# Hensel's lemma: lifting roots vs. lifting factorizations

There are various versions of Hensel's lemma in commutative algebra. One is about lifting roots of polynomials (Eisenbud, Theorem 7.3):

Let $R$ be a ring that is complete with respect to the ideal $\mathfrak{m}$ and let $f (x) \in R [x]$ be a polynomial. If $a$ is an approximate root of $f$ in the sense that $$f (a) \equiv 0 \pmod{f' (a)^2 \, \mathfrak{m}}$$ then there if a root $b$ of $f$ near $a$ in the sense that $$f (b) = 0 \quad \text{and} \quad b \equiv a \pmod{f' (a) \, \mathfrak{m}}.$$

And there is also a seemingly more general version about lifting factorizations of polynomials (Eisenbud, Theorem 7.18, Exercise 7.19-7.20):

Let $R$ be a Noetherian ring, complete with respect to an ideal $\mathfrak{m}$. Let $F (x) \in R [x]$ be a polynomial in one variable with coefficients in $R$ and let $f (x)$ be the polynomial over $R/\mathfrak{m}$ obtained by reducing the coefficients of $f$ mod $\mathfrak{m}$. If $f$ factors as $$f = g_1 g_2 \in (R/\mathfrak{m}) [x]$$ in such a way that $g_1$ and $g_2$ generate the unit ideal, and $g_1$ is monic, then there is a unique factorization $$F = G_1\,G_2 \in R [x]$$ such that $G_1$ is monic and $G_i$ reduces to $g_i$ mod $\mathfrak{m}$.

Eisenbud in his textbook gives the second version as an exercise, and explains how to prove it from scratch. He also writes that it can be "deduced in a page" from the first version, but unfortunately, he refers to Nagata's "Local rings" (1962) which I find quite hard to read.

Could anyone explain this implication? It is stated in several places, but I am interested in an intelligible one-page proof, as promised by Eisenbud.

Thank you.

• I too wouldn’t mind seeing such a proof. – Lubin Aug 24 '17 at 18:56
• I will not proof it rigorously, but the idea is that we can assume $g_1$ to be irreducible and then we have a root of $f$ in a field extension of $R/\mathfrak m$. By the first theorem, this gives rise to a root in a ring extension of $R$ and its minimal polynomial is the desired $G_1$. – MooS Aug 25 '17 at 6:27
• Probably the right method, @MooS, but nobody said anything about $R/\mathfrak m$ being a field, or even an integral domain. It was exactly the extreme level of generality of the quoted proposition that stopped me in my tracks. – Lubin Aug 25 '17 at 18:45
• @Lubin I think when Eisenbud refers to Nagata's book, he actually means the following implication: if $(R,\mathfrak{m})$ is a local ring (not necessarily complete) such that simple roots in $R/\mathfrak{m}$ may be lifted to roots in $R$, then factorizations in $R/\mathfrak{m} [x]$ may be lifted to factorizations in $R [x]$. Something similar to what is stated on p. 39 in math.lsa.umich.edu/~hochster/615W10/615.pdf (I came across these notes and they are much more readable than Nagatas's book..) But there should be a direct proof, something like MooS suggests. – AAA Aug 26 '17 at 18:00
• @Lubin I wonder why some textbooks about $p$-adic fields (e.g. Gouvêa) prove the two versions of Hensel's lemma separately. Maybe proving both from scratch (for the ring of integers in a complete nonarchimedian field) is more enlightening than going through the commutative algebra that gives the implication? – AAA Aug 26 '17 at 18:11