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This question already has an answer here:

This problem is paraphrased from an old version of an exam that
I will be taking, and I have no idea how one would do solve it.

Let $p$ be a prime number, let $F$ be a field of characteristic $p$, and let $c$ be a field element
such that $\:x^{\hspace{.025 in}p}\hspace{-0.04 in}+(-c)\:$ has no roots in $F$. $\;\;\;$ Show that $\:x^{\hspace{.025 in}p}\hspace{-0.04 in}+(-c)\:$ is irreducible over $F$.

How would one solve that problem?

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marked as duplicate by Jyrki Lahtonen, Amzoti, user1337, Amitesh Datta, Adriano Aug 17 '13 at 22:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @Zev: Different polynomial. That one is separable, this ain't. Might still be a duplicate though. Searching... $\endgroup$ – Jyrki Lahtonen Aug 17 '13 at 21:08
  • $\begingroup$ @Jyrki: Whoops, should have read closer. Thanks for pointing that out. $\endgroup$ – Zev Chonoles Aug 17 '13 at 21:09
  • $\begingroup$ @Zev: It IS still a duplicate. $\endgroup$ – Jyrki Lahtonen Aug 17 '13 at 21:18
  • $\begingroup$ @JyrkiLahtonen: Sorry, I can't vote to close anymore now that I've retracted. $\endgroup$ – Zev Chonoles Aug 17 '13 at 21:23
  • $\begingroup$ I see. Can't be helped, @Zev :-) $\endgroup$ – Jyrki Lahtonen Aug 17 '13 at 21:24
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Let $K$ be the splitting field of $f(x)=x^p-c$ over $F$, and $a$ a root. Thus, $a^p=c$ and so $x^p-c=(x-a)^p$ over $K$. Hence every factor of $f$ has the form $(x-a)^k$ for some $k\le p$. Suppose there is some $k\lt p$ with $g(x)=(x-a)^k\in F[x]$, since $k\lt p, (k,p)=1$, so there are integers $m,n$ with $mp+nk=1$, but this implies $f^m(x)g^n(x)=(x-a)^{mp+nk}=x-a\in F[x]$, contradiction.

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  • $\begingroup$ Very nice +1.... $\endgroup$ – DonAntonio Aug 17 '13 at 21:15
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    $\begingroup$ This is one of the ways I had in mind. IMHO an even simpler way is to look at the degree $k-1$ term of $g(x)$. That has coefficient $k\cdot a\in F$. As $0<k<p$ this implies that $a\in F$. Can't tell for sure which way Dylan Moreland had in mind in his hint :-) Well done anyway, +1. $\endgroup$ – Jyrki Lahtonen Aug 17 '13 at 21:22
  • $\begingroup$ @JyrkiLahtonen: Very nice! I like this idea of looking at certain coefficient! In fact, a similar trick is employed in this excellent answer of yours :) $\endgroup$ – Prism Aug 17 '13 at 23:17
  • $\begingroup$ Wait a minute: $\:$ $f$ and $g$ aren't units, but one of $\{m\hspace{.01 in},\hspace{-0.03 in}n\}$ must be negative, so how can the last line work? $\hspace{.32 in}$ $\endgroup$ – user57159 Aug 19 '13 at 6:40
  • $\begingroup$ You are right, it is a bit hasty. You probably need to insert a Euclidean Algorithm type argument like: $$p=qk+r,0\le r\lt k\Longrightarrow (x-a)^{p-qk}=(x-a)^r\in F[x].$$ From this it will follow that $(x-a)^{(p,k)}=(x-a)\in F[x].$ $\endgroup$ – walcher Aug 19 '13 at 10:16