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Let $F$ be a field of characteristic $\neq 2$. Let $a\neq b $ be in $F$. Suppose $\sqrt{a}+\sqrt{b}\in F$. Prove that $\sqrt{a}\in F$.

We have $(\sqrt{a}+\sqrt{b})^2=a+b+2\sqrt{ab}\in F$, so $\sqrt{ab}\in F$. Then $\sqrt{ab}(\sqrt{a}+\sqrt{b})=a\sqrt{b}+b\sqrt{a}\in F$. What then?

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Alternatively, write $(\sqrt{a}+\sqrt{b})^{-1}=(\sqrt{a}-\sqrt{b})/(a-b)$ which implies $\sqrt{a}-\sqrt{b}\in F$, then notice $\sqrt{a}$ is the average of $\sqrt{a}+\sqrt{b}$ and $\sqrt{a}-\sqrt{b}$. – anon Jan 18 '14 at 0:34
up vote 10 down vote accepted

Subtract $a(\sqrt{b}+\sqrt{a})$ for the next step.

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Oh that's fast, thanks! – JJ Beck Jan 18 '14 at 0:31

Hint $\ $ If a field $F$ has two $F$-linear independent combinations of $\ \sqrt{a},\ \sqrt{b}\ $ then you can solve for $\ \sqrt{a},\ \sqrt{b}\ $ in $F.\,$ For example, the Primitive Element Theorem works that way, obtaining two such independent combinations by Pigeonholing the infinite set $\ F(\sqrt{a} + r\ \sqrt{b}),\ r \in F,\ |F| = \infty,\,$ into the finitely many fields between F and $\ F(\sqrt{a}, \sqrt{b}),$ e.g. see PlanetMath's proof.

In the OP, note that $F$ contains the independent $\ \sqrt{a} - \sqrt{b}\ $ since

$$ \sqrt{a}\ - \sqrt{b}\ =\ \dfrac{\,\ a\ -\ b}{\sqrt{a}+\sqrt{b}}\ \in\ F $$

To be explicit, notice that $\ u = \sqrt{a}+\sqrt{b},\ \ v = \sqrt{a}-\sqrt{b}\in F\ $ so solving the linear system for the roots yields $\ \sqrt{a}\ =\ (u+v)/\color{#c00}2,\ \ \sqrt{b}\ =\ (v-u)/\color{#c00}2,\ $ both of which are clearly $\,\in F,\,$ since $\,u,\,v\in F\,$ and $\,\color{#c00}2\ne 0\,$ in $\,F,\,$ so $\,1/\color{#c00}2\,\in F.\,$ This works over any field where $\,2\ne 0\,,\,$ i.e. where the determinant (here $2$) of the linear system is invertible, i.e. where the linear combinations $\,u,v\,$ of the square-roots are linearly independent over the base field.

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