Decomposition of $\mathbb{Q}(\sqrt {2}) \otimes \mathbb{Q}(\sqrt {i}) $

Question

I should find the decomposition of $\mathbb{Q}(\sqrt {2}) \otimes _\mathbb{Q} \mathbb{Q}(\sqrt {i}) $

I expect it to be isomorph to Decomposition of $\mathbb{Q}(\sqrt {2}) \cdot \mathbb{Q}(\sqrt {i}) $ but cant find why. Anyone has an idea?

Answer 1

The trick is to write one field abstractly and the other concretely:

$$\mathbf Q(\sqrt{2}) \otimes \mathbf Q(i) \cong \mathbf Q[x]/(x^2-2) \otimes \mathbf Q(i) \cong (\mathbf Q[x] \otimes \mathbf Q(i))/(( x^2-2) \otimes \mathbf Q(i))$$ $$ \cong \mathbf Q(i)[x]/(x^2-2)$$

where I used, among other things, the fact that $\mathbf Q[x] \otimes E \cong E[x]$ for any field $E$ containing $\mathbf Q$. Since $x^2-2$ remains irreducible over $\mathbf Q(i)$,

$$\mathbf Q(i)[x]/(x^2-2) \cong \mathbf Q(i)(\sqrt{2}) = \mathbf Q(i,\sqrt{2})$$