# xp−cx^p-c has no root in a field FF if and only if xp−cx^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?

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

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

Now your question can be answered as follows:

Let $$K≥FK\geq F$$ be a field where $$xp−cx^p-c$$ factors into linear polynomials, say $$xp−c=(x−z1)⋯(x−zp)x^p-c=(x-z_1)\cdots(x-z_p)$$. Suppose $$xp−cx^p-c$$ is not irreducible in $$F[x]F[x]$$, then there are polynomials $$f(x),g(x)∈F[x]f(x),g(x)\in F[x]$$ of degree $$≥1\geq 1$$ such that $$xp−c=f(x)g(x)x^p-c=f(x)g(x)$$, then we may assume $$f(x)=(x−z1)⋯(x−zn)f(x)=(x-z_1)\cdots(x-z_n)$$, where $$degf(x)=n.

Put $$z=z1⋯znz=z_1\cdots z_n$$, then $$zz$$ is $$±\pm$$ the constant term of $$f(x)f(x)$$, so $$z∈Fz\in F$$, and $$zp=(z1⋯zn)p=zp1⋯zpn=cnz^p=(z_1\cdots z_n)^p=z_1^p\cdots z_n^p=c^n$$. As $$pp$$ is prime there are integers $$a,ba,b$$ such that $$1=ap+bn,1=ap+bn,$$ then $$(cazb)p=capzbp=capcbn=c,(c^az^b)^p=c^{ap}z^{bp}=c^{ap}c^{bn}=c,$$
but $$cazb∈Fc^az^b\in F$$, so $$xp−cx^p-c$$ has a root in $$FF$$.