# When is the polynomial $X^a+Y^b \in \mathbb{Q}[X,Y]$ irreducible?

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?

This follows from ($$*$$) $$x^m+c$$ being irreducible in $$\mathbb{Z}[x]$$ iff $$-c$$ is not $$p$$-th power for any prime $$p\mid m$$ and $$c \neq 4z^4$$ if $$4\mid m$$ for any integer $$z$$, see the general statement in When is $X^n-a$ is irreducible over F? . Here the field is $$F=\mathbb{Q}$$ (and clearly if integer $$-c$$ is power of rational number, then it is in fact a perfect power of an integer). Following proof (arguably not the simplest one) is based on this fact.

Lemma 1. If $$\gcd(m,n)=2^k$$, then there are infinitely many integers $$d$$ such that $$F(x,d)=x^m+d^n$$ is irreducible in $$\mathbb{Q}[x]$$.

Proof. We need infinitely many $$d$$ such that conditions in ($$*$$) are satisfied, we consider $$d=2^{4u}v$$ where $$v>0$$ is odd and square-free. Clearly $$-d$$ is not a perfect square, and it is also not an odd perfect power for $$p\mid m$$ ($$v$$ is square-free, so $$p$$ would have to divide $$n$$, impossible by assumption). Finally $$d^n=4z^4$$ is also impossible as exponent of $$2$$ is multiple of $$4$$ on the left hand side, but it is congruent to $$2 \bmod 4$$ on the right hand side.$$\square$$

Lemma 2. Let $$u(x,y)\in \mathbb{Q}[x,y]$$. If $$u(x,d)\in \mathbb{Q}[x]$$ is degree $$0$$ for infinitely many $$d \in \mathbb{Q}$$, then $$u(x,y)=v(y)$$ (i.e. $$u$$ is independent of $$x$$).

Proof. For the sake of contradiction assume there is a non-zero term at $$x^k$$, $$k>0$$ in $$u(x,y)\in (\mathbb{Q}[y])[x]$$, say $$a(y)x^k$$. Since $$u(x,d)$$ is zero degree infinitely often, we must have $$a(d)=0$$ infinitely often, but that makes $$a(y)$$ zero polynomial, hence $$a(y)x^k=0$$, contradiction.$$\square$$

Lemma 3. Let $$F(x,y)\in \mathbb{Q}[x,y]$$. If $$F(x,d)$$ is irreducible in $$\mathbb{Q}[x]$$ for infinitely many $$d\in \mathbb{Q}$$, then $$F(x,y)=f(y)g(x,y)$$ where $$g(x,y)$$ is irreducible in $$\mathbb{Q}[x,y]$$.

Proof. Take any factorization $$F(x,y)=f(x,y)g(x,y)$$ with $$g(x,y)$$ irreducible. Then since $$F(x,d)=f(x,d)g(x,d)$$ is irreducible in $$\mathbb{Q}[x]$$ for infinitely many $$d$$, lemma 2 implies $$f(x,y)=f(y)$$ or $$g(x,y)=g(y)$$. If $$f(x,y)=f(y)$$ then $$F(x,y)=f(y)g(x,y)$$ as desired. If $$g(x,y)=g(y)$$, it's easy to see that $$f(x,y)$$ cannot have non-trivial factorization where both factors depend on $$x$$ - indeed if $$f(x,y)=u(x,y)v(x,y)$$ non-trivial factorization and both $$u,v$$ depend on $$x$$, then $$F(x,y)=u(x,y)(v(x,y)g(y))$$ and by the above argument one of $$u(x,y)$$ or $$v(x,y)g(y)$$ must be independent on $$x$$, contradiction. Hence $$f(x,y)=u(x,y)v(y)$$ with $$u$$ irreducible and we have found $$F(x,y)=u(x,y)(v(y)g(y))$$ which is of desired form. $$\square$$

Combining together we get the final criterion.

Theorem. Let $$m,n$$ be positive integers. Then $$F(x,y)=x^m+y^n$$ is irreducible in $$\mathbb{Q}[x,y]$$ if and only if $$\gcd(m,n)=2^k$$ for integer $$k \geq 0$$.

Proof. "$$\Leftarrow$$" If $$\gcd(m,n)=2^k$$, by lemma 1 $$F(x,d)$$ is irreducible infinitely often, and by lemma 3 we must have $$F(x,y)=f(y)g(x,y)$$ with $$g(x,y)$$ irreducible. But there was nothing special in choice of $$x$$ in lemma 1, hence by symmetric argument $$F(x,y)=u(x)v(x,y)$$ with $$v(x,y)$$ irreducible. Comparing the two factorizations, we must have either $$g(x,y)\mid v(x,y)$$ or $$g(x,y)\mid u(x)$$.

If $$g(x,y)\mid v(x,y)$$, then since $$v(x,y)$$ is irreducible, we have in fact $$g(x,y)=qv(x,y)$$ for $$q\in \mathbb{Q}$$ (and clearly $$q\neq 0$$). The above implies $$f(y)q=u(x)$$, which in turn implies both $$f,u$$ must be constant polynomials, so $$F(x,y)=ag(x,y)=bv(x,y)$$ is irreducible.

If $$g(x,y)\mid u(x)$$, then $$g(x,y)=g(x)$$ and we have $$x^m+y^n=f(y)g(x)$$. Next consider the highest degree terms $$a_px^p$$ and $$b_qy^q$$ in $$f$$ and $$g$$ respectively. Clearly, we have $$a_pb_qx^py^q=x^m$$ or $$a_pb_qx^py^q=y^n$$. This immediately implies $$p=0$$ or $$q=0$$, i.e. $$f$$ or $$g$$ is a constant polynomial. Hence $$x^m+y^n$$ is either $$af(y)$$ or $$bg(x)$$, impossible.

"$$\Rightarrow$$" If $$\gcd(m,n)=2^k d$$ for $$d>1$$ odd, put $$m=dm',n=dn'$$ and notice $$x+y\mid (x^{m'})^d+(y^{n'})^d=F(x,y)$$ by well-known factorization $$x^d+y^d=\left(x+y\right)\sum \left(-1\right)^ix^{d-i}y^i$$ for $$d$$ odd. So $$F(x,y)$$ is reducible. $$\square$$

Note: One implication above can be generalized for example as follows:

Let $$F(x,y)=a(x)+b(y)$$ and $$a(x),b(x) \in \mathbb{Q}[x]$$. Furthermore the two conditions are satisfied:

1. $$F(x,d)$$ is irreducible in $$\mathbb{Q}[x]$$ for infinitely many $$d\in \mathbb{Q}$$
2. $$F(d,x)$$ is irreducible in $$\mathbb{Q}[x]$$ for infinitely many $$d\in \mathbb{Q}$$

Then $$F(x,y)$$ is irreducible in $$\mathbb{Q}[x,y]$$.

• Thank you for your helpful answer! I have no objection to the overview. I would like to go through your comments in detail. May 22, 2022 at 8:51
• Could I prove the Lemma 1 if I take $d=2^{4u}$ where $u$ and $m$ are relatively prime? May 23, 2022 at 7:44
• @certain_integral Yes, I think that works fine since $d^n=2^{4un}$ won't be $p$-th perfect power for any odd prime $p \mid m$.
– Sil
May 23, 2022 at 10:06
• Thank you! Also, I came up with another proof for my question by using the theorem written in the link you showed : $x^n−c$ is irreducible over $F$ $\Leftrightarrow$ $c \in F^p$ for all primes $p|n$ and $c \in −4F^4$ when $4 | n$. By thinking $x^m+y^n$ is in $\mathbb{Q}(y)[x]$, where $\mathbb{Q}(y)$ is the rational function field, I could derive that $x^m+y^n \in \mathbb{Q}[x,y]$ is irreducible $\Longleftrightarrow$ ${\rm GCD}(m,n)=2^l$. May 23, 2022 at 15:49
• @certain_integral That's a good point, not sure why I didn't do it this way... I've tried to formulate this in a separate answer now as well, this answer was more of a brute force approach
– Sil
May 26, 2022 at 21:17

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

1. 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.
2. Application of When is $X^n-a$ is irreducible over F? for $$F=\mathbb{Q}(y)$$ field of rational functions.
3. 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.
4. We must have a polynomial equality in fact, but then it follows it must be of form $$(ay^k)^p$$ for some $$a\in \mathbb{Q}$$.
5. Just comparing the coefficients and exponents on both sides.
6. For odd $$p$$ we can always choose $$a=-1$$, no choice works for $$p$$ even (i.e. $$p=2$$).
7. Just another way to say only prime $$p=2$$ is allowed to divide both $$m$$ and $$n$$.