Levitzky's theorem

In mathematics, more specifically ring theory and the theory of nil ideals, Levitzky's theorem, named after Jacob Levitzki, states that in a right Noetherian ring, every nil one-sided ideal is necessarily nilpotent.[1][2] Levitzky's theorem is one of the many results suggesting the veracity of the Köthe conjecture, and indeed provided a solution to one of Köthe's questions as described in (Levitzki 1945). The result was originally submitted in 1939 as (Levitzki 1950), and a particularly simple proof was given in (Utumi 1963).

Proof

This is Utumi's argument as it appears in (Lam 2001, p. 164-165)

Lemma[3]

Assume that R satisfies the ascending chain condition on annihilators of the form where a is in R. Then

  1. Any nil one-sided ideal is contained in the lower nil radical Nil*(R);
  2. Every nonzero nil right ideal contains a nonzero nilpotent right ideal.
  3. Every nonzero nil left ideal contains a nonzero nilpotent left ideal.
Levitzki's Theorem [4]

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.

gollark: Duckduckgoing it just turns up a lot of information on compiling *everything* like that, which is slow.
gollark: I thought there was a way to do this but I forgot it; can you compile a single dependency with a higher optimization level?
gollark: Oh, never mind, found it.
gollark: Thanks. Apparently that works. Is there a way to *cancel* that task from the function which spawns it?
gollark: I think I'm missing something then. It says```rusterror[E0373]: async block may outlive the current function, but it borrows `ws`, which is owned by the current function --> src/connection.rs:40:23 |40 | task::spawn(async { | _______________________^41 | | let mut interval = stream::interval(Duration::from_secs(10));42 | | while let Some(_) = interval.next().await {43 | | ws.send_string("Hi".to_string()); | | -- `ws` is borrowed here44 | | }45 | | }); | |_____^ may outlive borrowed value `ws````

See also

Notes
  1. Herstein 1968, p. 37, Theorem 1.4.5
  2. Isaacs 1993, p. 210, Theorem 14.38
  3. Lam 2001, Lemma 10.29.
  4. Lam 2001, Theorem 10.30.

References
  • Isaacs, I. Martin (1993), Algebra, a graduate course (1st ed.), Brooks/Cole Publishing Company, ISBN 0-534-19002-2
  • Herstein, I.N. (1968), Noncommutative rings (1st ed.), The Mathematical Association of America, ISBN 0-88385-015-X
  • Lam, T.Y. (2001), A First Course in Noncommutative Rings, Springer-Verlag, ISBN 978-0-387-95183-6
  • Levitzki, J. (1950), "On multiplicative systems", Compositio Mathematica, 8: 76–80, MR 0033799.
  • Levitzki, Jakob (1945), "Solution of a problem of G. Koethe", American Journal of Mathematics, The Johns Hopkins University Press, 67 (3): 437–442, doi:10.2307/2371958, ISSN 0002-9327, JSTOR 2371958, MR 0012269
  • Utumi, Yuzo (1963), "Mathematical Notes: A Theorem of Levitzki", The American Mathematical Monthly, Mathematical Association of America, 70 (3): 286, doi:10.2307/2313127, hdl:10338.dmlcz/101274, ISSN 0002-9890, JSTOR 2313127, MR 1532056
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