# $x^p-c$ has no root in a field $F$ if and only if $x^p-c$ is irreducible?

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?

• Have you done it in the case where the characteristic of $F$ is $p$? May 27, 2013 at 15:04
• yes with freshman's dream May 27, 2013 at 15:06
• possible duplicate of $x^p -a$ irreducible in a field of char 0 May 27, 2013 at 15:32
• 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. May 27, 2013 at 15:50

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 $$K[x]$$.

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.

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 $$F$$.

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

• this is a very nice proof but how did you get $na^{n-1}\in F$? Feb 12 at 2:46
• @nomadd $na^{n-1}$ is the coefficient of $X$ in $f(X)=(X-a)^n$ which belongs to $F[X]$. May 15 at 9:21
• The answerer could have considered the coefficient of $X^{n-1}$ in $f(X)=(X-a)^n$ and conclude that $na\in F$. Then use that $\gcd(n,p)=1$. May 15 at 9:23