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What I mean to ask is this:

given an irreducible cubic polynomial $P(X)\in \mathbb{Z}[X]$ is there always a quadratic $Q(X)\in \mathbb{Z}[X]$ such that $P(Q)$ is reducible (as a polynomial, and then necessarily the product of 2 irreducible cubic polynomials)?

I did quite some testing and always found a $Q$ that does the job. For example:

$P=aX^3+b,\quad Q=-abX^2,\quad P(Q)=-b(a^2bX^3-1)(a^2bX^3+1)$

$P=aX^3-x+1,\quad Q=-aX^2+X,\quad P(Q)=-(a^2X^3-2aX^2+X-1)(a^2X^3-aX^2+1)$

and a particular hard one to find:

$P=2X^3+X^2-X+4,\quad Q=-8X^2+5X+1,\quad P(Q)=(16X^3-18X^2+X+3)(64X^3-48X^2-11x-2)$

Could there be a formula for $Q$ that works for all cases?

It feels to me that this may have a really basic Galois theoretic proof or explanation, but I can't figure it out.

Update. Maybe a general formula for $Q$ is close. For $P=aX^3+cX+d$ taking $Q=-adX^2+cX$ works.

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  • $\begingroup$ Sorry, but why can't it be a product of an irreducible quadratic and an irreducible quartic? $\endgroup$
    – mathworker21
    Commented Jul 15, 2021 at 20:27
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    $\begingroup$ @mathworker21: Then $P(Q)$ would have a quadratic root $\alpha$ and therefore $P$ would have a quadratic root $Q(\alpha)$, contradicting the irreducibility of $P$ (which means that a root would generate a field of degree $3$ over $\mathbb{Q}$). $\endgroup$
    – Yaakov Baruch
    Commented Jul 15, 2021 at 20:33
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    $\begingroup$ There are a number of interesting results known about "composition factorization" of polynomials. I realize this is not what you're asking, but one could ask "which cubic polynomials $f$ and $g$ have the property that $f\cdot g$ has a composition factorization. In any case, here are two reference that you might find useful: Beardon, A. F. Ng, T. W. On Ritt's factorization of polynomials. J. London Math. Soc. (2) 62 (2000), no. 1, 127–138. [MR1771856] Beardon, A. F. Composition factors of polynomials. Complex Variables Theory Appl. 43 (2001), no. 3-4, 225–239. [MR1820924] $\endgroup$
    – Joe Silverman
    Commented Jul 15, 2021 at 22:12
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    $\begingroup$ The case $P=aX^3+cX+d$ can be transformed through variable changes and rescalings, into the case $P=aX^3+bX^2+d$, giving then $Q=-(81a^3d+6ab^3)X^2-3b^2X-b/(3a)$. While I believe that there are always particular solutions over $\mathbb{Z}$, it's possible that general solutions can only be found over $\mathbb{Q}$... $\endgroup$
    – Yaakov Baruch
    Commented Jul 15, 2021 at 23:09
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    $\begingroup$ Over $\mathbb{Q}$, for $P = ax^3+bx^2+cx+d$ you get $Q = \frac{-27da^2 + 9bca - 2b^3}{27a}x^2 + \frac{3ca - b^2}{3a}x - \frac{b}{3a}$ by eliminating the $x^2$ term and applying your formula. $\endgroup$
    – Aurel
    Commented Jul 16, 2021 at 7:52

1 Answer 1

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You should refer to Lemma 10 (page-233) in this paper by Schinzel where he proves that for any polynomial $F(x)$ of degree $d$ we have a polynomial $G(x)$ of degree $d-1$ such that their composition is reducible.

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    $\begingroup$ To save others the scrolling: Lemma 10 is on page 233 of the paper. $\endgroup$
    – Alex B.
    Commented Jul 16, 2021 at 12:02
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    $\begingroup$ @AlexB. Indeed... I wish I saw your comment earlier! $\endgroup$
    – Yaakov Baruch
    Commented Jul 16, 2021 at 14:21
  • $\begingroup$ Sorry @Yaakov Baruch I should have added page number. I will add right now. $\endgroup$
    – Hhhhhhhhhhh
    Commented Jul 16, 2021 at 14:23
  • $\begingroup$ The integrality of coefficients is achieved there by a suitable choice of some $k$; I'm not sure, but it seems to me that $k$ could be a chosen to be a rational function of the coefficients (maybe a very messy one, using sums of squares of many things to avoid using the max function), in which case the resulting $H$ (our $Q$) would be given by one formula in the coefficients. $\endgroup$
    – Yaakov Baruch
    Commented Jul 16, 2021 at 14:40
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    $\begingroup$ See also my paper with Bober, Fretwell, and Wooley, Theorem 3.2, to see that there are infinitely many such quadratic polynomials $G(x)$. (The method is certainly based on that of Schinzel.) For those with a background in abstract algebra, I believe our proof has more intuition behind it. $\endgroup$
    – Greg Martin
    Commented Jul 16, 2021 at 19:15

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