Is this tensor product of $ \mathbb Q(\sqrt { 2 }) $ a field?

Question

I am trying to prove that $\mathbb{Q}(\sqrt{2})\otimes_{\mathbb{Q}}\mathbb{Q}(\sqrt{2})$ is not a field, but I do not know which is the right isomorphism. I am thinking of $$\mathbb{Q}(\sqrt{2})[x]/(x^2-2)\cong\mathbb{Q}(\sqrt{2})\otimes_{\mathbb{Q}}\mathbb{Q}(\sqrt{2}).$$ Could someone help, please? Thank you so much.

Answer 1

The isomorphism you have listed is exactly right! One way of seeing this is by remembering for for two $K$-modules $A$ and $L$, we have

$$A/I \otimes_K L \cong (A \otimes_K L) / (I \otimes_K L)$$

(cf. this MSE question)

So, for us, we rewrite $$\mathbb{Q}(\sqrt{2}) \otimes_\mathbb{Q} \mathbb{Q}(\sqrt{2})$$ as $$\mathbb{Q}[x]/(x^2-2) \otimes_\mathbb{Q} \mathbb{Q}(\sqrt{2}),$$ and then apply the above identity to get $$\mathbb{Q}(\sqrt{2})[x]/(x^2 - 2)$$ which is not a field, since $x^2-2$ is not $\mathbb{Q}(\sqrt{2})$-irreducible.

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I hope this helps ^_^