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

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 $$F$$.

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