# Does any cubic polynomial become reducible through composition with some quadratic?

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.

• Sorry, but why can't it be a product of an irreducible quadratic and an irreducible quartic? – mathworker21 Jul 15 at 20:27
• @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 $\mathbb{Q}$). – Yaakov Baruch Jul 15 at 20:33
• 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] – Joe Silverman Jul 15 at 22:12
• 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 $\mathbb{Q}$... – Yaakov Baruch Jul 15 at 23:09
• 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. – Aurel Jul 16 at 7:52

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.
• 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. – Yaakov Baruch Jul 16 at 14:40
• 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. – Greg Martin Jul 16 at 19:15