Proof.
Let
(
a_{0},
a_{1},...,
a_{d}) be a
padic expansion of
a w.r.t.
p.
Let
k <
d + 1 be a positive integer.
By Proposition
13,
the element
a^{(k)} = a_{0} + a_{1}p + ^{ ... }a_{k1}p^{k1} 
(92) 
is a
padic approximation of
a at order
k.
By Proposition
11
there exists a polynomial
R[
y] such that
(a^{(k)} + a_{k}p^{k}) = (a^{(k)}) + ^{}(a^{(k)})a_{k}p^{k} + (a^{(k)} + a_{k}p^{k})(a_{k}p^{k})^{2}. 
(93) 
Since we have
a a^{(k)} + a_{k}p^{k}modp^{k+1} 

we deduce from Proposition
14
(a) (a^{(k)} + a_{k}p^{k})modp^{k+1} 
(94) 
Since
(
a) = 0 this shows that
(
a^{(k)} +
a_{k}p^{k}) is in the ideal generated by
p^{k+1}.
Similarly,
(
a^{(k)}) is in the ideal generated by
p^{k}.
Therefore we can divide
(
a^{(k)} +
a_{k}p^{k}) and
(
a^{(k)}) by
p^{k}, leading to
= + ^{}(a^{(k)})a_{k} + (a^{(k)} + a_{k}p^{k})(a_{k})^{2}p^{k}. 
(95) 
Now observe that
0 (a^{(k)} + a_{k}p^{k})(a_{k})^{2}p^{k}modp. 
(96) 
Let us denote by
the canonical homomorphism from
R to
R/
p.
Then we obtain
() =  (^{}(a^{(k)}))(a_{k}) 
(97) 
Now, since
a a_{0}mod
p holds we have
(^{}(a^{(k)})) (^{}(a_{0}))modp 
(98) 
Finally, since
^{}(
a_{0})
0 mod
p holds we can solve
Equatiion
97
for
(
a_{k}).
Theorem 13
Let
R be a commutative ring with identity element
and let
be a finitely generated ideal of
R.
Let
r be a positive integer.
Let
f_{1},...,
f_{n} R[
x_{1},...,
x_{r}] be
n
multivariate polynomials in the
r variables
x_{1},...,
x_{r}.
Let
a_{1},...,
a_{r} R be elements.
Let
U be the Jacobian matrix of
f_{1},...,
f_{n}
evaluated at
(
a_{1},...,
a_{r}).
That is,
U is the
n×
r matrix defined by
U = (u_{ij}) where u_{ij} = (a_{1},..., a_{r}) 
(99) 
We assume that the following properties hold
 for every
i = 1^{ ... }n we have
f_{i}(a_{1},..., a_{r}) 0 mod.
 the Jacobian matrix U is leftinvertible.
Then, for every positive integer
we can compute
a^{()}_{1},...,
a^{()}_{r} R such that
 for every
i = 1,..., n we have
f_{i}(a^{()}_{1},..., a^{()}_{r}) 0 mod^{},
 for every
j = 1,..., r we have
a^{()}_{j} a_{j}mod.
Proof.
We proceed by induction on
1.
For
= 1 the claim follows from the hypothesis of the theorem.
So let
1 be such that the claim is true.
Hence there exist
a^{()}_{1},...,
a^{()}_{r} R such that
f_{i}(a^{()}_{1},..., a^{()}_{r}) 0 mod^{}, i = 1,..., n 
(100) 
and
a^{()}_{j} a_{j}mod, i = 1,..., r 
(101) 
Since
is finitely generated, then so is
^{}
and let
g_{1},...,
g_{s} R such that
^{} = g_{1},..., g_{s} 
(102) 
Therefore, for every
i = 1,...,
n, there exist
q_{i1},...
q_{is} R
such that
f_{i}(a^{()}_{1},..., a^{()}_{r}) = q_{ik}g_{k} 
(103) 
For each
j = 1,...,
r we want to compute
B_{j} R such that
a^{(+1)}_{j} = a^{()}_{j} + B_{j} 
(104) 
is the desired
next approximation.
We impose
B_{j} ^{} so let
b_{j1},...,
b_{js} R be such that
B_{j} = b_{jk}g_{k} 
(105) 
Using Proposition
12
we obtain
where
u^{()}_{ij} is the Jacobian matrix
of
(
f_{1},...,
f_{n}) at
(
a^{()}_{1},...,
a^{()}_{r}).
Hence, solving for
a^{(+1)}_{1},..., a^{(+1)}_{r} R
such that
f_{i}(a^{(+1)}_{1},..., a^{(+1)}_{r}) 0 mod^{+1} 

leads to solving the system of linear equations:
q_{ik} + u^{()}_{ij}b_{jk} 0 mod 
(107) 
for
k = 1,...,
s and
i = 1,...,
n.
Now using
a^{()}_{j} a_{j}mod
for
j = 1,..., r we obtain
u^{()}_{ij} u_{ij}mod 
(108) 
Therefore the system linear equations
given by Relation (
107)
has solutions.