 \otimes 1$,$ (0,1) \mapsto 1 \otimes i$ extends to a surjective $\mathbb{R}$-module homomorphism $\varphi : \mathbb{R}^2 \rightarrow \mathbb{C} \otimes_\mathbb{C} \mathbb{C}$.

Now define $\psi$ : $\mathbb{C} \times \mathbb{C} \rightarrow \mathbb{R}^2$ by $\psi(a+bi,c+di) = (ac-bd, ad+bc)$. Evidently, $\psi$ is $\mathbb{C}$-balanced, and thus induces an $\mathbb{R}$-module homomorphism $\Psi$ : $\mathbb{C} \otimes_\mathbb{C} \mathbb{C} \rightarrow \mathbb{R}^2$. Moreover, $\Psi$ is surjective. Evidently, $\varphi$ and $\psi$ are mutual inverses. Thus $\mathbb{C} \otimes_\mathbb{C} \mathbb{C}$ is free of rank 2 as a left $\mathbb{R}$-module.

Now let $(a+bi) \otimes (c+di)$ be a simple tensor in $\mathbb{C} \otimes_\mathbb{R} \mathbb{C}$. Evidently, $(a+bi) \otimes (c+di) = a \otimes c + a \otimes di + bi \otimes c + bi \otimes di = ac(1 \otimes 1) + ad(1 \otimes i) + bc(i \otimes 1) + bd(i \otimes i)$. Certainly then every element of $\mathbb{C} \otimes_\mathbb{R} \mathbb{C}$ is an $\mathbb{R}$-linear combination of $1 \otimes 1, 1 \otimes i, i \otimes 1$, and $i \otimes i$.

We claim that these four elements form a basis for $\mathbb{C} \otimes_\mathbb{R} \mathbb{C}$ as a left $\mathbb{R}$-module. To see this, note that the universal property of free modules again induces a surjective $\mathbb{R}$-module homomorphism $\mathbb{R}^4 \rightarrow \mathbb{C} \otimes_\mathbb{R} \mathbb{C}$.

Evidently, the mapping $\mathbb{C} \times \mathbb{C} \rightarrow \mathbb{R}^4$ given by $(a+bi,c+di) \mapsto (ac,ad,bc,bd)$ is $\mathbb{R}$-balanced, and so induces a (clearly surjective) $\mathbb{R}$-module homomorphism $\mathbb{C} \otimes_\mathbb{R} \mathbb{C} \rightarrow \mathbb{R}^4$.

These mappings are mutual inverses. Thus $\mathbb{C} \otimes_\mathbb{R} \mathbb{C}$ is free of rank 4 as a left $\mathbb{R}$ module.

Over a commutative ring, (finitely generated) free modules have a unique rank. Thus we have $\mathbb{C} \otimes_\mathbb{C} \mathbb{C} \not\cong_R \mathbb{C} \otimes_\mathbb{R} \mathbb{C}$.

  • $\begingroup$ What is the universal property of free modules? It's not in my notes and the wikipedia article isn't very clear about it. $\endgroup$ – BananaCats Category Theory App Nov 29 '13 at 17:41
  • $\begingroup$ You can see the definition and properties of The universal property of free modules in the textbook: ABSTRACT ALGEBRA , Authors: David S.Dummit and Richard M.Foote, page 354. Hope it will help you. $\endgroup$ – An Khuong Doan Nov 30 '13 at 5:54

Recall that as $k$-vector spaces, we have $\dim(V\otimes_k W)=\dim(V)\dim(W)$. Now also recall that if $V$ is an $n$-dimensional $\mathbb{C}$-vector space, then it is also a $2n$-dimensional $\mathbb{R}$-vector space.

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    $\begingroup$ As an $R$-vector space, $C \otimes_R C$ has dim $4$. $\endgroup$ – BananaCats Category Theory App Nov 24 '13 at 17:36
  • $\begingroup$ @EnjoysMath: So how about the other tensor product $\mathbb{C} \otimes_{\mathbb{C}} \mathbb{C}$ ? $\endgroup$ – hardmath Nov 24 '13 at 17:40
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    $\begingroup$ Oh, it has dim $1$, so has dim $2$ as an $R$ vs. $\endgroup$ – BananaCats Category Theory App Nov 24 '13 at 17:43

Thinking of what your doing, it is more natural to show that they are not isomorphic as $\mathbb C$-vector spaces (even if the exercise does not say so). Indeed, you start with $\mathbb C$ and

  • in one hand, you consider it as a $\mathbb C$-vector space and extend the scalar to $\mathbb C$, so basically do nothing : this is the isomorphism $\mathbb C \otimes_{\mathbb C} \mathbb C \simeq \mathbb C$ (as $\mathbb C$-vector spaces),

  • in the other hand, you consider it as a $\mathbb R$-vector space and extend the scalar to the field $\mathbb C$ (you brutally decide that the scalar are now complex numbers, not just real numbers) : this is the isomorphism $\mathbb C \otimes_{\mathbb R} \mathbb C \simeq \mathbb C^2$ (as $\mathbb C$-vector spaces).

(Obviously $\mathbb C$-isomorphism leads $\mathbb R$-isomorphism and $\dim_{\mathbb R}(\mathbb C) \neq \dim_{\mathbb R}(\mathbb C^2)$, solving your exercise. But, in my humble opinion, this is not the important part of the exercise.)


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