This is Utumi's argument as it appears in ;Lemma Assume that
R satisfies the
ascending chain condition on
annihilators of the form \{r\in R\mid ar=0\} where
a is in
R. Then • Any nil one-sided ideal is contained in the lower nil radical Nil*(
R); • Every nonzero nil right ideal contains a nonzero nilpotent right ideal. • Every nonzero nil left ideal contains a nonzero nilpotent left ideal. ;Levitzki's Theorem Let
R be a right Noetherian ring. Then every nil one-sided ideal of
R is nilpotent. In this case, the upper and lower nilradicals are equal, and moreover this ideal is the largest nilpotent ideal among nilpotent right ideals and among nilpotent left ideals.
Proof: In view of the previous
lemma, it is sufficient to show that the lower nilradical of
R is nilpotent. Because
R is right Noetherian, a maximal nilpotent ideal
N exists. By maximality of
N, the
quotient ring R/
N has no nonzero nilpotent ideals, so
R/
N is a
semiprime ring. As a result,
N contains the lower nilradical of
R. Since the lower nilradical contains all nilpotent ideals, it also contains
N, and so
N is equal to the lower nilradical.
Q.E.D. ==See also==