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I am reviewing for my exam and i need your help to answer the following problem:

Let $F$ be a field, $E$ an extension of $F$ of degree 2, and $L$ an extension of $E$ of degree 2 . Let $\alpha \in L$ and $f(x) \in F[x]$ be the minimal polynomial of $\alpha$ over $F$.

Show that the degree of $f$ divides 4 .

Show that there is a extension $K$ of $F$ whose degree over $F$ divides 8 such that $f$ splits in $K$.

Thanks

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  • $\begingroup$ What are your thoughts about the problem? Any advance towards the solution? Where are you stuck at? $\endgroup$
    – egreg
    Oct 30, 2017 at 23:49
  • $\begingroup$ unfortunately, i didn't find the right answer, please help me, i will be greatful if you do so. $\endgroup$
    – aymen
    Oct 30, 2017 at 23:59

2 Answers 2

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You can do this by hand. Without loss $$E=F(\sqrt{a})$$ where $a\in F$

and

$$L=E(\sqrt{p+q\sqrt{a}})$$

its now easy to see that

$$L(\sqrt{p-q\sqrt{a}})$$ is a splitting field.

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For the the first part:

$[L:E]=2$ and $[E:F]=2$. So $$[L:F]=[L:E][E:F]=4.$$ The mininmal polynomial of $\alpha\in L$ over $F$ is $f$. Then $$[F(\alpha):F]=\deg (f)$$

But we know that $$[L:F]=[L:F(\alpha)][F(\alpha):F],$$ From the second and the third equtaion it follows that $\deg(f)$ divides $[L:F]=4$.

Fact 1:

Let $[L(\theta):L]=2$. Then the minimal polynomial of $\theta$ over $L$ splits over $L(\theta)$.

Proof: Let $p(x)=x^2+bx+c$ be the minimalpolynomial of $\theta$ over $F$. Let $\theta'$ be the other root $p$. Then $\theta+\theta'=-b$, so $\theta'=-b-\theta\in F(\theta)$.

For the second part:

There are two possibilities: $\deg (f)=2$ or $\deg (f)=4$. If $\deg(f)=2$, then (by Fact 1) $f$ splits over $F(\alpha)$ and we are done.

Assume that $\deg (f)=4$. Let $g$ be the minimal polynomial of $\alpha$ over $E$ (so that $\deg (g)=2$). Then $f=gh$ for some $h\in E[x]$ with $\deg(h)=2$. Then (by fact 1) $g$ splits over $K$. If $h$ splits over $K$ then, we are done. Otherwise, define $L=K(\beta)$ where $\beta$ is a root of $g$. Then $[L:K]=2$ and $h$ splits in $L$ (again by Fact 1). So $f=gh$ splits in $L$ and $[L:F]=[L:K][K:F]=8$; and we are done.

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  • $\begingroup$ and how to Show that there is a extension $K$ of $F$ whose degree over $F$ divides 8 such that ff splits in $K$?. $\endgroup$
    – aymen
    Oct 31, 2017 at 0:39
  • $\begingroup$ @aymen Take a look now. $\endgroup$ Oct 31, 2017 at 1:20

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