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I've been thinking about Hensel's lemma, specifically the singular root case, and trying to build some intuition around the edge cases. In thinking about this, the following question crossed my mind.

For any given odd prime $p$, does there necessarily exist a polynomial $P \in \mathbb{Z}[x]$ such that $P$ has exactly one root $\text{mod} \ p$, say $r$, and this root $r$ does not lift to a root $ \text{mod} \ p^2$, so that $P$ has no roots $ \text{mod} \ p^2$. (Note by Hensel's lemma, any such polynomial must satisfy $P'(r) \equiv 0 \ \text{mod} \ p$ and $P(r) \not\equiv 0 \ \text{mod} \ p^2$).

If we replace "exactly one" with "at least one", something like $P(x) = (p^2x+p)(px+2)$ should work. Here, every $x \in \mathbb{Z}$ is a root mod $p$ but there are no roots $\text{mod} \ p^2$. If we allow $p=2$, $P(x) = x^2 + 1$ works. We have exactly one root $\text{mod} \ 2$ but no roots $\text{mod} \ 4$. However, $P(x) = x^p + 1$ doesn't quite work for odd primes.

I would guess that there is a simple construction that's eluding me (but I suppose it's also possible that the claim does not hold).

J. W. Tanner
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MathematicsStudent1122
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1 Answers1

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Take $P(x) = x^p - p$. Modulo $p$, this is just $x^p$ and has the unique root $a\equiv 0\pmod{p}$.

Now, modulo $p^2$, if $a\equiv 0\pmod{p}$ then $a^p\equiv 0\pmod{p^2}$, and since $p\not\equiv 0\pmod{p^2}$, then $a$ cannot be a root of $P(x)$ modulo $p^2$. And if $a\not\equiv 0\pmod{p}$ then it is not a root since $a^p\equiv a\pmod{p}$.

See also this previous post for a version of Hensel’s Lemma in the case that $f’(a)\equiv 0\pmod{p^j}$, indicating some situations in which the root lifts.

Arturo Magidin
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