0
$\begingroup$

If $\overline{\mathbb{Q}}$ denotes the algebraic closure of $\mathbb{Q}$, and $n$ is a positive integer, why is $\mathbb{Q}^n \otimes_\mathbb{Q} \overline{\mathbb{Q}} = \overline{\mathbb{Q}}^n$?

I.e. why is the tensor product over $\mathbb{Q}$ of $\mathbb{Q}^n$ with the algebraic closure of $\mathbb{Q}$ isomorphic to the algebraic closure of $\mathbb{Q}$ to the $n$?

$\endgroup$
1
  • 4
    $\begingroup$ Next time, please kindly make your posts human readable by using latex. $\endgroup$
    – Alex B.
    Apr 25, 2011 at 2:48

1 Answer 1

6
$\begingroup$

The two properties of tensors that you need are 1) it commutes with direct sum and 2) $A \otimes_R R = A$. Just expand out the first factor, $\mathbb Q^n$, and apply (2).

$\endgroup$

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.