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: $\mathrm{char}(F) = p$ and $\mathrm{char}(F) \ne p$.]

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

  • $\begingroup$ Have you done it in the case where the characteristic of $F$ is $p$? $\endgroup$ May 27, 2013 at 15:04
  • 1
    $\begingroup$ yes with freshman's dream $\endgroup$
    – user79709
    May 27, 2013 at 15:06
  • $\begingroup$ possible duplicate of $x^p -a$ irreducible in a field of char 0 $\endgroup$
    – vonbrand
    May 27, 2013 at 15:32
  • $\begingroup$ Ok, so since we now assume that the characteristic of $F$ is not $p$, there exists an extension of $F$ which has a primitive $p$'th root of unity $\xi$. If $b$ is any $p$'th root of $c$ in some extension of $F$, then $\xi b$ is a primitive $p$'th root of $c$, and so there exists an extension of $F$ of degree $p$ in which the polynomial $x^p - c$ has a root. But this means that $c$ is a root of some field lying between $F$ and this larger field, which means that this intermediate field has degree dividing $p$, so it must be $F$ since we are assuming that $x^p - c$ is reducible. $\endgroup$ May 27, 2013 at 15:50

2 Answers 2


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 $F$.

  • 6
    $\begingroup$ Why is $\;z^p=c^n\;$ ? I just can't see it...and clearly, even! $\endgroup$
    – Timbuc
    Apr 19, 2014 at 20:11
  • 6
    $\begingroup$ @Timbuc because $z=z_1\cdots z_n$, however $z_i^p=c$ for $i=1,\ldots,n$, so that $z^p=(z_1\cdots z_n)^p=z_1^p\cdots z_n^p=c^n$. $\endgroup$ Apr 20, 2014 at 20:37
  • $\begingroup$ Any idea why there was a hint to consider the characteristic $p$ separately? $\endgroup$
    – Alexey
    Jan 2, 2019 at 20:04
  • $\begingroup$ See also here on $\large \ c^n = z^p\, \Rightarrow\ c = (c^a z^b)^p\,$ for $\,a,b\in\Bbb Z\ \ $ $\endgroup$ Nov 21, 2019 at 1:41

Let $\mathrm{char}(F)=p$ and $P(X)=X^p-c$ be reducible in $F[X]$ and $L=K(a)$, where $a$ is a root of $P(X)$. Note that in $L[X]$ can write $P(X)=(X-a)^p$. Let $f(X)$ be an irreducible factor of degree $1\leq n<p$ in $F[X]$. Now, $\gcd(P(X),f(X))=f(X)$ and gcd is independent of the field extension, so $f(X)=(X-a)^n$. Hence, $a^n, na^{n-1} \in F$. Since $n$ is coprime to $p$, it follows $a^{n-1}\in F$ and $a \in F$.

  • $\begingroup$ this is a very nice proof but how did you get $na^{n-1}\in F$? $\endgroup$
    – nomadd
    Feb 12, 2022 at 2:46
  • $\begingroup$ @nomadd $na^{n-1}$ is the coefficient of $X$ in $f(X)=(X-a)^n$ which belongs to $F[X]$. $\endgroup$
    – user26857
    May 15, 2022 at 9:21
  • $\begingroup$ $a^p=c\in F$ and $a^n\in F$. Since $\gcd(p,n)=1$ we get $a\in F$. $\endgroup$
    – user26857
    Feb 28 at 8:15

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