In Cassels-Frolich, one can read this theorem (page 57): Let $K$ be a finite finite separable extension of the valued field $(k,v)$. Let $\overline k$ be the completion of $k$ and $(K_j)_{1\le j\le r}$ be the set of completions of $K$ for a valuation $v_j$ of $K$ above $v$. Then, $\overline k\otimes_k K\simeq\prod_{j=1}^rK_j$ algebraically and topologically (product topology on RHS). I do not understand well this theorem: $5=4^2\pmod{11}$, so by Hensel Lemma, $\sqrt5\in\mathbb Q_{11}$. The valuation $v_{11}$ of $\mathbb Q_{11}$ has two extensions $w_1$ and $w_2$ in $K=\mathbb Q(\sqrt5)$. Then, by Cassels-Frolich, $\mathbb Q_{11}\otimes_{\mathbb Q}K=K_{w_1}\times K_{w_2}$. But $\mathbb Q_{11}\otimes_{\mathbb Q}K=\mathbb Q_{11}(\sqrt 5)$ and it looks like that $\mathbb Q_{11}(\sqrt 5)$ belongs to $K_{w_1}$ since $\sqrt5\in K\subset K_{w_1}$ and $\mathbb Q_{11}$ belongs to it too. The same happens for $K_{w_2}$. Did I do a mistake anywhere? Thanks in advance for any help.
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7$\begingroup$ 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 $\cong \mathbf R \otimes_{\mathbf Q} \mathbf Q[x]/(x^2-2) \cong \mathbf R[x]/(x^2-2) = \mathbf R[x]/(x-\sqrt{2})(x+\sqrt{2}) \cong \mathbf R \times \mathbf R$. $\endgroup$– KConradCommented Dec 14, 2019 at 2:46
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$\begingroup$ 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 $K_{w_1}\simeq\mathbb Q_{11}$? $\endgroup$– joaopaCommented Dec 14, 2019 at 2:56
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2$\begingroup$ 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 $\mathbf Q_{11}$. $\endgroup$– KConradCommented Dec 14, 2019 at 6:59
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