Proof
Let and be a prime ideal, then for some . Thus
- ,
since is an ideal, which implies or . In the second case, suppose for some , then thus or and, by induction on , we conclude , in particular . Therefore is contained in any prime ideal and .
Conversely, we suppose
and consider the set
:=\lbrace J\subseteq R\mid J{\text{ is an ideal and }}f^{m}\notin J{\text{ for all }}m\in \mathbb {Z} _{>0}\rbrace }
which is
non-empty, indeed
.
is
partially ordered by
and any chain
has an upper bound given by
, indeed:
is an ideal
[Note 1] and if
for some
then
for some
, which is impossible since
; thus any chain in
has an upper bound and we can apply
Zorn's lemma: there exists a maximal element
. We need to prove that
is a prime ideal: let
, then
since
is maximal in
, which is to say, there exist
such that
, but then
, which is absurd. Therefore if
,
is not contained in some prime ideal or equivalently
and finally
.