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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have just started to learn about this today, and though i understand this is probably a very simple question, i'm still quite not sure about it:

is $\mathbb{C}\otimes_{\mathbb{R}}\mathbb{C}=\mathbb{H}$ (when this is a tensor product of algebras)?

my feeling is that the answer is yes. however, I am trying to build the product algebra with the regular construction, and can't seem to get it.

I would greatly appreciate a detailed answer, and not a hint, as i'm very new to this. tahnks alot!

share|cite|improve this question
up vote 5 down vote accepted

The short answer is that you don't have $\mathbb{C}\otimes_{\mathbb{R}}\mathbb{C}=\mathbb{H}$ because the left hand side is commutative but the $\mathbb R$- algebra $\mathbb H$ is not.
The more complete answer is that $\mathbb{C} \otimes_{\mathbb R} \mathbb{C}$ is isomorphic to $\mathbb{C} \times \mathbb{C}$ as an $\mathbb R$ -algebra : do you see why?
[Hint: write $\mathbb C \otimes_{\mathbb R} \mathbb{C}=\mathbb{R}[X]/(X^2+1) \otimes_{\mathbb R} \mathbb{C}=\mathbb C[X]/(X^2+1) $ and use the Chinese remainder theorem, noticing that the ideal $(X^2+1)\subset \mathbb C[X]$ equals $(X+i)(X-i)=(X+i)\cap(X-i) ]$

Warning Despite a widespread misconception $\mathbb H$ has no canonical structure of $\mathbb C$-algebra (but an infinity of non-canonical ones!), whereas $\mathbb{C} \otimes_{\mathbb R} \mathbb{C}$ has two such structures: the one coming from the $\mathbb C$ on the left of $\otimes_{\mathbb R}$ and the one coming from the right.

share|cite|improve this answer

This algebra can't be isomorphic to the quaternions because its multiplication is commutative whereas that of the quaternions isn't.

share|cite|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.