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 pZ be a prime; let F be a field and let cF. Then xpc is irreducible in F[x] if and only if xpc
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?

Answer

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 KF where f(x) has a root; as f(x) is irreducible in F[x], a principal ideal domain, then f(x) is a maximal ideal of F[x], hence K=F[x]/f(x) is a field, ˉx is a root of f(x), and it is easy to see how to embed F into K. Now given a polynomial f(x)F[x] it is clear how to construct a field KF such that f(x) factors into linear polynomials in K[x].

Now your question can be answered as follows:

Let KF be a field where xpc factors into linear polynomials, say xpc=(xz1)(xzp). Suppose xpc is not irreducible in F[x], then there are polynomials f(x),g(x)F[x] of degree 1 such that xpc=f(x)g(x), then we may assume f(x)=(xz1)(xzn), where degf(x)=n<p.

Put z=z1zn, then z is ± the constant term of f(x), so zF, and zp=(z1zn)p=zp1zpn=cn. As p is prime there are integers a,b such that 1=ap+bn, then (cazb)p=capzbp=capcbn=c,
but cazbF, so xpc has a root in F.

Attribution
Source : Link , Question Author : user79709 , Answer Author : Camilo Arosemena-Serrato

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