# Tensor product algebra $\mathbb{C}\otimes_\mathbb{R} \mathbb{C}$

I want to understand the tensor product $\mathbb C$-algebra $\mathbb{C}\otimes_\mathbb{R} \mathbb{C}$. Of course it must be isomorphic to $\mathbb{C}\times\mathbb{C}.$ How can one construct an explicit isomorphism?

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Take obvious basis vectors in tensor product map them into basis vectors of direct sum – Norbert Aug 12 '12 at 18:46
I mean an isomorphism of $\mathbf{C}$-algebras with respect to the left $\mathbf{C}$ in the tensor product... – Mikhail Borovoi Aug 12 '12 at 18:54
You should write this in your question. – Norbert Aug 12 '12 at 18:57
Why do you say "of course it must be isomorphic to..."? Because of dimension reasons or because of galois theory? – mland Aug 12 '12 at 21:29

An explicit isomorphism of $\mathbb C$-algebras is given (on generators) by $\mathbb C\otimes _\mathbb R \mathbb C\stackrel {\cong }{\to} \mathbb C\times \mathbb C: z\otimes w \mapsto (z\cdot w,z\cdot\bar w)$.
Here $\mathbb C \otimes _\mathbb R \mathbb C$ is considered as a $\mathbb C$-algebra through its first factor: $z_1\cdot (z\otimes w)=z_1 z\otimes w$
On the isomorphism $\mathbf{C}\otimes_\mathbf{R} \mathbf{C}\cong \mathbf{C}\times \mathbf{C}$: math.stackexchange.com/a/118275/3217 – Georges Elencwajg Aug 13 '12 at 6:18
Write $\mathbb C=\mathbb R[x]/\langle x^2+1\rangle$ for one of the copies. Then, using a universal property of tensor products, $$\mathbb C\otimes_{\mathbb R} \mathbb C \;\approx\; \mathbb R[x]/\langle x^2+1\rangle \otimes_{\mathbb R}\mathbb C \;\approx\; \mathbb C[x]/\langle (x+i)(x-i)\rangle \;\approx\; \mathbb C[x]/\langle x+i\rangle \oplus \mathbb C[x]/\langle x-i\rangle$$ the last isomorphism via Sun-Ze's theorem (a.k.a. "Chinese Remainder Theorem"). That last isomorphism can be made explicit by choice of polynomials $A(x),B(x)$ such that $A(x)\cdot (x+i) + B(x)\cdot (x-i)=1$.
Edit: this treats "right" $\mathbb C$-algebra, but/and reversing the roles gives the same outcome as "left" algebra.