*Proof*.
Let

be an element such that

.
Lift

to an algebraic integer

, and let

be the characteristic polynomial of

over

.
Using Proposition

9.3.4 we see that

reduces to a multiple of the minimal polynomial

of

(by the Proposition the coefficients of

are in

, and

satisfies

).
The roots of

are of the form

, and
the element

is also a root of

, so it is of the form

.
We conclude that the generator

of

is
in the image of

, which proves the theorem.

*Proof*.
By definition

for all , so it suffices to show that
if

, then there exists

such that

. If

, then

, so

. Since both
are maximal ideals, there exists

with

, i.e.,

. Thus

.