Irreducibility of $f(x)=x^n-p^m$

Question

I have a question. Let $f(x)=x^n-p^m$ where $m$ and $n$ are coprime and $p$ is prime integer. How can I show that $f(x)$ is irreducible over $\Bbb Q$?

Answer 1

Hats off to Lord Shark the Unknown for providing the key that unlocks the puzzle-door on this one; taking a tip from his comment on the question, I have tried below to flesh out an answer. I wanted to see how things work out in detail, so . . .

First of all, let's consider the case

$\gcd(n, m) = d > 1; \tag 1$

then since

$d \mid m, \; d \mid n, \tag 2$

we may write

$n = n_1 d, \; m = m_1 d, \tag 3$

and we have

$X^n - p^m = X^{n_1 d} - p^{m_1 d} = (X^{n_1})^d - (p^{m_1})^d$ $= (X^{n_1} - p^{m_1}) \displaystyle \sum_0^{d - 1} (X^{n_1})^{d - 1 - i} (p^{m_1})^i, \tag4$

which demonstrates that $X^n - p^m$ is in fact reducible in the event that $\gcd(n, m) > 1$; thus the condition $\gcd(n, m) = 1$ is not in fact inconsequential, insofar as does distinguish 'twixt reducible and irreducible cases of $X^n - p^m$, provided of course the question of the problem is answerable in the affirmative, which we shall indeed herein establish.

We begin with $m = 1$; here we trivially have $\gcd(n, m) = 1$, and the Eisenstein criterion with prime $p$ directly applies to show $X^n - p$ is irreducible over $\Bbb Q$. However, Eisenstein does not apply to $X^n - p^m$ when $m \ge 2$; therefore we seek another line of analysis.

We consider the field extension of $\Bbb Q$ formed by adjoining $\sqrt [n] p$; that is, the field $\Bbb Q(\sqrt [n] p)$; since $X^n - p$ is irreducible over $\Bbb Q$, we have from elementary considerations that

$\Bbb Q(\sqrt[n] p) \simeq \Bbb Q[X] / \langle X^n - p \rangle, \tag 5$

where

$\langle X^n - p \rangle = (X^n - p) \Bbb Q[X] \tag 6$

is the principal ideal generated by $X^n - p$ in $\Bbb Q[x]$. Furthermore it follows from (5) that

$[\Bbb Q(\sqrt[n] p): \Bbb Q] = n; \tag 7$

that is, the dimension of $\Bbb Q(\sqrt[n] p)$ as a vector space over $\Bbb Q$ is $n$.

We next observe that

$\sqrt[n]{p^m} = (\sqrt[n] p)^m \in \Bbb Q(\sqrt[n] p), \tag 8$

from which it is seen that

$\Bbb Q(\sqrt[n] {p^m}) \subset \Bbb Q(\sqrt[n] p); \tag 9$

we now show that under the hypothesis

$\gcd(n, m) = 1 \tag{10}$

we also have

$\sqrt[n] p \in \Bbb Q(\sqrt[n]{p^m}); \tag{11}$

for, as is well-known, (10) implies that there are $a, b \in \Bbb Z$ such that

$an + bm = 1; \tag{12}$

then

$\sqrt[n] p = (\sqrt[n] p)^1 = (\sqrt[n] p)^{an + bm} = (\sqrt[n] p)^{an} (\sqrt[n] p)^{bm}$ $= ((\sqrt[n] p)^n)^a ((\sqrt[n] p)^m)^b = p^a (\sqrt[n] {p^m})^b \in \Bbb Q(\sqrt[n]{p^m}); \tag{13}$

since, then

$\sqrt[n] p \in \Bbb Q(\sqrt[n]{p^m}), \tag{14}$

it follows that

$\Bbb Q(\sqrt[n] p) \subset \Bbb Q(\sqrt[n]{p^m}), \tag{15}$

and thus, via (9),

$\Bbb Q(\sqrt[n] p) = \Bbb Q(\sqrt[n]{p^m}), \tag{16}$

and now from (7) we see that

$[\Bbb Q(\sqrt[n]{p^m}): \Bbb Q] = n \tag{17}$

as well.

Now if $X^n - p^m$, $m > 1$, were reducible over $\Bbb Q$, we could write

$X^n - p^m = r(X) s(X), \tag{18}$

with

$r(X), s(X) \in \Bbb Q[X], 0 < \deg r, \deg s < n, \tag{19}$

then

$r(\sqrt[n]{p^m}) s(\sqrt[n]{p^m}) = (\sqrt[n]{p^m})^n - p^m = p^m - p^m = 0, \tag{20}$

so that at least one of

$r(\sqrt[n]{p^m}), s(\sqrt[n]{p^m}) = 0; \tag{21}$

then in the light of (19), we may conclude that

$[\Bbb Q(\sqrt[n]{p^m}): \Bbb Q] < n, \tag{22}$

since $\sqrt[n]{p^m}$ satisfies a polynomial of degree less than $n$. This of course stands in contrediction to (17); therefore,

$X^n - p^m \tag{23}$

is irreducible over $\Bbb Q$.

Answer 2

This is a nice example for using Newton polygons, tool from which criteria such as Eisenstein criterion follow. In this case the Newton polygon of $$ f(x)=x^n-p^m $$ with respect to prime $p$ is a line segment with end points $(0,m)$ and $(n,0)$. Now if there are no other integer points on this line, then by Dumas criterion the $f(x)$ is irreducible in $\mathbb{Q}[x]$ . Since you have $(m,n)=1$, there are no such integer points, and the polynomial is indeed irreducible.

Basically you can translate this into generalized version of Eisenstein's criterion in terms of prime powers, such as:

If $f(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0 \in \mathbb{Q}[x]$, such that $a_0a_n\neq 0$. Then if there is a prime $p$ such that $p \nmid a_n$, $p^k\mid a_i$ for $i<n$, $p^{k+1} \nmid a_0$, and $(k,n)=1$, then $f(x)$ is irreducible in $\mathbb{Q}[x]$.

Choosing $k=1$ gives you the familiar Eisenstein's criterion, while in your case you need $k=m$.