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As the title explained,if α is algebraic over F,then there is a polynomial p(x) such that p(α)=0 with all coefficient in F.But how can we just use this to show that every linear combination of α with coefficient in F(i.e.F(α)) is a root of some polynomial with coefficient in F?

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    Hint: Since $\alpha$ is algebraic, you can find a finite basis for $F(\alpha)$ over $F$. What if some element of $F(\alpha)$ wasn't algebraic over $F$? – 2017-02-07

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Hints:

1) As commented, $\;\dim_{\Bbb F}\Bbb F(\alpha)=n<\infty\;$

2) Take $\;\beta\in\Bbb F(\alpha)\;$ , and let $\;B:=\{1,\beta,\beta^2,\ldots,\beta^n\}\subset\Bbb F(\alpha)\;$ . Then $\;B\;$ has $\;n+1\;$ elements so it is linearly dependent over $\;F\;$ ...

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    DonAntonio:If I understand correctly,what you mean by β is one linear combination of α with coefficient in F.But how can you conclude that the set B is linearly dependent(i.e.a polynomial with β being a zero)?This is exactly what I am asking.The only thing we know in assumption is there is a polynomial p(x) such that p(α)=0. – 2017-02-07
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    @AndrewArmstrong (1) Yes, of course: $\;\beta\;$ is some lin. combination of powers of $\;\alpha\;$ over $\;\Bbb F\;$ , but *who cares about that*? (2) It is **a vector** in $\;\Bbb F(\alpha)\;$ and so are $\;1,\beta^2,...$ etc., so $\;n+1\;$ vectors in an $\;n\,-$ dimensioanl linear space **must be** linearly dependent.... – 2017-02-07
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    DonAntonio:Thank you.This explanation is quite clear and intuitive.So we just prove the existence theoretically but in fact it is sometimes hard to construct a polynomial for every linear combinations and we don't need to construct.Is this make sence? – 2017-02-07
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    @AndrewArmstrong Indeed so. The simple but powerful tools from linear algebra allow us to deduce the existence of a *trivial* $\;\Bbb F\,-$ linear combination of $\;1,\beta,...,\beta^n\;$, which is the same as the existence of a **non-zero** polynomial in $\;\Bbb F[x]\;$ which has $\;\beta\;$ as a root. Observe that this polynomial obtained via linear algebra does **not necessarily** is an irreducible one...and we, of course, don't care about that, either. – 2017-02-07
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    DonAntonio:So we have justshown that every finite extension is algebraic yeah?Thank you again,as I have stucked on here several days! – 2017-02-07