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Let $a, b \in \mathbb{Q}$, with $a\neq b$ and $ab\neq 0$, and $n$ a positive integer.

Is the polynomial $\bigl(X(X-a)(X-b)\bigr)^{2^n} +1$ irreducible over $\mathbb{Q}[X]$?

I know that $\bigl(X(X-a)\bigr)^{2^n} +1$ is irreducible over $\mathbb{Q}[X]$, but I have a hard time generalizing my proof with three factors.

PS: This is not homework (and may even be open).

user84673
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    Why is this tagged "cyclotomic polynomial"? – Zev Chonoles Jul 02 '13 at 02:44
  • @Zev, because the proof I have for 2 factors uses cyclotomic polynomials. – user84673 Jul 02 '13 at 02:49
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    Cool question, got me thinking. Would you mind explaining how you know that ${(X(X - a))}^{2^n} + 1$ is irreducible over Q? – Robert Lewis Jul 02 '13 at 02:51
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    Robert, not sure I can detail enough in the comment box, but if $a=0$ it's trivial (cyclotomic), suppose $a\neq 0$, let $z \in \mathbb{C}$ a root, and $\mathbb{Q}[z]$ is a field extension of $\mathbb{Q}[z(z-a)]$. Looking at the minimal polynomial of $z$ over $\mathbb{Q}$ we show that $\mathbb{Q}[z(z-a)] \neq \mathbb{Q}[z]$ and we conclude with the fact that $\mathrm{dim}_{\mathbb{Q}}\mathbb{Q}[z(z-a)]=2^n$. – user84673 Jul 02 '13 at 03:09
  • @user84673: can you do anything with the case $b = 0$, i.e. with ${{(X^2(X - a))}^{2^n}} + 1$? – Robert Lewis Jul 02 '13 at 06:39
  • @user84673: Could you explain why $z \not\in \mathbb{Q}(\zeta)$, or, what is the same, why $x^2-ax-\zeta$ is irreducible over $\mathbb{Q}(\zeta)$, where $\zeta= e^{2\pi i/2^n}$? – ladisch Jul 02 '13 at 09:08
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    This is far from an open question since it is a special case of an old conjecture of Schur proved long time ago by Seres. If I understand well you want to use Capelli's Lemma in order to prove the irreducibility. You said (but didn't show us!) that you were succesful for the product of two distinct polynomials $X$ and $X-a$. I'm eager to see a proof on this line at least for some particular cases like these. (I don't know a proof of Schur conjecture using Capelli's Lemma, if there is one.) –  Jul 02 '13 at 17:59
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    Schur's conjecture was the following: if $f(X)=X^{2^n}+1$, $n\ge 1$, and $g(X)=(X-a_1)\cdots(X-a_m)$ with $a_i\in\mathbb Z$ distinct, then $f(g(X))$ is irreducible over $\mathbb Q$. –  Jul 02 '13 at 19:05
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    Thanks YACP for the positive answer and reference. I started the the outline of the proof for the case $X(X-a)$, 3rd comment from the top. Here is the end of the outline. Let $P$ be the minimal polynomial mentioned, and suppose that the field extension is trivial. Denote $b=(a/2)^{2^{n-1}}$ and $c=b^2+1$, with some calculation involving $P(X)P(a-X)$ you get $b^4=(c-1)(c+1)$, and show that it is a contradiction. – user84673 Jul 02 '13 at 19:15

2 Answers2

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This is at least a partial answer for $a,b$ integers. As stated in comments, Schur's conjecture was: if $f(X)=X^{2^n}+1$ and $g(X)=(X-a_1)\cdots(X-a_m)$ with $a_i\in\mathbb Z$ distinct, then $f(g(X))$ is irreducible over $\mathbb{Q}$. I. Seres proved this statement in 1963 in Über die Irreduzibilität gewisser Polynome.

Note: Soon after, in Irreducibility of Polynomials he proved the following generalization (though it does not directly apply to our problem, it requires at least 6 distinct integer roots):

Let $\{a_k\}$ be a sequence of rational integers $a_1<a_2<\dots<a_n$ with $n \geq 6$. Let $f(z)$ be the monic polynomial over the rational field $\Gamma_1$ which contains the unit $\gamma$ among its roots. Let $Q(x)$ be a monic polynomial with rational integer coefficients, of degree less than $n$. Then the polynomial $$ F(x)=f(Q(x)\prod_{k=1}^n(x-a_k)) $$ is irreducible over the rational field.

Here $\gamma$ is arbitrary nonreal unit of the $m$-th cyclotomic field $\Gamma_m$.

Sil
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    This question has been answered in comments, so moved it from unanswered queue. I invite anyone to extend/improve this, especially some more details to the first article/proof would be welcome - the article is in German so I did not dare to translate it... – Sil Sep 03 '24 at 18:55
  • Also I wonder if there is a way to use the Seres' result with integer roots and apply it for rational roots, I just don't see a way now. – Sil Sep 03 '24 at 19:30
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Call $f$ your polynomial. Then $f$ is not irreducible if and only if there exists a root $x\in\mathbb{Q}$ of $f$. If a such root exists then, for $n\geq 1$ we have $$0\leq(x(x-a)(x-b))^{2^n}=-1.$$

I leave the case $n=0$ to you.

user26857
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Vincenzo Zaccaro
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