Using the lemma, one can "lift" a root r of the polynomial f modulo p^k to a new root s modulo p^k+1 such that r ≡ s mod p^k (by taking m = 1; taking larger m follows by induction). In fact, a root modulo p^k+1 is also a root modulo p^k, so the roots modulo p^k+1 are precisely the liftings of roots modulo p^k. The new root s is congruent to r modulo p, so the new root also satisfies ${\displaystyle f'(s)\equiv f'(r)\not \equiv 0{\bmod {p}}.}$ So the lifting can be repeated, and starting from a solution r_k of ${\displaystyle f(x)\equiv 0{\bmod {p}}^{k}}$ we can derive a sequence of solutions r_k+1, r_k+2, ... of the same congruence for successively higher powers of p, provided that ${\displaystyle f'(r_{k})\not \equiv 0{\bmod {p}}}$ for the initial root r_k. This also shows that f has the same number of roots mod p^k as mod p^k+1, mod p^k+2, or any other higher power of p, provided that the roots of f mod p^k are all simple.


What happens to this process if r is not a simple root mod p? Suppose that


Then ${\displaystyle s\equiv r{\bmod {p}}^{k}}$ implies ${\displaystyle f(s)\equiv f(r){\bmod {p}}^{k+1}.}$ That is, ${\displaystyle f(r+tp^{k})\equiv f(r){\bmod {p}}^{k+1}}$ for all integers t. Therefore, we have two cases:


Example. To see both cases we examine two different polynomials with p = 2:


${\displaystyle f(x)=x^{2}+1}$ and r = 1. Then ${\displaystyle f(1)\equiv 0{\bmod {2}}}$ and ${\displaystyle f'(1)\equiv 0{\bmod {2}}.}$ We have ${\displaystyle f(1)\not \equiv 0{\bmod {4}}}$ which means that no lifting of 1 to modulus 4 is a root of f(x) modulo 4.


${\displaystyle g(x)=x^{2}-17}$ and r = 1. Then ${\displaystyle g(1)\equiv 0{\bmod {2}}}$ and ${\displaystyle g'(1)\equiv 0{\bmod {2}}.}$ However, since ${\displaystyle g(1)\equiv 0{\bmod {4}},}$ we can lift our solution to modulus 4 and both lifts (i.e. 1, 3) are solutions. The derivative is still 0 modulo 2, so a priori we don't know whether we can lift them to modulo 8, but in fact we can, since g(1) is 0 mod 8 and g(3) is 0 mod 8, giving solutions at 1, 3, 5, and 7 mod 8. Since of these only g(1) and g(7) are 0 mod 16 we can lift only 1 and 7 to modulo 16, giving 1, 7, 9, and 15 mod 16. Of these, only 7 and 9 give g(x) = 0 mod 32, so these can be raised giving 7, 9, 23, and 25 mod 32. It turns out that for every integer k ≥ 3, there are four liftings of 1 mod 2 to a root of g(x) mod 2^k.

Related concepts[edit]

Completeness of a ring is not a necessary condition for the ring to have the Henselian property: Goro Azumaya in 1950 defined a commutative local ring satisfying the Henselian property for the maximal ideal m to be a Henselian ring.


Masayoshi Nagata proved in the 1950s that for any commutative local ring A with maximal ideal m there always exists a smallest ring A^h containing A such that A^h is Henselian with respect to mA^h. This A^h is called the Henselization of A. If A is noetherian, A^h will also be noetherian, and A^h is manifestly algebraic as it is constructed as a limit of étale neighbourhoods. This means that A^h is usually much smaller than the completion  while still retaining the Henselian property and remaining in the same category.