# $\mathbb C$-dimension of vector space $\mathbb C\otimes_{\mathbb R}\mathbb C$

Let $\mathbb R$ be the field of real numbers, $\mathbb C$ be the field of complex numbers. Consider $\mathbb C\otimes_{\mathbb R}\mathbb C$ as a $\mathbb C$-vector space via $a(b\otimes c) := ab \otimes c,$ for $a, b, c \in \mathbb C$. Compute the $\mathbb C$-dimension of this vector space.

I guess the dimension is 1 and a basis is $\{1\otimes 1\}$. Is it correct? Thank you in advance!

• Hint: what's the dimension as a vector space over $R$? – Mark Dickinson Feb 9 '16 at 19:51
• How would you get $1\otimes i$ as a multiple of that vector? – Tobias Kildetoft Feb 9 '16 at 19:52
• @TobiasKildetoft By def. $i(1\otimes 1) = i\otimes 1 = 1\otimes i$, right? – David Li Feb 9 '16 at 19:53
• No, you are tensoring over the reals, not over the complex numbers. – Tobias Kildetoft Feb 9 '16 at 19:53
• You mean, $x\otimes y \neq y\otimes x$ and bases are $\{1\otimes 1, 1\otimes i\}$ – David Li Feb 9 '16 at 19:55

Sometimes it is better to look at a more general situation. In fact, this makes it easier to see what is really going on.

1) Let $V$ be a $K$-vector space and let $L/K$ be a field extension. Then $L \otimes_K V$ carries the structure of a vector space over $L$ via (linear extension of) $\alpha(\beta \otimes x) = \alpha\beta \otimes x$.

2) If $\{b_1,\dotsc,b_n\}$ is a $K$-basis of $V$, then $\{1 \otimes b_1,\dotsc,1 \otimes b_n\}$ is an $L$-basis of $L \otimes_K V$.

3) We have $\dim_L(L \otimes_K V) = \dim_K(V)$.

In 1) you have to check that this scalar multiplication exists at all. Use the universal property of the tensor product ( = definition of the tensor product) for this.

For the proof of 2) there are several methods. I suggest that you prove more generally that the tensor product commutes with direct sums in each variable (since this is useful anyway); then the claim follows immediately.

Of course, 3) follows from 2).

In particular, we see that $\{1 \otimes 1, 1 \otimes i\}$ is a $\mathbb{C}$-basis of $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$.