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I am trying to prove that the bivariate polynomial $ x^m y^n + x^p y^q+ 1 $ is irreducible in $ \mathbb{C}[x, y] $, where $ m, n, p, q \in \mathbb{Z}_+ $, $m$ is coprime to $p$ , $n$ is coprime to $q$ , and $mq-np\neq 0$.

For special cases of the question (such as $m=n=1,p=2,q=3$), we can directly use exhaustive search and the method of undetermined coefficients to determine its irreducibility. However, for the general case, the method of undetermined coefficients no longer works.

How can we solve this problem, or are there any relevant papers that could be valuable for reference?

Thank you so much!

Tengel
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    Pick $m=p, n=q$, gcd$(m,n)=1$, and the larger coefficient odd. Then the corresponding polynomial is never irreducible. Simply because we can write it as a univariate polynomial in $xy$ with odd degree (which never is irreducible). – Severin Schraven Dec 19 '24 at 22:08
  • @SeverinSchraven You mean $m = n$, $p = q$, and the degree parity is irrelevant since we factor over $\Bbb C$ :-) – Amateur_Algebraist Dec 19 '24 at 22:11
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    @Amateur_Algebraist Ahh yes, I thought we were over the reals. Of course, I got the equalities wrong :( – Severin Schraven Dec 19 '24 at 22:17
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    Not really that relevant but the condition $(m-p)^2+(n-q)^2\ne 0$ is unnecessary. This is true iff $m=p$ and $n=q$, but then $mq - np = 0$ which is already disallowed. – Philip Speegle Dec 19 '24 at 23:15
  • @Philip Speegle How to prove the irreducibility of this polynomial when $mq-np\neq0$? Can you please provide some proof ideas or related papers? Thank you very much! – Tengel Dec 20 '24 at 10:59

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Easy counterexample to the original question (without the condition $mq-np\not =0$). It shows that such polynomials can be reducible even over $\Bbb Z$: $$x^5 y^5 + xy + 1 = (x^3y^3 - x^2y^2 + 1)(x^2y^2 + xy + 1)$$

(thanks, Sage! I just generated the tuples of coefficients using list(Tuples(range(1,10),4)), filtered them and then factor()'ed the polynomials over the symbolic ring SR. No thinking was required. Now if $mq-np = 0$, by @SeverinSchraven's observation the polynomial in question would be a univariate polynomial in disguise, thus always reducible.)

Amateur_Algebraist
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