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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?

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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. –  Peter Mueller Nov 15 '12 at 13:26
    
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 $x_0$.) –  René Nov 16 '12 at 1:50
    
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 $\{ x^2+1, 2x^2+1, 3x^2+1, 6x^2+1 , x^5, x^5+1 \}$. –  René Nov 16 '12 at 2:04
    
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. –  Will Sawin Nov 16 '12 at 2:21
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