# Hensel’s lemma for integers

Let $f(x)$ be a polynomial with integer coefficients, $p$ a prime number, and $n$ a positive integer.  Assume that an integer $a$ (and naturally its whole residue class modulo $p^{n}$) satisfies the congruence

 $\displaystyle f(x)\;\equiv\;0\;\;\pmod{p^{n}}.$ (1)

The solution  $x=a$  of (1) may be refined in its residue class modulo $p^{n}$ to a solution  $x=a\!+\!rp^{n}$  of the congruence

 $\displaystyle f(x)\;\equiv\;0\;\;\pmod{p^{n+1}}.$ (2)

This refinement is unique modulo $p^{n+1}$ iff  $f^{\prime}(a)\not\equiv 0\,\pmod{p}$.

Proof.  Now we have  $f(a)=sp^{n}$.  We have to find an $r$ such that

 $f(a\!+\!rp^{n})\;\equiv\;0\;\;\pmod{p^{n+1}}.$

The short Taylor theorem requires that

 $f(a\!+\!rp^{n})\;\equiv\;f(a)+rf^{\prime}(a)p^{n}\;\;\pmod{r^{2}p^{2n}}$

where  $2n\geq n\!+\!1$, whence this congruence can be simplified to

 $sp^{n}+rf^{\prime}(a)p^{n}\;\equiv\;0\;\;\pmod{p^{n+1}}.$

Thus the integer $r$ must satisfy the linear congruence

 $s+rf^{\prime}(a)\;\equiv\;0\;\;\pmod{p}.$

When  $f^{\prime}(a)\not\equiv 0$,  this congruence has a unique solution modulo $p$ (see linear congruence); thus we have the refinement  $a^{\prime}=a\!+\!rp^{n}$  which is unique modulo $p^{n+1}$.

When  $f^{\prime}(a)\equiv 0$  and  $s\not\equiv 0\;\pmod{p}$,  the congruence evidently is impossible.

In the case  $f^{\prime}(a)\equiv s\equiv 0\;\pmod{p}$  the congruence (2) is identically true in the residue class of $a$ modulo $p^{n}$.   □

## References

Title Hensel’s lemma for integers HenselsLemmaForIntegers 2013-04-08 19:35:26 2013-04-08 19:35:26 pahio (2872) pahio (2872) 7 pahio (2872) Definition msc 11A07