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Let $L/K$ be a finite field extension with $[L:K] = 2^k$ and $f \in K[X]$ be a polynomial of degree 3 that has a root in $L$. Show that $f$ has a root in $K$.

I was thinking about proving this with induction but I have trouble with the base case:

$k=1$: Without loss of generality we can assume that $f$ is monic, so $f = X^3 + aX^2 + bX + c$. Then $[L:K] = 2$ and since $2$ is prime, there exists an $\alpha \in L$ such that $L = K(\alpha)$. Now the Minimal polynomial $g$ of $\alpha$ over $K$ is of degree 2 and since it is also monic, it must be a divider of $f$ (I'm not sure about this one). So $f = g\cdot h$ with $\deg(h) = 1$. Since $h$ is a linear polynomial over $K$ it must have a root in $K$.

I'm not sure if this was correct since I didn't really need that $f$ has a root in $L$? Any hints would be greatly appreciated.

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    $\begingroup$ $[L:K(\alpha)][K(\alpha):K]=[L:K]$ $\endgroup$
    – J. W. Tanner
    Nov 30, 2020 at 16:34

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If $f$ has no solution over $K$ and a root $a$ in $L$, then $f$ is irreducible in $K$ and $L$ contains a subfield $K(a)$ isomorphic to $K[X]/(f)$. Such a field has degree $3$ and cannot be contained in $L$, since $3$ doesn't divide $[L:K]$

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    $\begingroup$ You say that $K(\alpha)$ has degree $3$ but why exactly is that? How can we be sure that $f$ is the minimal polynomial of $a$ over $K$? $\endgroup$
    – Shuster
    Nov 30, 2020 at 16:49
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    $\begingroup$ Because it is irreducible and $f(a)=0$. $\endgroup$
    – Julian Quast
    Nov 30, 2020 at 17:18
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    $\begingroup$ You're right, thanks! $\endgroup$
    – Shuster
    Nov 30, 2020 at 17:26

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