Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
Take obvious basis vectors in tensor product map them into basis vectors of direct sum –  userNaN 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. –  userNaN 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

2 Answers 2

up vote 11 down vote accepted

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 $

share|improve this answer
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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.