Hungerford's book of algebra has exercise $6$ chapter $3$ section $6$ [Probably impossible with the tools at hand.]:

Let $p \in \mathbb{Z}$ be a prime; let $F$ be a field and let $c \in F$. Then $x^p - c$ is irreducible in $F[x]$ if and only if $x^p - c$ has no root in $F$. [Hint: consider two cases: char$(F) = p$ and char$(F)$ different of $p$.]

I have attempted this a lot. Anyone has an answer?

  • $\begingroup$ Have you done it in the case where the characteristic of $F$ is $p$? $\endgroup$ – Tobias Kildetoft May 27 '13 at 15:04
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    $\begingroup$ yes with freshman's dream $\endgroup$ – user79709 May 27 '13 at 15:06
  • $\begingroup$ possible duplicate of $x^p -a$ irreducible in a field of char 0 $\endgroup$ – vonbrand May 27 '13 at 15:32
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    $\begingroup$ This is about finite fields, while the alleged duplicate is for characteristic $0$. Am I missing something? $\endgroup$ – user1729 May 27 '13 at 16:18
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    $\begingroup$ @user1729 I don't see how the hint has anything to do with finite fields (not all fields of positive characteristic are finite). $\endgroup$ – Tobias Kildetoft May 27 '13 at 16:37

Perhaps the simplest tool I can think of is the following:

Let $F$ be a field and $f(x)$ an irreducible polynomial over $F$, then there is a field $K\geq F$ where $f(x)$ has a root; as $f(x)$ is irreducible in $F[x]$, a principal ideal domain, then $\langle f(x)\rangle$ is a maximal ideal of $F[x]$, hence $K=F[x]/\langle f(x)\rangle$ is a field, $\bar{x}$ is a root of $f(x)$, and it is easy to see how to embed $F$ into $K$. Now given a polynomial $f(x)\in F[x]$ it is clear how to construct a field $K\geq F$ such that $f(x)$ factors into linear polynomials in $K[x]$.

Now your question can be answered as follows:

Let $K\geq F$ be a field where $x^p-c$ factors into linear polynomials, say $x^p-c=(x-z_1)\cdots(x-z_p)$. Suppose $x^p-c$ is not irreducible in $F[x]$, then there are polynomials $f(x),g(x)\in F[x]$ of degree $\geq 1$ such that $x^p-c=f(x)g(x)$, then we may assume $f(x)=(x-z_1)\cdots(x-z_n)$, where $\deg f(x)=n<p$.

Put $z=z_1\cdots z_n$, then $z$ is $\pm$ the constant term of $f(x)$, so $z\in F$, and $z^p=(z_1\cdots z_n)^p=z_1^p\cdots z_n^p=c^n$. As $p$ is prime there are integers $a,b$ such that $1=ap+bn,$ then $$(c^az^b)^p=c^{ap}z^{bp}=c^{ap}c^{bn}=c,$$ but $c^az^b\in F$, so $x^p-c$ has a root in $F$.

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    $\begingroup$ Very nice and elementary way to do it. $\endgroup$ – Tobias Kildetoft May 27 '13 at 15:59
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    $\begingroup$ Why is $\;z^p=c^n\;$ ? I just can't see it...and clearly, even! $\endgroup$ – Timbuc Apr 19 '14 at 20:11
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    $\begingroup$ @Timbuc because $z=z_1\cdots z_n$, however $z_i^p=c$ for $i=1,\ldots,n$, so that $z^p=(z_1\cdots z_n)^p=z_1^p\cdots z_n^p=c^n$. $\endgroup$ – Camilo Arosemena-Serrato Apr 20 '14 at 20:37
  • $\begingroup$ Oh, dear! So clear, indeed...thank you, Camilo.+1 $\endgroup$ – Timbuc Apr 20 '14 at 22:54
  • $\begingroup$ Any idea why there was a hint to consider the characteristic $p$ separately? $\endgroup$ – Alexey Jan 2 '19 at 20:04

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