1. BASIC NOTIONS 5

Proof. For V finite dimensional and R complete, the proposition follows from

[AM69, Proposition 10.13]. If R is Artin,

mi

= 0 for i suﬃciently large, so the

inverse limit in (1.3) stabilizes. The topology of R V is therefore discrete and

R V

∼

= R V as (discrete) topological R-modules.

Lemma 1.9. Let U and V be (discrete) vector spaces, and R = (R, m) a local

complete Noetherian ring with residue field . Then there is a natural one-to-one

correspondence

(1.7) Φ : Lin

(

U, R V

)

∼

=

←→ LinR

c

(

R U , R V

)

: Ψ,

where Lin(−, −) denotes, as usual, the space of -linear maps and LinR(−,

c

−) the

space of continuous R-linear maps. Moreover, (1.7) restricts, for each k ≥ 0, to

the isomorphism

Φk : Lin

(

U,

mk

⊗ V

)

∼

=

←→ LinR

c

(

R U ,

mk

⊗ V

)

: Ψk.

Proof. The lemma, of course, follows from the universal property of R U in

the category of complete topological R-modules, but we include a direct proof here.

Let us define first the correspondences Φ and Ψ.

For a -linear map φ : U → R V denote by

˜

φ : R ⊗ U → R ⊗ V its R-linear

extension given by

˜(r

φ ⊗ u) := rφ(u), for u ∈ U and r ∈ R. Clearly,

˜(mi

φ ⊗ U) =

miφ(U)

⊂

mi(R

⊗ V ),

so, by (1.5),

˜(mi

φ ⊗ U) ⊂ mi ⊗ V . The map

˜

φ therefore induces a map

R ⊗ U

mi ⊗ U

−→

R ⊗ V

mi ⊗ V

of the quotients which, combined with the isomorphisms

R ⊗ U

mi ⊗ U

∼

=

R/mi

⊗ U and

R ⊗ V

mi ⊗ V

∼

=

R/mi

⊗ V

gives a map

˜

φ

i

:

R/mi

⊗ U →

R/mi

⊗ V . Define finally, in the notation (1.4) for

elements of the inverse limits,

Φ(φ)(x1,x2,x3,...) :=

(˜1(x1),

φ

˜

φ 2(x2),

˜

φ 3(x3),...).

It is easy to verify that Φ(φ) : R U→ R V is a well-defined continuous map.

The definition of the inverse correspondence Ψ is even simpler. Given an R-

linear map f : R U→ R V , Ψ(f) : U → R V is the composition

Ψ(f) : U

ι

→ R U

f

−→ R V .

It is clear that Ψ◦Φ = . It therefore remains to prove that Ψ is a monomorphism.

So assume that Ψ(f) = 0 for a continuous f : R U → R V and prove that

then f = 0. Since R ⊗ V is dense in R V , it is enough to show that f(r ⊗ v) = 0

for r ∈ R and v ∈ V . But this is obvious, since

f(r ⊗ v) = rf(1 ⊗ v) = rf(ι(v)) = rΨ(f)(v) = 0.

We leave the proof that the isomorphisms Φ resp. Ψ restrict to Φk resp. Ψk as an

exercise.