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1h
revised Non-bijective endomorphisms of finite groups
edited title
1h
revised A generalization of upper nilradical
deleted 27 characters in body
1h
revised A generalization of upper nilradical
deleted 27 characters in body
2h
answered A generalization of upper nilradical
16h
comment Complex numbers - What am I missing?
@LiziPizi If you are willing to sit hours for others to do the problem, then you ought to be willing to spend (probably at most 10 minutes) writing what you did.
17h
comment Complex numbers - What am I missing?
Hmm, since you spent all that time working on the problem, you should reproduce it so that we can see where you're stuck. Surely it won't take you long since you have all of the work you've done so far right in front of you.
17h
comment Is this limit of an infinte product infinity, undefined or something else
Good answer. In reflection, there's no choice but to work in the extended real line because otherwise we would be looking at a sequence of undefined quantities. So it's the most natural environment for this problem.
18h
comment Is this limit of an infinte product infinity, undefined or something else
@wythagoras it is strange you're drawing a distinction between a limit at infinity to an undefined limit. A limit at infinity is already undefined. We are able to use 'at infinity' suggestively on certain undefined limits.
19h
comment Proof that $C^\infty(0,1)$ is a subring
What does the notation $M((0,1),\Bbb R)$ mean?
20h
comment Proof that $C^\infty(0,1)$ is a subring
If you know what additively and multiplicatively closed mean, and you know the two operations of the superring, then there is nothing in your way. If you have a specific question during this process , that would be a better question
1d
comment About $R/I$ Where $I$ is a Prime Ideal
+1 complete and clear!
1d
answered Stabilization of all even/odd terms of sequence of iterated centralizers.
1d
comment A finite von Neumann regular ring is unital and has $ab = 1$ if $ba = 1$
@moonlight Yeah, finiteness is certainly a lot easier to use than the chain conditions. My guess is that it is simply for simplicity. Perhaps there is a clever trick in conjunction with VNR too.
1d
comment A finite von Neumann regular ring is unital and has $ab = 1$ if $ba = 1$
It holds for rings (with identity) satisfying much looser chain conditions. It holds for right Noetherian rings, and even rings satisfying the ACC on annihilators. A ring with the property that $xy=1$ implies $yx=1$ is called a Dedekind finite ring (or directly finite ring.) (Thought you might find this of interest @moonlight )
1d
comment Is a subring contained in the centralizer of its centralizer?
Whenever the defining relationship of some set like this is symmetric, then you'll get that the original set is in the double application of your special operator. Another example is the annihilator operator for subsets in a commutative ring.
Sep
2
comment Determining North-South Line Via Watch Method: Theory & Reason
Hi: there are several reasons your post won't be well received. 1) the question isn't self contained (you make people hunt through unnumbered slides to read) 2) it's not really a mathematical question: it is a physics or astronomy question 3) you're asking too many questions. 4) you labeled it as a discussion (although it isn't really and you could easily remedy that )
Sep
2
comment Show that $k[x,y]/(xy-1)$ is not isomorphic to a polynomial ring in one variable.
+1 That's a nice way to reduce the problem!
Sep
2
comment Show that $k[x,y]/(xy-1)$ is not isomorphic to a polynomial ring in one variable.
@Learner Sorry if I misunderstood the hypotheses you intended. Hope you get an answer you like.
Sep
2
comment Show that $k[x,y]/(xy-1)$ is not isomorphic to a polynomial ring in one variable.
@JyrkiLahtonen That's OK by me. These things can go either way sometimes.
Sep
1
comment What do we call the ring homomorphism $R \rightarrow \mathrm{End}_{\mathbf{Ab}}(X)$ associated with an $R$-module $X$?
I was just thinking that representors of right modules would be fundamentally different since they map $R$ via a ring anti-homomorphism, so I wasn't sure you'd call them the same thing.