Timeline for continuous isomorphism of $p$-adic field

4 events

when toggle format	what		by	license	comment
Dec 14, 2019 at 6:59	comment	added	KConrad		You wrote $\mathbf Q_{11}(\sqrt{5})$. The polynomial $x^2 - 5$ splits over $\mathbf Q_{11}$, so $\mathbf Q_{11}(\sqrt{5}) = \mathbf Q_{11}$ just as $\mathbf R(\sqrt{5}) = \mathbf R$. Both $K_{w_1}$ and $K_{w_2}$ are $\mathbf Q_{11}$.
Dec 14, 2019 at 2:56	comment	added	joaopa		Thanks for the explanation. But I do not understand the sentence "and the right side is 1-dimensional over $\mathbb Q_{11}$". Did you mean $K_{w_1}$ is $1$-one dimensional over $\mathbb Q_{11}$. So $K_{w_1}\simeq\mathbb Q_{11}$?
Dec 14, 2019 at 2:46	comment	added	KConrad		It is false that $\mathbf Q_{11} \otimes_{\mathbf Q} K = \mathbf Q_{11}(\sqrt{5})$: the left side is 2-dimensional over $\mathbf Q_{11}$ (since $K$ is 2-dimensional over $\mathbf Q$) and the right side is 1-dimensional over $\mathbf Q_{11}$ since $\sqrt{5} \in \mathbf Q_{11}$. Review how to correctly compute tensor products of fields. For example, $\mathbf R \otimes_{\mathbf Q} \mathbf Q(\sqrt{2})$ is not $\mathbf R(\sqrt{2})$: it is $\cong \mathbf R \otimes_{\mathbf Q} \mathbf Q[x]/(x^2-2) \cong \mathbf R[x]/(x^2-2) = \mathbf R[x]/(x-\sqrt{2})(x+\sqrt{2}) \cong \mathbf R \times \mathbf R$.
Dec 14, 2019 at 2:04	history	asked	joaopa	CC BY-SA 4.0