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Jun
26
answered why is a simple ring not semisimple?
Jun
26
comment why is a simple ring not semisimple?
"When will the result hold?" What result?
Jun
26
comment Graphs associated with rings and modules
There is also something called the "annihilator graph" of a ring, and a graph for the central idempotents.
Jun
26
comment Please give an example of a ring that does NOT have a multiplicative identity but contains a subring that does have an identity..
@GeoffreyCritzer $\Bbb Z[x]$ is certainly a domain, and no nontrivial ideal will have an identity. Yes, you could use the multiples of $1+x^2$, or even the multiples of $x$ for that matter.
Jun
26
awarded  Nice Question
Jun
26
revised Example of a commutative, local, dual ring with nilradical $N$ such that $ann(N)\nsubseteq N$
edited title
Jun
26
revised Semilocal commutative ring with two or three maximal ideals
added 337 characters in body
Jun
26
comment Radical of An Ideal In $Z_n$
the zero ideal is an easy edge case that could be mentioned.
Jun
25
comment How bad must be a ring to allow cyclic artinian modules that are not noetherian?
@Pipicito It's unclear how to answer your question "how bad must a ring be" when you are already aware of examples of the phenomenon. Can you elaborate on what you're looking for?
Jun
25
comment Semilocal commutative ring with two or three maximal ideals
@JohnBrevik That one is particularly interesting since it's also a domain, and therefore not semiperfect. The four idempotents of $R/J(R)$ can't lift to the two idempotents in $R$.
Jun
25
revised Semilocal commutative ring with two or three maximal ideals
added 136 characters in body
Jun
25
answered Semilocal commutative ring with two or three maximal ideals
Jun
25
comment Semilocal commutative ring with two or three maximal ideals
I didn't realize you were looking for a way to determine $n$, sorry. I took the decomposition for granted. Regards
Jun
25
comment Semilocal commutative ring with two or three maximal ideals
The maximal ideals of $A$ correspond exactly to those of $A/J(A)$, and since the latter is just a product of fields ($n$ of them, say) then it is elementary that there are $n$ maximal ideals.
Jun
24
comment Show $\mathbb{Q}[x,y]/\langle x,y \rangle$ is Not Projective as a $\mathbb{Q}[x,y]$-Module.
@user237522 (1) OK, yes, I have shown it for the projective case. (2) Interesting, I am not very well versed in flatness. I know in non-Noetherian domains you can have $I^2=I$, so your inclusion of Noetherianness is very necessary...
Jun
24
comment Show $\mathbb{Q}[x,y]/\langle x,y \rangle$ is Not Projective as a $\mathbb{Q}[x,y]$-Module.
@user237522 (1) I didn't mention anything about flatness. (2) If A and B are nonzero ideals of a domain, then $\{0\}\neq AB\subseteq A\cap B$, so you can't form a direct sum of the two ideals.
Jun
24
awarded  Nice Answer
Jun
24
comment Noether & Schmeidler- Hurwitz-Ideals
Nobody is asking you to "draw any consequences," mainly I am asking you to give readers a little help on what specific topic you are reading about. Most people will be totally unable to read the contents of your link, so don't you think you could help out?
Jun
24
comment Noether & Schmeidler- Hurwitz-Ideals
I just meant that if you might make more comments on the subject matter instead of relying so heavily on the German text.
Jun
24
comment Noether & Schmeidler- Hurwitz-Ideals
Any way you could add more English context? I can read much German, but maybe I have some clues.