*Proof*.
Let

be an element such that

.
Lift

to an algebraic integer

, and let

be the characteristic polynomial of

over

.
Using Proposition

14.1.4 we see that

reduces to the minimal polynomial

of

(by the Proposition the coefficients of

are in

, and

satisfies

, and the
degree of

equals the degree of the minimal polynomial
of

). 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

, we have

, so
since both are maximal ideals, there exists

with

,
i.e.,

. Thus

.