I wonder if the following fact is true:

Pick $l\in \mathbb N$ a number and let $f,g\in \mathbb Z_p[x]$ be monic polynomials with coefficients in the ring of $p$-adic integers such that $f\equiv g \pmod{p^l}$ and they are irreducible mod $p^l$. Then the roots of $f$ generate the same field as the roots of $g$.

Can someone help me proving this or finding a counterexample?

  • $\begingroup$ What's an irreducible polynomial modulo $p^l$ for $l>1$? For instance, $2X+2$ is irreducible modulo $4$? $\endgroup$ – user26857 Jun 27 '16 at 10:53
  • $\begingroup$ @user26857 the polynomials in question are monic. $\endgroup$ – Mathmo123 Jun 27 '16 at 11:12
  • $\begingroup$ @Mathmo123 Still don't get it: what's an irreducible polynomial over a non-integral domain? I suppose there is a definition that works for all polynomials. $\endgroup$ – user26857 Jun 27 '16 at 12:11
  • $\begingroup$ @user26857 The product of two monic polynomials cannot be zero. It therefore makes sense to define a polynomial as reducible if it has a non-trivial monic factor. $\endgroup$ – Mathmo123 Jun 27 '16 at 12:14

This is false. For example, $X^2+2$ and $X^2+6$ are both equal and irreducible mod $4$.

However, since $\mathbb Q_2$ does not contain a square root of $3$, their roots give different extensions of $\mathbb Q_2$.

  • $\begingroup$ Thank you! A curiosity: is this related to the fact that the field generated by the roots is wildly ramified? $\endgroup$ – Joel92 Jun 27 '16 at 12:18
  • $\begingroup$ Yes very much so. There is a unique unramified extension of any given degree, so the extensions certainly have to be ramified. Moreover, any tamely ramified extension is of the form $\mathbb Q_p(\sqrt[n]{mp})$ where $n$ and $m$ are prime to $p$. In particular, if $X^n-mp\equiv X^n-kp\pmod{p^k}$, then $mk^{-1}\equiv 1 \pmod p$ ($k$ is invertible since $p\nmid k$), and therefore $\sqrt[n]{mk^{-1}}\in \mathbb Q_p$ by Hensel's lemma (since $p\nmid n$). It follows that $X^n-mp$ and $X^n-kp$ define the same extension. $\endgroup$ – Mathmo123 Jun 27 '16 at 15:46

Check out Lang's Algebraic Number Theory section II.2. In a nutshell, if two polynomials are $p$-adically close then their roots are close as well, and by Krasner's lemma the fields they generate over $\mathbb Q_p$ will be the same.

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    $\begingroup$ Yes, but I need something more. Krasner Lemma gives a neighbourhood of $f$ such that every polynomial in it is irreducible and the roots generate the same fields as $f$. My question is a bit different: I have a neighbourhood of $f$ such that every polynomial in it is irreducible. Is it true that the roots of all these polynomials generate the same fields as the roots of $f$? $\endgroup$ – Joel92 Jun 23 '16 at 14:51
  • $\begingroup$ Oh, that's different. So you have an open ball with every element inside being irreducible? My first idea would be to use Krasner to cover your set with balls each containing polynomials that generate the same field. Partition the set of these balls by which field the polynomials inside generate and for each field take the union of all balls belonging to it. Then your original set is the disjoint union of these open sets, each containing only polynomials generating a particular field. But your original set is connected, so there is only one set. $\endgroup$ – Lance Sackless Jun 23 '16 at 15:30
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    $\begingroup$ I don't think that our set is connected... $\endgroup$ – Joel92 Jun 23 '16 at 16:00
  • $\begingroup$ Ah, you're right. My bad. I'm out of ideas then :/ $\endgroup$ – Lance Sackless Jun 23 '16 at 16:06
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    $\begingroup$ Of course they’re not connected. The only connected sets in this situation are singletons. $\endgroup$ – Lubin Jun 23 '16 at 20:57

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