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Let $F\subset E$ be a field extension, $S\subset E$ and elements in $S$ are all algebraic over $F$. Prove $F(S)$ is an algebraic extension over $F$. And then all algbraic elements of $E$ form a field extension over $F$.

My try:
1. To prove $F(S)$ is an algebraic extension over $F$, we need to prove every elemnents in $F(S)$ is algebraic.
2. To prove every elemnents in $F(S)$ is algebraic. And since elements in $F(S)$ is of form of product of $\sum_{s\in S} f_s s,f_s\in F$, we need to prove each kind of product of $\sum_{s\in S} f_s s,f_s\in F$ is algebraic.
3. To prove each kind of product of $\sum_{s\in S} f_s s,f_s\in F$ is algebraic, we only need to prove for any two algebraic elements $\alpha,\beta$, $\alpha\cdot\beta$, $\alpha+\beta$ is algebraic over $F$.
I already know how to prove the sum and product are algebraic. Now I am not sure these steps could finish the proof. Please give a verification.

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    $\begingroup$ First prove that all the algebraic elements over F that are in E form a field. You know how to prove the sum and product of algebraic elements are algebraic now you need to show the inverse of an algebraic element is algebraic. Once you show the collection of all algebraic elements over F in E form a field then F(S) is trivially an algebraic extension. $\endgroup$ – sykh Oct 13 '18 at 15:32

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