Define a finite set of polynomials over a field $K$ to cover $K$ if the images of the polynomials, viewed as functions from $K$ to itself, have union the whole set.

Define a minimal cover to be a finite set of polynomials that cover a field, but such that no proper subset covers that field.

Can you classify all minimal covers of $\mathbb Q_p$?

A minimal cover of a number field must consist of just a single, linear polynomial. Indeed, for most $y$, for each polynomial $f$ in the set, $f(x)-y$ is an irreducible polynomial in $x$ by Hilbert irreducibility. $y$ is in the image of some $f$ so some irreducible polynomial $f(x)-y$ has a root, so is linear, so $f(x)$ is linear. A single linear polynomial covers the set so no other polynomials are needed. This is a special case of this argument.

This does not hold in $\mathbb Q_p$. This is because $\mathbb Q_p^\times/\left(\mathbb Q_p^\times\right)^n$ is finite, so if $S$ is a set of coset representatives for $\mathbb Q_p^\times/\left(\mathbb Q_p^\times\right)^n$, then $\{ s x^n | s \in S\}$ is a finite set of nonlinear polynomials that cover $f$. It is also a minimal cover.

Is this the only kind of minimal cover of $\mathbb Q_p$, up to translation and other obvious things?

The only result I have in this direction is that if you take a cover of $\mathbb Q_p$, the leading terms of all the polynomials also cover $\mathbb Q_p$. This is because every coset of $\left(\mathbb Q_p^\times\right)^n$ (where $n$ is a multiple of the degree of each polynomial) has elements which are very large in the $p$-adic norm, and when $f(x)$ is very large in the $p$-adic norm $x$ must be very large, and when $x$ is very large, the terms other than the leading term do not change what coset $f(x)$ is in, so there must be one polynomial whose leading term can reach each coset.

A vaguely related question:

What interesting fields, other than finite extensions of $\mathbb Q_p$, admit nontrivial/interesting minimal covers?

  • $\begingroup$ If there are positive examples $f(X)$ to the MO-question at mathoverflow.net/questions/6820/…, then there would exist quite bizarre minimal covers of length $2$ of the form $f(X)$, $f(X+c)$ over some weird fields. $\endgroup$ – Peter Mueller Nov 15 '12 at 13:26
  • $\begingroup$ No, there are many finite covers. There is a converse to the result you mention: if a set of polynomials $f_i$ is such that their leading terms form a cover, then the $f_i$ cover all elements outside a bounded set $S$ of $\mathbf{Q}_p$. (Easy application of Hensel's lemma, which I think you also used to prove your result.) The remaining set $S$ can just be covered with any non-constant $f$ together with a finite number of its translates. (Again using a version of Hensel: for any polynomial $f \in \mathbf{Q}_p[x]$ and any $x_0$ with $f'(x_0)\neq 0$, $f$ is a local homeomorphism around $x_0$.) $\endgroup$ – RP_ Nov 16 '12 at 1:50
  • $\begingroup$ An example: one verifies that $x^2+1, 2x^2+1, 3x^2+1$ and $6x^2+1$ cover $\mathbf{Q}_3 \setminus \mathbf{Z}_3$. Now since $x^5$ contains $1 + \mathbf{Z}_3$ and $2 + \mathbf{Z}_3$, a cover of $\mathbf{Q}_3$ is given by $\{ x^2+1, 2x^2+1, 3x^2+1, 6x^2+1 , x^5, x^5+1 \}$. $\endgroup$ – RP_ Nov 16 '12 at 2:04
  • $\begingroup$ The first four cover $\mathbb Q_3$: first subtract 1, then factor, but it's easy to massage it to create an actual gap. I think that pretty much kills the problem. $\endgroup$ – Will Sawin Nov 16 '12 at 2:21

In your example covering is an almost partition (the only common point is $0$). If you allow overlapping, covers can be rather complicated. For exmaple, in $\mathbb Q_5$: $$ x^2, \ 2x^2, \ x^3, \ 5x^6, \ 2\cdot 5x^6, \ 5^5x^6, \ 2\cdot 5^5x^6. $$

Now I will describe monomial minimal covering. We assume $p>2$. If we have monomial minimal covering $\{a_1x^{n_1}, \dots, a_kx^{n_k}\}$, then we should check covering only on $\mathbb Q^*_p/(\mathbb Q^*_p)^D$, where $D=lcm(n_1, \dots, n_k)$. Since $$\mathbb Q^*_p=\mathbb Z\times C_{p-1}\times \mathbb Z_p,$$ then $$\tag{1} \mathbb Q^*_p/(\mathbb Q^*_p)^D=C_D\times C_{gcd(D, p-1)}\times C_{p^{v_p(D)}}. $$ Analogously, image of $x^n$ in this group is a subgroup of the form $$C_n\times C_{gcd(n, p-1)}\times C_{p^{v_p(n)}},$$ Where every factor is a subgroup of the corresponding factor in (1). If we interesting in image of $ax^n$, then it is a coset of corresponding subgroup. Now we have pure combinatorial question about covering group of cosets of special forms. If we see on first factor (which determines what occurs in the other two) then we can see classical combinatorial question: cover vertices of regular $D$-gon by smaller regular polygons. In my example, I cover 6-gon by 3-gon, 2-gon (i.e. diametr) and two 1-gons (i.e. vertices).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.