Let us consider the set of all polynomials with the following properties:
i) all coefficients are integer;
ii) the leading coefficient equals one;
iii) all zeros are real and simple and belonging to $[-1.99,1.99] $.
Is this set finite or infinite?
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6$\begingroup$ It is finite. This is a well known problem and an analog of Kronecker's theorem: the integer monic polynomials having all their roots in $[-2,2]$ are exactly the products of the minimal polynomials of algebraic integers of the form $2\cos(2\pi \, m/d)$. The finiteness then follows once you know also that all these numbers having $(m,d) = 1$ for a given $d$ are algebraic conjugates. $\endgroup$– Vesselin DimitrovCommented Oct 28, 2014 at 4:57
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1$\begingroup$ As to the proof of the statement, it is an application of pigeonholing. Write the roots as $2\cos(2\pi \alpha)$ and consider for $n \in \mathbb{N}$ the monic polynomials with simple roots at $2\cos(2\pi n \alpha)$. Show that it is an integer polynomial with coefficients bounded uniformly in $n$, hence some two such polynomials must coincide, and the conclusion follows. $\endgroup$– Vesselin DimitrovCommented Oct 28, 2014 at 5:00
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$\begingroup$ @ Vesselin Dimitrov: Can you kindly elaborate your comment as an answer, giving references? Note that the interval under consideration is a proper subset of $[-2,2]$. $\endgroup$– user64494Commented Oct 28, 2014 at 5:21
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$\begingroup$ I think this is related to a concept called the monic integer transfinite diameter. $\endgroup$– Gerry MyersonCommented Oct 28, 2014 at 5:34
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1$\begingroup$ @GerryMyerson: Yes, and also to the usual transfinite diameter of $[-2,2]$ being $1$. Fekete's theorem states more generally that a compact set $K \subset \mathbb{C}$ stable under complex conjugation contains only finitely many Galois orbits of algebraic integers if its transfinite diameter is smaller than $1$, while conversely, every open neighborhood of $K$ contains infinitely many such orbits if the transfinite diameter exceeds $1$. $\endgroup$– Vesselin DimitrovCommented Oct 28, 2014 at 9:31
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1 Answer
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It is finite. This follows by combining two separate results:
- If $p \in \mathbb{Z}[x]$ is monic integer having all its complex roots lying in $[-2,2]$, then all these roots are of the form $2\cos(2\pi q)$ with $q \in \mathbb{Q}$.
For the proof of this, write $p = \prod_{i=1}^d (x - 2\cos(2\pi \, t_i))$ and observe that the sequence of degree $d$ monic polynomials $p_n := \prod_{i=1}^d(x - 2\cos(2\pi n t_i))$ is also in $\mathbb{Z}[x]$ and that, of course, it has bounded coefficients (in terms of $d$ alone, independently of $n$). Thus there must be two $p_n = p_{n'}$ with $n \neq n'$; considering their roots, we get the claim with $q(n-n') \in \mathbb{Z}$.
- For a given $d \in \mathbb{N}$, the monic polynomial $P_d$ with simple roots at the set $\{2\cos(2\pi m/d) \mid (m,d) = 1\}$ is in $\mathbb{Z}[x]$ and irreducible.
This follows from the corresponding irreducibility property of the cyclotomic polynomial $\Phi_d(z)$, which can be written as $z^{\deg{P_d}} P_d(z+1/z)$.
Combining these two statements, you can see that your set consists precisely of the products of pairwise different minimal polynomials of those $2\cos(2\pi / d)$ that do not exceed $1.99$. There are only finitely many such $d$.
answered Oct 28, 2014 at 5:35
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$\begingroup$ @ Vesselin Dimitrov: You wrote "Combining these two statements, you can see that your set consists precisely of the products of pairwise different minimal polynomials of those $2\cos(2π/d)$ that do not exceed 1.99. There are only finitely many such d". However, "consists" does not mean "coinsides". Can you explain that place in detail? I still don't see any references. $\endgroup$ Commented Oct 28, 2014 at 5:43
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2$\begingroup$ @user64494: Isn't the coincidence clear? if a polynomial has all roots lying in [-1.99,1.99], then Vesselin Dimitrov explains why it is a product of certain special polynomials $Q_d$, all distinct (the $d$'s are such that $\cos(2\pi/d)\le 1.99$ so there are finitely many of them). Conversely for a product of distinct $Q_d$'s satisfying the property, you can check that the roots are simple and all lie in [-1.99,1.99], so you're done, right? $\endgroup$ Commented Oct 28, 2014 at 7:07
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$\begingroup$ Vote down is mine. Unfortunately, Vesselin Dimitrov does not explain why it is a product of certain special polynomials$ Q_d$ $\endgroup$ Commented Oct 28, 2014 at 8:55
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$\begingroup$ @ Vesselin Dimitrov. Can you kindly base your ungrounded statement "Combining these two statements, you can see that your set consists precisely of the products of pairwise different minimal polynomials of those $2\cos(2π/d)$ that do not exceed 1.99"? I still don't see any references in your edited answer. $\endgroup$ Commented Oct 28, 2014 at 8:59
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2$\begingroup$ @user64494: The answer has not been edited, I had simply deleted it after you made your first comment above, as I didn't see how I could help you, and anyway felt that my original comment already answered the question. References to what? This is a very special case of a well known theorem of Fekete; if you read German, you can see Uber die Verteilung der Wurzeln bei gewissen algebraischen Gleichungen mit ganzzahligen Koeffizienten (Math. Zeitschrift, 1923). As to my ungrounded statement, I am sorry, but I do not know what I can add to this explanation. $\endgroup$ Commented Oct 28, 2014 at 9:21