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!

  • $\begingroup$ Hint: what's the dimension as a vector space over $R$? $\endgroup$ – Mark Dickinson Feb 9 '16 at 19:51
  • 3
    $\begingroup$ How would you get $1\otimes i$ as a multiple of that vector? $\endgroup$ – Tobias Kildetoft Feb 9 '16 at 19:52
  • $\begingroup$ @TobiasKildetoft By def. $i(1\otimes 1) = i\otimes 1 = 1\otimes i$, right? $\endgroup$ – David Li Feb 9 '16 at 19:53
  • 5
    $\begingroup$ No, you are tensoring over the reals, not over the complex numbers. $\endgroup$ – Tobias Kildetoft Feb 9 '16 at 19:53
  • $\begingroup$ You mean, $x\otimes y \neq y\otimes x$ and bases are $\{1\otimes 1, 1\otimes i\}$ $\endgroup$ – 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 $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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