TITLE: When is the polynomial $X^a+Y^b \in \mathbb{Q}[X,Y]$ irreducible?
QUESTION [8 upvotes]: I am studying the irreducibility of the polynomial $X^a+Y^b$.
I proved that if $X^a+Y^b \in \mathbb{C}[X,Y]$, the irreducibility is equivalent to that $a$ and $b$ is relatively prime. I also proved that $X^a+Y^b \in \mathbb{R}[X,Y]$, it is equivalent to that the ${\rm GCD}(a,b)$ is equal to $1$ or $2$.
I expect that when $X^a+Y^b \in \mathbb{Q}[X,Y]$, if and only if there is a non-negative integer $l$ such that ${\rm GCD}(a,b)=2^l$, the polynomial is irreducible. However, I don't have any ideas for the proof. Would you tell me any ideas or hints?

REPLY [2 votes]: Previous answer of mine was more of a brute force approach, here is slightly more systematic way inspired by your observation in comments. We have
\begin{align}
x^m+y^n &\text{ irreducible in }\mathbb{Q}[x,y] \\
&\stackrel{1}{\iff} x^m+y^n \text{ irreducible in }\mathbb{Q}(y)[x]\\
&\stackrel{2}{\iff} -y^n \not\in \mathbb{Q}(y)^p \text{ for prime }p\mid m \text{ and }y^n\not\in 4\mathbb{Q}(y)^4 \text{ when } 4\mid m\\
&\stackrel{3}{\iff} -y^n \not\in \mathbb{Q}(y)^p \text{ for prime }p\mid m\\
&\stackrel{4}{\iff} -y^n \neq a^py^{kp} \text{ for all primes }p\mid m \text{ and } k\in\mathbb{N},a\in \mathbb{Q}\\
&\stackrel{5}{\iff} -1\neq a^p \text{ or }n\neq kp \text{ for all primes }p\mid m \text{ and } k\in\mathbb{N},a\in \mathbb{Q}\\
&\stackrel{6}{\iff} p=2 \text{ or } p\nmid n \text{ for all primes }p\mid m\\
&\stackrel{7}{\iff} \gcd(m,n)=2^k, k\geq 0\\
\end{align}
where

Show that non-trivial factorization of polynomial $x^m+y^n$ in one ring gives non-trivial factorization in the other one and vice-versa.
Application of When is $X^n-a$ is irreducible over F? for $F=\mathbb{Q}(y)$ field of rational functions.
Using that $y^n=4r(y)^4$, $r(y)\in \mathbb{Q}(y)$ implies $1=4a^4$ for  some $a\in \mathbb{Q}$, i.e. $a=\pm 1/\sqrt{2} \in \mathbb{Q}$, impossible.
We must have a polynomial equality in fact, but then it follows it must be of form $(ay^k)^p$ for some $a\in \mathbb{Q}$.
Just comparing the coefficients and exponents on both sides.
For odd $p$ we can always choose $a=-1$, no choice works for $p$ even (i.e. $p=2$).
Just another way to say only prime $p=2$ is allowed to divide both $m$ and $n$.