A celebrated theorem of Jacobson states that
Theorem. Let be a ring, not necessarily containing
. If, for each
there exists a positive integer
such that
, then
is commutative.
This is a very strong and difficult result (although not very useful in practice). However, we can obtain some special cases via elementary means.
Proposition 1. Let be a ring such that for each
we have
. Then
is commutative.
Proof. Let . Then
, i.e.,
. Again,
, i.e.,
. Thus
, i.e.,
for each
. Thus
, as desired.
The next case is already considerably harder.
Proposition 2. Let be a ring such that for each
we have
. Then
is commutative.
Proof. Let . Then
shows that
,
and shows that
.
Hence
for all .
Plugging into
gives
, i.e.,
for each
.
Plugging into
gives
, i.e.,
for each
. Replacing
by
gives
, i.e.,
.
Also, multiplying by
first on the left and then on the right and then subtracting the two gives
.
From the last two paragraphs we conclude that for all
.
Corollary. Let be a ring such that for each
we have
for some
. Then
is commutative.
Proof. Note that if for some
then
. Hence the result follows by Proposition 2.