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

∼

=

,