1. BASIC NOTIONS 13
Lemma 1.22. Let (U, · , e) be an associative unital -algebra with a complete
linear topology given by a descending sequence {Ui}i≥1 of ideals such that
(1.25) Ua · Ub Ua+b, for each a, b 1.
Then each element v U such that v e U1 is invertible. In other words, the
subset
(U1 + e) := {v V | v e U1}
of U is a group.
Proof. Let α := v e so that v = e + α. For any k 1 consider the element
vk
−1
:= e α +
α2

α3
+ · · · +
(−1)kαk.
By (1.25),
αi
Ui for each i 1 and, since U is complete, the sequence {vk
−1}k≥1
converges to an element v−1 U. One clearly has, for each k 1,
e vk
−1v
= e vvk
−1
Uk+1,
which implies that e
v−1v
= e
vv−1

i≥1
Ui. Since complete linear spaces
are Hausdorff,
i≥1
Ui = {0}, therefore
v−1v
=
vv−1
= e.
By (1.10), condition (1.25) implies the uniform continuity of the multiplication
· : U × U U. Notice also that all ideals Ui, i 1, must be proper. Indeed, if
Ui = U for some i, then certainly U = U1 for the filtration is descending. Thus,
by (1.25),
Ua · U1 = Ua · U = Ua Ua+1.
This implies that Ua = U for each a 1, so
i≥1
Ui = U = {0}, which contradicts
the completeness of U.
Proof of Proposition 1.21. We apply Lemma 1.22 to the unital associative
algebra LinR(R
c
V, R V ) of continuous R-linear maps, with the multiplication
given by the (point-wise) composition, and the identity map : R V R V
as the unit. It is easy to see that the descending filtration
(1.26) LinR(R
c
V, R V )i := LinR(R
c
V,
mi
V ), i 1,
satisfies (1.25) and that, for φ LinR(R
c
V, R V ), the induced map φ : V V
is the identity if and only if
φ LinR(R
c
V, R V )1.
Lemma 1.22 now produces an inverse map
φ−1.
Another useful consequence of Lemma 1.22 is the following standard
Lemma 1.23. Let R = (R, m) be a local, not necessary complete, Noetherian
ring with residue field . For each n 0, the quotient an :=
R/mn+1
is a local
Artin ring with the maximal ideal
m/mn+1.
Proof. Since the completion of a local Noetherian ring is again local Noether-
ian [AM69, Proposition 10.16] and since
R/m

=
R/m,
we may assume that R is complete. It is clear that m/mn+1 is an ideal in R/mn+1.
Since
R/mn+1
m/mn+1

=
R
m

=
,
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