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?


2 Answers 2


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 $\mathbb{Q}[x,y]$.

  • $\begingroup$ Thank you for your helpful answer! I have no objection to the overview. I would like to go through your comments in detail. $\endgroup$ May 22, 2022 at 8:51
  • $\begingroup$ Could I prove the Lemma 1 if I take $d=2^{4u}$ where $u$ and $m$ are relatively prime? $\endgroup$ May 23, 2022 at 7:44
  • $\begingroup$ @certain_integral Yes, I think that works fine since $d^n=2^{4un}$ won't be $p$-th perfect power for any odd prime $p \mid m$. $\endgroup$
    – Sil
    May 23, 2022 at 10:06
  • 1
    $\begingroup$ 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 $\Longleftrightarrow$ ${\rm GCD}(m,n)=2^l$. $\endgroup$ May 23, 2022 at 15:49
  • $\begingroup$ @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 $\endgroup$
    – 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}


  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 $n$.

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