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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!

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2 Answers

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.

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

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