Proof.
Write

with

and

.
First suppose

is exactly divisible by a power of

,
so for some

we have

but

.
Then
If

is the largest power of

that divides

, then

,

,

satisfy the conclusion of the lemma.
By unique factorization of integers, there is a smallest multiple
of
such that
is exactly divisible by
. Now apply the
above argument with
and
replaced by
and
.
Proof.
The first two properties of Definition
16.2.1 are
immediate. For the third, we first prove that if

then
Assume, without loss, that

and
that both

and

are nonzero.
Using Lemma
16.2.5 write

and

with

or

possibly negative. Then
Since

it follows that

.
Now suppose

. Then
so
hence

.