3

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.

WhatsUp
  • 22,614
John Dieck
  • 49
  • 1
    Over $\mathbb{Q}(\sqrt{2})$, $x^2 - 2$ factors as $(x - \sqrt{2})(x + \sqrt{2})$. Then you can apply the Chinese remainder theorem. – Viktor Vaughn Dec 04 '19 at 02:19
  • FYI: I happen to have written an answer on MathOverFlow: https://mathoverflow.net/questions/280081/tensor-product-of-field-extensions/280090#280090 The example given there might be useful to you. – WhatsUp Dec 04 '19 at 02:26

1 Answers1

4

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.

I hope this helps ^_^

Chris Grossack
  • 42,257