Let $f(x) \in K[x]$ an irreducible polynomial of $K[x]$ of degree $n$.
Let $K\leq F$ a field extension with $[F:K]=m$.
If $(n,m)=1$ show that $f(x)$ stays irreducible also as a polynomial of $F[x]$.
Could you give me some hints how to show this??
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3Let $\alpha$ be a zero of $f$ in some extension of $K$. Consider $K(\alpha)$ and $F(\alpha)$. – Daniel Fischer Oct 23 '14 at 17:38
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Do you mean that $a$ is algebraic over $K$?? – Mary Star Oct 23 '14 at 17:40
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1$\alpha$ is algebraic over $K$ since it is by assumption a zero of $f$. – Daniel Fischer Oct 23 '14 at 17:41
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Ok... Can we use that $$Irr(a,F) \mid Irr(a,K) \Rightarrow deg Irr(a,F) \leq deg Irr(a,K) $$ Or does this not stand?? – Mary Star Oct 23 '14 at 17:47
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That doesn't help. You need to use that $\gcd(\deg f, [F:K]) = 1$. – Daniel Fischer Oct 23 '14 at 17:55
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@DanielFischer If I may ask, are you typing an answer? – Git Gud Oct 23 '14 at 17:57
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@GitGud You may ask, and no, I'm not typing an answer. Go ahead and give a good hint if you wish. – Daniel Fischer Oct 23 '14 at 17:57
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@DanielFischer Could you explain me how I could use that?? – Mary Star Oct 23 '14 at 18:00
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Let $\alpha$ be a root of $f(x)$ and let $m_\alpha(x)$ be the minimal polynomial of $\alpha$ over $K$.
It's easy to conclude that $n=\deg\left(m_\alpha(x)\right)=[K(\alpha)\colon K]$.
It holds that $K\preceq K(\alpha)\preceq F(\alpha)$ and $K\preceq F\preceq F(\alpha)$.
Since all these extensions are finite it follows that
$$\begin{align} [F(\alpha)\colon K]&=[F(\alpha)\colon K(\alpha)]\overbrace{[K(\alpha)\colon K]}^{n},\\ [F(\alpha)\colon K]&=[F(\alpha)\colon F]\underbrace{[F\colon K]}_{m}. \end{align}$$
From $\gcd(m,n)=1$, it follows that $mn$ divides $[F(\alpha)\colon K]$ and consequently $[F(\alpha)\colon K]\ge mn$ which in turn implies that $$m[F(\alpha)\colon F]=[F(\alpha)\colon K]\ge mn.$$ Thus $[F(\alpha)\colon F]\ge [K(\alpha)\colon K]$.
Given this inequality, I'll let you think how to relate it to certain minimal polynomials.
Git Gud
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Could you explain me why it follows that $mn$ divides $[F(a):K]$ when $\gcd(m,n)=1$ ?? – Mary Star Oct 23 '14 at 18:44
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1@MaryStar It's a typical elementary number theory result. $\forall a,b,c((\gcd(a,b)=1\land a|c\land b|c)\implies ab|c)$. See for example here. – Git Gud Oct 23 '14 at 20:18
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Oh, I see... So, does $$[F(a):F] \geq [K(a):K]$$ mean that $$deg Irr(a,F) \geq deg Irr(a,K)$$ ?? – Mary Star Oct 23 '14 at 20:25
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In that way do we conclude that the minimal polynomial at both case, over $K$ and over $F$, is the same?? – Mary Star Oct 23 '14 at 20:48
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When we have that $deg Irr(a,F) \geq degIrr(a,K)$, why does it follows that the minimal polynomial at both case, over K and over F, is the same?? – Mary Star Oct 23 '14 at 20:56
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1That follows from what you wanted to prove initially, that is that $f(x)$ (or equivalently $m_\alpha(x)$) is irreducible over $F$. – Git Gud Oct 23 '14 at 21:04
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1If $f(x)$ was $F$-reducible, then the minimal polynomial of $\alpha$ over $F$ would have degree smaller than $\deg(f(x))=n=[K(\alpha)\colon K]$, contradicting the last inequality. – Git Gud Oct 24 '14 at 00:15
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