MULTIVARIATE HENSEL’S LEMMA FOR COMPLETE RINGS
YIFEI ZHAO
Contents
1. Completion
2. The inverse function theorem
3. Proof of Hensel’s lemma and the implicit function theorem
References
1
6
7
8
In this set of notes, we prove that a complete ring satisfies the multivariate Hensel’s lemma (Theorem
1.11). The result can be seen as a formal version of the implicit function theorem. We make this connection
precise in §3. The section on completion follows the Stacks project [4], since most textbooks on this subject
assumes Noetherian property. Our exposition of Hensel’s lemma largely follows Eisenbud’s book [2, Chapter
7], although he only proves the single-variable version, remitting multivariate Hensel’s lemma to exercises.
1. Completion
1.1. Inverse limit definition. Let R be a (commutative, unital) ring, and a be an ideal of R. The a-adic
completion of R is defined as the R-algebra.
R̂ = lim R/an .
←−
Let M be an R-module. Then the a-adic completion of M is the R̂-module
M̂ = lim M/an M.
←−
n
The collection of maps M → M/a M induces a map M → M̂ . It fits into the following exact sequence:
\
0→
an M → M → M̂ .
n≥1
Lemma 1.1. Let Mn , Nn , and Pn (n ≥ 1) be inverse systems of R-modules fitting into an exact sequence
(of inverse systems):
0 → Mn → Nn → Pn → 0.
Then the sequence
0 → lim Mn → lim Nn → lim Pn
←−
←−
←−
is exact. Furthermore, if Mn is a surjective system, i.e. the maps Mn → Mn−1 are all surjective, then the
sequence
0 → lim Mn → lim Nn → lim Pn → 0
←−
←−
←−
is exact.
Proof. An element of lim Mn can be seen as a sequence (xn ) where each xn ∈ Mn , and the image of xn in
Mn−1 is precisely xn−1 . Hence lim Mn is the kernel of the morphism
Y
Y
ϕM :
Mn →
Mn , ϕM (x1 , x2 , · · · ) = (x1 − x̄2 , x2 − x̄3 , · · · )
n≥1
n≥1
where x̄n+1 denotes the image of xn+1 in Mn . The snake lemma shows that
0 → Ker(ϕM ) → Ker(ϕN ) → Ker(ϕP ) → Coker(ϕM )
1
2
YIFEI ZHAO
is an exact sequence. Now, if Mn is a surjective system, then ϕM is surjective, so Coker(ϕM ) = 0.
1.2. Cauchy sequence definition. We can alternatively define a-adic completion via a topology on R.
We set a base of neighborhoods of each x by x + ak . This is the a-adic topology on R; it makes R into a
topological ring.
Definition 1.2. The topological a-adic completion of R is the set of Cauchy sequences (x1 , x2 , · · · ) in R, i.e.
sequences with elements in R and
∀k ∈ N,
∃N ∈ N,
such that
m, n ≥ N =⇒ xm − xn ∈ ak
modulo the equivalence relation: (xn ) ∼ (yn ) if
∀k ∈ N,
∃N ∈ N,
such that n ≥ N =⇒ xn − yn ∈ ak
We temporarily denote this quotient set by R̃. It is naturally equipped with an R-algebra structure.
Lemma 1.3. There is an R-algebra isomorphism
R̃ ∼
= R̂
Proof. Given a Cauchy sequence (x1 , x2 , · · · ) ∈ R̃ and any k ∈ N, pick sufficiently large N so that
xN = xN +1 = · · ·
mod ak .
Send (xn ) to the element in R/ak defined by xN . This process defines a compatible system of maps R̃ → R/ak ,
which induces a map R̃ → R̂.
Conversely, an element in R̂ is given by a sequence (x̄1 , x̄2 , · · · ) where each x̄k ∈ R/ak and its image in
R/ak−1 is exactly x̄k−1 . Pick a lift xk ∈ R of each x̄k . Then
xk = xk−1
mod ak−1
for each k. So (x1 , x2 , · · · ) is a Cauchy sequence. This process induces a well-defined map R̂ → R̃. We leave
it to the reader to check that these two maps are mutual inverses.
Similarly, M̂ can also be defined via the a-adic topology which associates to each x ∈ M a base of
neighborhoods x + ak M .
1.3. Complete rings and modules. R (respectively M ) is complete with respect to a if the natural map
R → R̂
(respectively M → M̂ )
is an isomorphism. It is not true in general that the completion R̂ (or M̂ ) is complete (with respect to aR̂).
Lemma 1.4. M̂ is complete with respect to aR̂ if and only if the sequence
0 → an M̂ → M̂ → M/an M → 0
is exact for each n.
The map M̂ → M/an M is always surjective. So the content of exactness is that the kernel of this map is
given by an M̂ .
Proof. Let Kn be the kernel of M̂ → M/an M . Note that Kn contains an M̂ , so we have exact sequences
0 → Kn /an M̂ → M̂ /an M̂ → M/an M → 0
It suffices to show that the Kn /an M̂ form a surjective system, since Lemma 1.1 will then give an exact
sequence
ˆ
0 → lim Kn /an M̂ → M̂ → M̂ → 0,
←−
ˆ
so M̂ ∼
= M̂ if and only if each Kn /an M̂ = 0.
Indeed, since M̂ → M/an+1 M is surjective, after multiplying by an , the map
an M̂ → an M/an+1 M
MULTIVARIATE HENSEL’S LEMMA FOR COMPLETE RINGS
3
is still surjective. Take x ∈ Kn . Then x maps to an M by definition. So there is some y ∈ an M̂ such that
x − y maps to an+1 M . This means that x − y ∈ Kn+1 . In other words, Kn = Kn+1 + an M̂ . It follows that
the map
Kn+1 /an+1 M̂ → Kn /an M̂
is surjective.
In particular, if M̂ is complete, we have an isomorphism
M̂ /an M̂ ∼
= M/an M
for each n. Define the associated graded module of M by
M
Gra (M ) =
an M/an+1 M, with
deg(an M/an+1 M ) = n
n≥0
By an application of the five-lemma, we see that GraR̂ (M̂ ) ∼
= Gra (M ) if M̂ is complete.
Proposition 1.5. Suppose a is finitely generated. Then M̂ is complete with respect to aR̂; in particular, R̂
is complete with respect to aR̂.
Proof. By Lemma 1.4, we just need to show that the kernel Kn of the surjection M̂ → M/an M is contained
in an M̂ . Indeed, let x ∈ Kn . It is represented by a sequence
x = (0, · · · , 0 = x̄n , x̄n+1 , x̄n+2 , · · · )
where each x̄m ∈ an M/am M (m ≥ n). Choose a lift xm ∈ an M of each x̄m . Then
δm = xm+1 − xm ∈ am M
and x can be written as the sum of the convergent power series
x = xn + δn + δn+1 + · · ·
Suppose a is generated by elements f1 , · · · , fr . We may write xn and each δm as sums:
X
X
xn =
f J xn,J , δm =
f J δm,J
|J|=n
|J|=n
where xn,J ∈ M , δm,J ∈ am−n M and f J = f1j1 · · · frjr . Therefore,
X
X
x=
f J (xn,J + δn,J + δn+1,J + · · · ) =
f J xJ
|J|=n
|J|=n
where each xJ = xn,J + δn,J + · · · lies in M̂ . So x ∈ an M̂ , as required.
Examples of complete rings include
(i) The p-adic integers: Zp . By definition,
Zp = lim Z/pn Z
←−
It is complete since the ideal (p) is principal (so finitely generated).
(ii) Formal power series rings: A[[x1 , · · · , xn ]] where A is any ring. By definition,
A[[x1 , · · · , xn ]] = lim A[x1 , · · · , xn ]/(x1 , · · · , xn )k
←−
It is complete with respect to the ideal a = (x1 , · · · , xn ). This ring is Noetherian if and only if A is1.
1For the “only if” direction, mod out a. The “if” direction is the formal Hilbert basis theorem; see [1] or [2].
4
YIFEI ZHAO
1.4. More on completion. A map of R-modules f : M → N induces a map fˆ : M̂ → N̂ of a-adic
completions. This construction is functorial. We study this functor in more details in this subsection.
Proposition 1.6. Suppose fˆ : M̂ → N̂ is the map induced by f : M → N . Then
(i) fˆ is surjective if the map M/aM → N/aN is surjective. In particular, completion preserves surjection.
(ii) fˆ is injective if the map Gra (M ) → Gra (N ) is injective.2
Before proving Proposition 1.6, we first need a version of Nakayama lemma that does not assume finite
generation.
Lemma 1.7 (“Nilpotent Nakayama”). Let a be a nilpotent ideal of R, i.e. an = 0 for some n. Suppose
N ⊂ M are R-modules with M = aM + N . Then M = N .
Proof. The hypothesis means that M/N = a(M/N ). Hence M/N = an (M/N ) = 0.
Proof of Proposition 1.6.
(i) There is a commutative diagram
M/an M
/ M/aM
N/an N
/ N/aN
where all but the left vertical arrow are surjective a priori. So the diagonal arrow M/an M → N/aN
is surjective. On the other hand, N/aN is the quotient of N/an N by the nilpotent ideal a/an in the
ring R/an . Hence, M/an M → N/an N is also surjective by Lemma 1.7.
Now, write down exact sequences
0 → Kn → M/an M → N/an N → 0.
It again suffices, by Lemma 1.1, to show that the Kn form a surjective system. An application of the
snake lemma boils this down to showing that
an M/an+1 M → an N/an+1 N
is surjective. Indeed, each y ∈ an N is of the form
r
X
ai yi , for ai ∈ an ,
y=
yi ∈ N.
i=1
But M → N/aN is surjective, so each
yi = f (xi ) +
si
X
aij yij ,
for xi ∈ M,
aij ∈ a,
yij ∈ N.
j=1
This shows that
y=
X
ai f (xi ) + ai aij yij .
i,j
The first term on the right hand side is in the image of an M , while the second term lies in an+1 N .
(ii) Suppose x ∈ M̂ is mapped to zero under fˆ. Write x = (x̄1 , x̄2 , · · · ), where each x̄n ∈ M/an M is
represented by some xn ∈ M . The hypothesis is that f (xn ) ∈ an N for all n. If x1 ∈
/ aM , then the
element
(x̄1 , 0, 0, · · · ) ∈ Gra (M )
is nonzero, and maps to zero in Gra (N ). This is impossible, so x1 ∈ aM , and x̄1 = 0.
We now assume by induction that x̄1 , · · · , x̄n = 0. Since xn = xn+1 mod an M , we must have
xn+1 ∈ an M . If xn+1 ∈
/ an+1 M . Then
(0, · · · , 0, x̄n+1 , 0, 0, · · · ) ∈ Gra (M )
is nonzero, and maps to zero in Gra (N ). This is again impossible, so xn+1 ∈ an+1 M , and x̄n+1 = 0. 2The point here is that we do not impose any finiteness condition. In the case where R is Noetherian, completion is an exact
functor on the category of finite R-modules. This is the Artin-Rees lemma; see [1] or [2].
MULTIVARIATE HENSEL’S LEMMA FOR COMPLETE RINGS
5
The rest of this subsection is devoted to general results on completions that will not be used in the sequel.
Corollary 1.8. If M is a finite R-module, then the natural map M ⊗R R̂ → M̂ is surjective.
Proof. There is a surjection R⊕r → M . By Proposition 1.6, the map R̂⊕r → M̂ is surjective. This map also
arises as the composition
∼
R̂⊕r −
→ R⊕r ⊗R R̂ → M ⊗R R̂ → M̂ .
So the last map in the chain is again surjective.
We say M is separated in a-adic topology if
\
an M = 0.
n≥1
This is equivalent to M being Hausdorff in the a-adic topology. Since the intersection of all an M is precisely
the kernel of M → M̂ , the module M is separated if and only if M → M̂ is injective.
Corollary 1.9. If R is complete with respect to a, then any finite, separated R-module M satisfies M ∼
= M̂ .
Proof. The map
∼
M−
→ M ⊗R R̂ → M̂
is injective by hypothesis, and surjective by Corollary 1.8.
Here is another version of Nakayama lemma that does not assume finite generation.
Corollary 1.10 (“Complete Nakayama”). Suppose R is complete with respect to a, and M is a separated
R-module. If the images of m1 , · · · , mn ∈ M generate M/aM , then m1 , · · · , mn generate M .
Proof. Let N be the submodule of M generated by m1 , · · · , mn . There is a commutative diagram

/ M
N
_
N̂
/ M̂
where the left vertical arrow is an isomorphism because N is separated and finite, so Corollary 1.9 applies
to N . Since N → M/aM is surjective, the bottom arrow is surjective by Proposition 1.6(i). Hence every
arrow in the diagram is an isomorphism.
1.5. Hensel’s lemma. Our objective is to show that complete rings satisfy the multivariate Hensel’s lemma.
Let f1 , · · · , fn ∈ R[x1 , · · · , xn ]. Use f to denote the n-tuple (f1 , · · · , fn ), and Ja (f ) to denote the
Jacobian matrix evaluated at a ∈ Rn :
∂fi
(a)
Ja (f ) =
∂xj
1≤i,j≤n
Theorem 1.11. Suppose R is complete with respect to some ideal a. If there is some a ∈ Rn satisfying
f (a) = 0 mod det(Ja (f ))2 a,
then there is some b ∈ Rn such that f (b) = 0 and
b=a
mod det(Ja (f ))a
Furthermore, b is unique if det(Ja (f )) is not a zero-divisor.
Note that no Noetherian hypothesis is required for R.
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YIFEI ZHAO
1.6. Variants. There are other statements that go under the name of Hensel’s lemma. In fact, for any local
ring (R, m) with residue field k, the following statements are equivalent:
(i) For every monic f ∈ R[x] and every simple root ā ∈ k of the reduced polynomial f¯ ∈ k[x], there exists
some a ∈ R such that
f (a) = 0
and
a = ā mod m.
(ii) For every monic f ∈ R[x] and any factorization f¯ = ḡ h̄ into coprime polynomials ḡ, h̄ ∈ k[x], there
exist polynomials g, h ∈ R[x] such that
f = gh
and
g = ḡ, h = h̄ mod m
(iii) For any étale ring map ϕ : R → S and prime q of S lying over m with residue field k, there exists a
section τ : S → R of ϕ with q = τ −1 (m).
(iv) Any finite R-algebra is a finite product of local rings.
The local ring (R, m) satisfying these properties is called a Henselian local ring. For proofs of these statements, see [4, Tag 04GG]. Note that (i) is a (very) special case of Theorem 1.11.3
2. The inverse function theorem
2.1. The inverse function theorem. We first study maps from a formal power series ring R[[x1 , · · · , xn ]].
They are uniquely determined by their actions on xi , when the target is complete.
Proposition 2.1. Let R be a ring (not necessarily complete). Let S be an R-algebra that is complete with
respect to some ideal n. Given f1 , · · · , fn ∈ n, there is a unique R-algebra map
ϕ : R[[x1 , · · · , xn ]] → S
sending each xi to fi . Furthermore,
(i) ϕ sends each power series g(x) to g(f ).
(ii) ϕ is surjective if the induced map
R → S/n
is surjective and f1 , · · · , fn generate n.
(iii) ϕ is injective if the induced map
Gr(ϕ) : R[x1 , · · · , xn ] → Grn (S)
is injective and f1 , · · · , fn generate n.4
Proof. Let m be the ideal (x1 , · · · , xn ) of R[[x1 , · · · , xn ]]. For each k ∈ N, there is a unique ring map
ϕk : R[[x1 , · · · , xn ]] → S/nk
sending xi to the class of fi , since ϕk factors through
R[[x1 , · · · , xn ]]/mk ∼
= R[x1 , · · · , xn ]/(x1 , · · · , xn )k .
The collection ϕk induces the unique map ϕ since S is the inverse limit of S/nk .
(i) We ought to show that the image of g(x) in S agrees with g(f ) after reducing by nk . This is clear.
(ii) The fact that f1 , · · · , fn generate n shows that as an R[x1 , · · · , xn ]-submodule of S,
n = aS
where a is the ideal (x1 , · · · , xn ) of R[x1 , · · · , xn ]. Therefore, the R[x1 , · · · , xn ]-module map
R[x1 , · · · , xn ] → S
satisfies the hypothesis of Proposition 1.6(i). So the induced map on completions is surjective.
3I wonder whether the multivariate Hensel’s lemma holds for every Henselian ring?
4In (iii), the hypothesis that f , · · · , f generate n can be removed; see [2, Theorem 7.16c].
n
1
MULTIVARIATE HENSEL’S LEMMA FOR COMPLETE RINGS
7
(iii) Gr(ϕ) is exactly the map of associated graded modules of R[x1 , · · · , xn ] and S, over the ring R[x1 , · · · , xn ]
and the ideal a = (x1 , · · · , xn ). Hence Proposition 1.6(ii) applies.
The following result is the inverse function theorem for formal power series:
Let R be a ring, and m be the ideal (x1 , · · · , xn ) of R[[x1 , · · · , xn ]].
Theorem 2.2. Suppose f1 , · · · , fn ∈ m. If ϕ is the R-linear endomorphism
ϕ : R[[x1 , · · · , xn ]] → R[[x1 , · · · , xn ]]
sending xi to fi . Then ϕ is an isomorphism if and only if det(J0 (f )) is a unit.
Proof. The endomorphism ϕ induces a map of graded rings:
Grm (ϕ) : R[x1 , · · · , xn ] → R[x1 , · · · , xn ]
given by x 7→ J0 (f ) · x (where x is understood as a column vector), i.e. each
n
X
∂fi
xi 7→ f¯i =
(0)xj
∂x
j
j=1
Suppose ϕ is an isomorphism. Then Grm (ϕ) must be surjective on the degree-1 piece of R[x1 , · · · , xn ].
Hence each xi is a linear combination of f¯1 , · · · , f¯n . This implies that we have an n-by-n matrix T with
coefficients in R such that
T · J0 (f ) = I.
Hence det(J0 (f )) is a unit.
Conversely, suppose det(J0 (f )) is a unit. Then J0 (f ) is invertible, so Grm (ϕ) is an isomorphism. In
particular, f¯1 , · · · , f¯n generate m/m2 over R. In order to apply Proposition 2.1 to conclude that ϕ is an
isomorphism, we only need to show that f1 , · · · , fn generate the ideal m of R[[x1 , · · · , xn ]]. Note that
(i) m is contained in the Jacobson radical of R[[x1 , · · · , xn ]]. This is because 1 − g is invertible for every
g ∈ m; its inverse is given by 1 + g + g 2 + · · · .
(ii) m can be regarded as a finite R[[x1 , · · · , xn ]]-module. Indeed, it is generated by x1 , · · · , xn .
The fact that f¯1 , · · · , f¯n generate m/m2 now shows that f1 , · · · , fn generate m, by Nakayama lemma.
3. Proof of Hensel’s lemma and the implicit function theorem
3.1. The proof. Hensel’s lemma follows easily from Proposition 2.1 and the inverse function theorem.
Proof of Theorem 1.11. Let e = det(Ja (f )). Then we have an n-by-n matrix T with coefficients in R such
that T · Ja (f ) = Ja (f ) · T = eI. In particular,
f (a + eT · x) = f (a) + Ja (f ) · eT · x + e2 g(T · x)
= f (a) + e2 (x + g(T · x))
for some g ∈ m2 . The map ϕ : R[[x1 , · · · , xn ]] → R[[x1 , · · · , xn ]] determined by
ϕ : x 7→ x + g(T · x)
has Jacobian I. By the inverse function theorem 2.2, it is an isomorphism. The inverse ϕ−1 is also R-linear.
In other words,
f (a + eT · ϕ−1 (x)) = f (a) + e2 x
Now, the hypothesis shows that f (a) = e2 c, where each component of c lies in the ideal a. Since R is
complete with respect to a, there is an R-linear map ψ : R[[x1 , · · · , xn ]] → R determined by
ψ : x 7→ −c
Applying ψ to the above equation, we obtain
f (a + eT · ψϕ−1 (x)) = e2 ψ(c + x) = 0.
Thus b = a + eT · ψϕ−1 (x) is the desired element.
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YIFEI ZHAO
We now assume e is not a zero-divisor, and prove the uniqueness of b. Note that
T · f (a + ex) = T · f (a) + T · Ja (f ) · ex + e2 T · g(x)
= T · f (a) + e2 (x + T · g(x))
Like before, the map ϕ̃ : R[[x1 , · · · , xn ]] → R[[x1 , · · · , xn ]] determined by
ϕ̃ : x 7→ x + T · g(x)
is an isomorphism. Now, suppose b = a + er and b0 = a + er0 satisfy f (b) = f (b0 ) = 0. Then the above
formula, along with cancellation by e2 , shows that
b + T · g(b) = b0 + T · g(b0 )
Let β, β 0 : R[[x1 , · · · , xn ]] → R be determined by sending x to b and b0 , respectively. We have just seen that
the two compositions
β
R[[x1 , · · · , xn ]]
ϕ̃
/ R[[x1 , · · · , xn ]]
&
8R
β0
agree on x. The uniqueness of Proposition 2.1 then shows that β ϕ̃ = β 0 ϕ̃. But ϕ̃ is an isomorphism, so
β = β 0 . Thus b = b0 .
3.2. The implicit function theorem. Hensel’s lemma should really be understood as saying that the ring
R “satisfies the implicit function theorem,” i.e. the variables x1 , · · · , xn afford a unique parametrization
by elements in R when the Jacobian Ja (f ) is invertible. The standard form of implicit function theorem is
precisely the case where R is a ring of power series:
Let k be a field.
Theorem 3.1. Suppose f1 , · · · , fn ∈ k[y1 , · · · , ym ; x1 , · · · , xn ] satisfy f (0; a) = 0, and
∂fi
det
(0; a)
6= 0
∂xj
1≤i,j≤n
Then there are unique power series g1 , · · · , gn ∈ k[[y1 , · · · , ym ]] such that f (y; g(y)) = 0 and g(0) = a.
Proof. Let R = k[[y1 , · · · , ym ]]. It is complete with respect to the unique maximal ideal a = (y1 , · · · , ym ).
Denote by f˜1 , · · · , f˜n the images of f1 , · · · , fn in R[x1 , · · · , xn ]. The hypothesis f (0; a) = 0 means precisely
that
f̃ (a) = 0 mod a
while the non-vanishing of the determinant means that
det(Ja (f̃ )) 6= 0
mod a
In other words, det(Ja (f̃ )) is a unit in the ring R. Hensel’s lemma 1.11 provides a unique g ∈ Rn with
the property that f̃ (g) = 0 and g = a mod a. These properties translate directly to f (y; g(y)) = 0 and
g(0) = a.
References
[1] Atiyah, Michael Francis, and Ian Grant Macdonald. Introduction to commutative algebra. Vol. 2. Reading: Addison-Wesley,
1969.
[2] Eisenbud, David. Commutative Algebra: with a view toward algebraic geometry. Vol. 150. Springer Science & Business
Media, 1995.
[3] Hartshorne, Robin. Algebraic geometry. Vol. 52. Springer Science & Business Media, 1977.
[4] de Jong, Aise Johan. “Stacks Project.” Open source project (2010).