# Proving that $\mathbb{C}\otimes_\mathbb{R}\mathbb{C}\cong \mathbb{C}\times \mathbb{C}$ [duplicate]

Possible Duplicate:
Tensor product $\mathbf{C}\otimes_\mathbf{R} \mathbf{C}$

I have to prove $\mathbb{C}\otimes_\mathbb{R}\mathbb{C}\cong \mathbb{C}\times \mathbb{C}$.

By definition of the tensor product, I want to find a map $f:\mathbb{C}\times \mathbb{C}\rightarrow\mathbb{C}\times \mathbb{C}$ such that:

1) $f(a+a',b)=f(a,b)+f(a',b)$.

2) $f(a,b+b')=f(a,b)+f(a,b')$.

3) $f(ar,b)=f(a,rb)$.

With the universal property that if $T$ is an abelian group such that there is a mapping $g:\mathbb{C}\times \mathbb{C}\rightarrow T$ that satisfies $1,2,$ and $3$, then $g$ factors through $f$.

I was thinking about how to define $f:\mathbb{C}\times \mathbb{C}\rightarrow \mathbb{C}\times \mathbb{C}$ such that property three holds. The problem that I am having is how to find a function such that $f(ar,b)=f(a,rb)$. Could someone maybe help me how to define $f$ and then I could try to go from there and prove that it works.

Thanks

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## marked as duplicate by Hurkyl, Qiaochu YuanDec 11 '12 at 2:28

Do you really mean just as real vector spaces? Or, rather, as rings or (sums/products of) fields? In the latter case, still $\mathbb C\otimes_{\mathbb R} \mathbb C\approx \mathbb R[x]/\langle x^2+1\rangle \otimes \mathbb C\approx \mathbb C[x]/\langle x^2+1\rangle\approx \mathbb C[x]/\langle x-i\rangle \times \mathbb C[x]/\langle x+i\rangle \approx \mathbb C \times \mathbb C$. – paul garrett Dec 11 '12 at 0:47
I meant it as modules over a the field $\mathbb{R}$ (that is, as vector spaces). – Daniel Montealegre Dec 11 '12 at 1:03

We can use the algebra of modules:

$$\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \cong \mathbb{C} \otimes_{\mathbb{R}} ( \mathbb{R} \oplus \mathbb{R} ) \cong (\mathbb{C} \otimes_{\mathbb{R}} \mathbb{R}) \oplus (\mathbb{C} \otimes_{\mathbb{R}} \mathbb{R}) \cong \mathbb{C} \oplus \mathbb{C}$$

An explicit isomorphism can be obtained by following the diagram. Taking real and imaginary parts for isomorphism $\mathbb{C} \mapsto \mathbb{R} \oplus \mathbb{R}$ (and using the canonical choice for the other isomorphisms) leads to

$$z \otimes_\mathbb{R} w \mapsto z \otimes_\mathbb{R} (\Re{w}, \Im{w}) \mapsto (z \otimes_\mathbb{R} \Re w, z \otimes_\mathbb{R} \Im w) \mapsto (z \cdot \Re w, z \cdot \Im w)$$

($\Re$ and $\Im$ are for real part and imaginary part)

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Any two finite dimensional vector spaces over the same field of the same dimension are isomorphic. Since you are tensoring over $\Bbb{R}$ the left hand side has real dimension $2 \times 2$, will the right hand side has real dimension $2 + 2$. Since these are both $4$ they are isomorphic as real vector spaces.

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but the $f$ that you have above is supposed to satisfy $f(ar,b)=f(a,rb)$, but it is not true that $(ar,b)=(a,rb)$ – Daniel Montealegre Dec 11 '12 at 0:58
@DanielMontealegre Yes you are right. I will delete that part of my answer. – user38268 Dec 11 '12 at 1:10

What about $f(a+bi,z)=(az,bz)$?

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