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

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 $n$?

share|cite|improve this question
Next time, please kindly make your posts human readable by using latex. – Alex B. Apr 25 '11 at 2:48

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).

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.