Irreducibility of $x^n + p^m \in \mathbb{Q}[x]$ where $m, n$ coprime.

Question

Let $p$ be a prime number. From this question, we know that $x^n - p^m$ is irreducible over $\mathbb{Q}[x]$, where $m, n \in \mathbb{Z}_+$ are coprime. And from this question, we know that $x^n + p^{n + 1}$ is irreducible over $\mathbb{Q}[x]$.

Now pick up $m,n \in \mathbb{Z}_+$ relatively prime. I wonder if $ x^n + p^m $ is always irreducible over $\mathbb{Q}[x]$. Or more generally, for all $a \in \mathbb{Q}_+$ such that $ x^n - a \in \mathbb{Q}[x] $ irreducible, can we deduce that $$ x^n + a^m $$ irreducible over $\mathbb{Q}[x]$?

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For $x^n - a^m$, one may use field theory to obtain its irreduciblity: Let $\alpha := \sqrt[n]{a}$, then we have $\mathbb{Q}(\alpha^m) \subset \mathbb{Q}(\alpha)$, and $$ \mathbb{Q}(\alpha) = \mathbb{Q} \left( \alpha^{nx + my} \right) = \mathbb{Q} ( \alpha^{my} ) \subset \mathbb{Q}( \alpha^m ). $$ Apriori, $ \deg_{\mathbb{Q}} (\alpha^m) = \deg_{\mathbb{Q}} (\alpha) = n $. For $x^n + a^{n + 1}$, as a special case of $x^n + a^m$, this trick works (cf. this answer).

For $ x^n + a^m $, I want to follow the same trick. But things turns a little bit different. One root of $x^n + a^m$ is $\alpha^m \zeta_{2n}$, where $\zeta_{2n} := e^{2 \pi i / 2n}$ means the $(2n)$-th primitive root of unity. I tried to use Galois theory to deduce that $$ \mathbb{Q} (\alpha \zeta_{2n}) = \mathbb{Q} \left( \alpha^m \zeta_{2n} \right), $$ thus I considered the splitting field of $x^{2n} - a^2$, i.e. $K := \mathbb{Q} \left( \alpha, \zeta_{2n} \right)$. Now what we need to show has been reduced to $$ \operatorname{Gal} \left( K | \mathbb{Q} \left( \alpha \zeta_{2n} \right) \right) = \operatorname{Gal} \left( K | \mathbb{Q} \left( \alpha^m \zeta_{2n} \right) \right). $$ For $\sigma \in \operatorname{Gal} \left( K | \mathbb{Q} \left( \alpha \zeta_{2n} \right) \right)$, we know that $ \sigma(\alpha) = \alpha \zeta_{2n}^{2k} $, $\sigma(\zeta_{2n}) = \zeta_{2n}^\ell$, where $0 \le k < n$ and $\gcd(\ell, n) = 1$. But I cannot deduce that $\sigma$ fixes $\alpha^m \zeta_{2n}$ from it.

Answer 1

I wonder if $ x^n + p^m $ is always irreducible over $\mathbb{Q}[x]$. Here is a counterexample. We have $$ x^8+2^6=(x^4 + 4x^2 + 8)(x^4 - 4x^2 + 8). $$

Answer 2

The answer is yes. If you know a bit about Newton polygons and/or local fields, one of the answer of the first post you linked should also apply, turn first to the polynomial $x^n - p^m$:

The Newton polygon at $p$ of your polynomial is a single slope joining $(n,0)$ to $(0,m)$. The coefficient of the slope is then $\frac{m}{n}$ which is irreducible hence the polynomial is also irreducible (actually in $\mathbf{Q}_p$ but that is stronger than what we want). Indeed, assuming you know the fact (that you can actually show) that slopes' coefficients are valuation of the roots, and their $x$-lenght the number of such roots, a factorisation $P=QR$ with $Q$ non constant yields that $$\frac{u}{deg(Q)} = \frac{m}{n}$$ where $u$ is the difference of the valuations of the coefficient of $Q$, hence $n|deg(Q)$, but recall that $n=deg(P)$ so that we actually have equality since $deg(P)\ge deg(Q)$

Now for a general $a\in\mathbf{Q}_+$, assuming that $n$ is fixed, this reasonning implies that whenever $a=\prod_i p_i^{e_i}$ (allowing $e_i$ to be negative) is such that $n$ is coprime to at least one $e_i$, then indeed the polynomial $x^n - a^m$ is irreducible when $m$ is coprime to $n$.
Actually it is sufficient to have $\gcd(n,(e_i)_i)=1$. Indeed, assume the condition holds, and suppose for the sake of contradiction that we can factorise non trivially $x^n-a = PQ$ (which is a sufficient case, by setting $a'= a^m$). For any prime $p$ appearing in a, the Newton polygon remains a single slope, hence all roots of $x^n-a$ have the same valuation $e_i/n$. Now the factorisation is happening in $\mathbf{Q}$ hence the constant coefficient of $P$ and $Q$ have integral valuation at each prime, thus the degree of $P$ verifies that $deg(P)\cdot e_i/n \in \mathbf{Z}$, or in other words $n|deg(P)e_i$ for all $i$, hence $\frac{n}{\gcd(n,deg(P))} | e_i$ which contradicts our hypothesis.
For the minus case, if the $\gcd$ is not 1, say equal to $l$ we have a trivial factorisation by writing $X^n - a = (X^{n/l})^l - (a^{1/l})^l$ hence we have an equivalence.

For the plus case, the sufficient condition still holds, which means that since you supposed $X^n-a$ irreducible, the $\gcd$ condition holds, hence the polynomial $X^n+a^m$ is also irreducible by what I've written above.