It can easily be shown that if a complex polynomial $P$ leaves invariant $\mathbb{Z}$ ($P(\mathbb{Z}) \subseteq \mathbb{Z}$) then it must be a linear combination (with integer coefficients) of Hilbert polynomials $H_k$, i.e. polynomials of the form : $$ H_k(X) : = \frac{X(X-1)\cdots (X-k+1)}{k!} $$

Now, what happens when $P$ stabilizes an entire lattice, say the Gaussian integers $\mathbb{Z}[i] $, i.e $P(\mathbb{Z}[i] ) \subseteq \mathbb{Z}[i] $ ?

Pretty clearly, the set $\mathcal{A}$ of all such polynomials is an additive sub-group of $\mathbb{Q}[i][X]$ (and even a $\mathbb{Z}[i]$ module). Such polynomials can be expressed as linear combinations (with coefficients in $\mathbb{Z}[i]$) of the polynomials $H_k$ (same proof as in the case of $\mathbb{Z}$). It is also straightforward to see that $\mathbb{Z}[i] [X]$ is contained in $\mathcal{A}$. However equality does not arise, since the polynomial $ \widetilde{H}_2(X): = (1+i)\frac{X(X-1)}{2} = (1+i)H_2(X)$ does not lie in $\mathbb{Z}[i] [X]$ and yet verifies the property of leaving invariant the Gaussian integers.

Question : Can we give a precise description of such polynomials, say an explicit basis of $\mathcal{A}$ (seen as a module) ?


Your question is related to the study of (generalized) numerical polynomials: If $R$ is an integral domain and $K$ the field of fractions of $R$, then the set ${\rm Int}(R) := \{f \in K[x]: f(R) \subseteq R\}$ is a subdomain of $K[x]$, whose elements are called the numerical polynomials over $R$ (in one variable $x$).

The domain ${\rm Int}(R)$ has been the subject of a great deal of research: Original work in the area was entirely centered on the case where $R$ is the ring of integers, and was motivated by interpolation problems in the early days of calculus. It was only in 1919 that A. Ostrowski and G. Pólya first considered numerical polynomials in their own right, though focused on the case where $R$ is the ring of integers of a number field. In particular, they could show, in this context, that ${\rm Int}(R)$ has a regular basis $(f_k)_{k \ge 0}$ as an $R$-module if and only if the products of prime ideals of $R$ of every given norm are principal, which is certainly true if $R$ is a PID ("regular" means that $\deg f_k = k$ for all $k$): Their proof is constructive, so the answer to your question ("Can we give a precise description of such polynomials etc.?") is yes.

For further details and results, you may want to have a look to P.-J. Cahen and J.-L. Chabert's monograph, Integer-valued polynomials, Math. Surveys Monogr. 48, Amer. Math. Soc., 1997. More specifically, see Remark II.1.5(ii) for an "explicit basis".

  • $\begingroup$ I would not call this basis explicit. It is constructed inductively, and studying concrete properties of this basis (say, the growth of coefficients) is non-trivial. $\endgroup$ – Fedor Petrov Jul 24 '17 at 6:07
  • $\begingroup$ @FedorPetrov. I wouldn't either, that's why I wrapped the term explicit basis in quotation marks. However, I'm not sure that what the OP is asking for is possible if taken literally (whatever it means to have a closed form for a basis), and on the other hand, the construction in Remark II.1.5(ii) of Cahen & Chabert's book qualifies, for me, as "a precise description of such polynomials" (I'm quoting the OP). $\endgroup$ – Salvo Tringali Jul 24 '17 at 6:13
  • $\begingroup$ Thank you @SalvoTringali ! do you have any idea on how to describe the generators of the ideals $\mathfrak{J}_n$ (using the notations of the book you quoted)? (say induction relation, or growth of the modules, as FedorPetrov pointed out) I have shown that for instance $ \mathfrak{J}_2 = \frac {1+i}{2}\mathbb{Z}[i] $ $\endgroup$ – MarcoSan Jul 24 '17 at 20:09
  • $\begingroup$ @MarcoSan If your question means, "Do you know of an algorithm to compute the coefficients $d_n$ in Remark II.1.5(ii)?", the answer is, "No, not off the top of my head." But if you insist to have a "more explicit description" of a basis, then take a look at Proposition II.3.14 and Exercise 24 on p. 48 of the book: This is perhaps closer to the spirit of what you're asking for, and may convince you that the kind of "nice closed form for a basis" you're implicitly hoping for is not so likely. $\endgroup$ – Salvo Tringali Jul 24 '17 at 21:48

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