# Why is this tensor product equal to $\overline{\mathbb{Q}}^n$?

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 $n$?

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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).