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I saw that if $E/F$ and $F/k$ are finite extensions then $E/k$ is also finite and wondered if it also holds when they are algebraic.

I tried to choose $\alpha\in E$, and we know that since it's algebraic over $F$ there exists $f\in F[x]$ such that $f(\alpha)=0$.

Then I think I must use the fact that the coefficients of $f$ are also algebraic to get some polynomial in $k[x]$ which vanishes at $\alpha$. But I don't get how could I do it.

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    The algebraic extensions are an example of distinguished class of field extensions. See here: https://en.wikipedia.org/wiki/Algebraic_extension – 2017-02-13
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    every finite extension is algebraic extension – 2017-02-13
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    @Mustafa I am talking about when they are algebraic and not necessarily finite, as in the title of the question. – 2017-02-13

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To complete the argument, write $f(x) = \beta_nx^n + \beta_{n-1}x^{n-1} + \dots + \beta_0$, where $\beta_0,\dots,\beta_{n} \in F$. Then $\beta_0,\dots,\beta_n$ are algebraic over $k$, so $L = k(\beta_0,\dots,\beta_n)$ is a finite extension of $k$. But now $f \in L[x]$, so that $\alpha$ is algebraic over $L$ and $L(\alpha)/L$ is finite. Since finite extensions of finite extensions are finite, $L(\alpha)/k$ is finite, hence algebraic. Therefore, $\alpha$ is algebraic over $k$.

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    Oh, that is it. I forgot the characterization of finite extensions as algebraic and finitely generated. – 2017-02-13