I've decided to repost this question, which originally appeared on MSE, here. It is part of my series of open problems for enthusiasts and, while I understand this crowd is focused on professionals, this is the first open problem I've posted which I have no idea where to even begin. Any commentary by an interested professional would probably be hugely educational for me...

A complex polynomial is called Hurwitz if each one of its roots has a negative real part. Such polynomials are of central importance in linear control theory. If $p$ an $q$ are two polynomials, we denote their composition by $p\circ q$, which is defined as the polynomial $r$ such that $r(x) = p(q(x))$ for all complex $x$. Our first question is immediate:

  1. When is the composition of two Hurwitz polynomials itself Hurwitz?

Suppose $r$ is Hurwitz and admits the representation $r = p_1\circ p_2\circ \dots \circ p_n$ ($n \geq 2$) for Hurwitz polynomials $p_i$ of degree strictly less than $r$. In this case we say $r$ is Hurwitz factorizable and call this representation the Hurwitz factorization of $r$ with factors $p_i$. The set is nonempty--in fact, it's uncountably infinite--for it contains $x^4 +2ax^2 + a^2 +a = (x^2+a)\circ(x^2+a)$ for all $a > 0$. If a Hurwitz polynomial does not possess a Hurwitz factorization, it is called irreducible.

Furthermore, if one or more factors of a Hurwitz factorization are themselves Hurwitz factorizable, another representation of the first factorization is possible using the factors of the second. If this process of factoring is continued, the degree condition implies it must terminate after a finite number of steps (if the degree condition is not enforced it need not! Consider $x+a = (x+a/2)\circ (x+a/2)$.) Thus this process ends exactly when every factor is irreducible. This invites question

  1. Given an arbitrary Hurwitz polynomial,

    (a) determine whether or not it is irreducible and, if not,

    (b) provide an algorithm to calculate an irreducible Hurwitz factorization.

  • $\begingroup$ From the way your question 2(b) is phrased, "calculate its irreducible Hurwitz factorization," you seem to be implying that the factorization is unique. Is that obvious? $\endgroup$ Apr 27, 2017 at 19:54
  • $\begingroup$ @JoeSilverman No. In fact I would conjecture the factorization is not unique in general. I have fixed the question to make it more clear. $\endgroup$
    – JMJ
    Apr 27, 2017 at 19:59

1 Answer 1


Are you aware of results on composition factorization of complex polynomials that does not require the Hurwitz condition. For example:

Composition factors of polynomials, A. F. Beardon, Complex Variables, Theory and Application 43 (2001), 225-239.

The abstract says, in part: A complex polynomial is composite if it can be expressed as the composition of two nonlinear polynomials; otherwise it is prime. It is easy to see that any polynomial $p$ can be expressed as a composition, say $p=p_1\circ p_2\circ\cdots\circ p_r$, where the $p_j$ are prime. This decomposition into prime polynomials is not unique; however, in 1926 Ritt showed that the set of degrees of the $p_j$ is uniquely determined by $p$. Here we show that the vector of degrees determines the $p_j$ uniquely.

Forward and backward referencing Beardon's paper should give you an overview of what's known about composition factorization of polynomials in general, and possibly some of the tools can be applied under the more restrictive condition that the factors be Hurwitz, as in your question.


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