Irreducibility of polynomials x^p – a

Posted: May 13, 2022 in Basic Algebra, Fields
Tags: ,
0

Problem. Let F be a field, a \in F, and let p be a prime number. Show that the polynomial f(x):=x^p-a is irreducible in F[x] if and only if f has no root in F.

Solution. Since there’s nothing to prove if a =0, we will assume that a \ne 0. If f has a root b \in F, then f(x)=(x-b)g(x) for some g(x) \in F[x] and so f is not irreducible in F[x]. Conversely, suppose that f is not irreducible in F[x]. Let K \supseteq F be a splitting field of f, and let b \in K be a root of f. Then b \ne 0 and x^p=a=b^p, which give (b^{-1}x)^p=1. So v is a root of f if and only if b^{-1}v is a root of x^p-1. Therefore if u_1, \cdots , u_p are all the roots of x^p-1, then bu_1, \cdots , bu_p are all the roots of f. Now, since f is reducible in F[x], there exist g(x),h(x) \in F[x] such that

Since every root of g is a root of f, the roots of g are bu_{i_1}, \cdots , bu_{i_n}, for some 1 \le i_1, \cdots i_n \le p, and hence

b^nu_{i_1} \cdots u_{i_n}=\prod_{j=1}^nbu_{i_j}=(-1)^nc_n.

Therefore, since b is a root of x^p-a and each u_{i_j} is a roots of x^p-1, we get that

a^n=b^{np}=((-1)^nc_n)^p. \ \ \ \ \ \ \ \ \ (*)

Now, \gcd(n,p)=1 because n < p and p is a prime number. So rn+sp=1 for some integers r,s, and thus, by (*),

a=a^{rn+sp}=(a^n)^ra^{sp}=(((-1)^nc_n)^p)^ra^{sp}=((-1)^{nr}c_n^ra^s)^p.

Hence (-1)^{nr}c_n^ra^s \in F is a root of f. \ \Box

Leave a Reply