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May
21
comment Example of a commutative, local, dual ring with nilradical $N$ such that $ann(N)\subsetneq N$
@Pierre-GuyPlamondon Oh, I see what you mean. I actually forgot that fields are exceptional in my discussion since the annihilator of N isn't proper. You're right that I have to eliminate them manually. Among nonfields, Krull dimension > 0 is necessary, and that is what I had been thinking of. Thanks for catching that before it went too far. In questions with a lot of hypotheses, it's very annoying to have left out a detail like that.
May
21
revised Example of a commutative, local, dual ring with nilradical $N$ such that $ann(N)\subsetneq N$
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May
21
revised Example of a commutative, local, dual ring with nilradical $N$ such that $ann(N)\subsetneq N$
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May
21
comment Example of a commutative, local, dual ring with nilradical $N$ such that $ann(N)\subsetneq N$
@Pierre-GuyPlamondon What you said is true, but a field has Krull dimension zero, so it doesn't satisfy the requirement of having nonzero Krull dimension.
May
20
comment Algebraic closure for rings
Oh I see what you meant now: a domain has its fraction field and then its algebraic closure. I only followed the thought halfway.
May
20
comment Algebraic closure for rings
the integral closure requires me to choose some auxiliary B Why wouldn't you use the full ring of quotients in the role of the field of fractions? Of course, there are still some bad elements... but that supplies a canonical overring.
May
20
revised Zorn's lemma converse? (Context: Maximal proper subgroups)
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May
20
answered Zorn's lemma converse? (Context: Maximal proper subgroups)
May
19
comment $R$ is a finite ring and for every $a \in R$, there exists a natural number $n(a)$ such that $a^{n(a)}=a$
@MarianoSuárez-Alvarez So it can be proved that an Artinian rng with Jacobson radical zero has an identity? That would be pretty useful...
May
19
revised $R$ is a finite ring and for every $a \in R$, there exists a natural number $n(a)$ such that $a^{n(a)}=a$
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May
19
revised $R$ is a finite ring and for every $a \in R$, there exists a natural number $n(a)$ such that $a^{n(a)}=a$
rolled back to a previous revision
May
19
comment $R$ is a finite ring and for every $a \in R$, there exists a natural number $n(a)$ such that $a^{n(a)}=a$
@darijgrinberg The original paper (Structure Theory for Algebraic Algebras of Bounded Degree, Theorem 11) does deal with rings without identity such as nil rings and radical rings. Since he seems to be specifying when identity is necessary, it seems like the theorem also applies to rings without identity. However I haven't traced the proof yet to be certain...
May
19
comment $R$ is a finite ring and for every $a \in R$, there exists a natural number $n(a)$ such that $a^{n(a)}=a$
@Mojtaba-CH That's what I intended with "generates the ideal as a right ideal," but looking at this after a night of sleep, I realize this may not be one of the equivalent characterizations of von Neumann regularity for rings without identity. I'll keep looking at the problem.
May
19
revised $R$ is a finite ring and for every $a \in R$, there exists a natural number $n(a)$ such that $a^{n(a)}=a$
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May
19
comment $R$ is a finite ring and for every $a \in R$, there exists a natural number $n(a)$ such that $a^{n(a)}=a$
@darij found a better way now, I think.
May
19
revised $R$ is a finite ring and for every $a \in R$, there exists a natural number $n(a)$ such that $a^{n(a)}=a$
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May
19
answered $R$ is a finite ring and for every $a \in R$, there exists a natural number $n(a)$ such that $a^{n(a)}=a$
May
18
asked Example of a commutative, local, dual ring with nilradical $N$ such that $ann(N)\subsetneq N$
May
18
comment If $I$ is a finitely generated semisimple left $R-$module, show that $I=Re$ for an idempotent $e \in I$.
@user26857 That is a very nice text.
May
16
awarded  Revival