rschwieb
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 23h comment Upper Nilradical of a Ring @s.harp For future reference, there is no (significant volume of) ring theory literature that I'm aware of that ever refers to right or left ideals as simply "ideals". Two sidedness is always implied without the right-left adjectives. Might be good to beat this in mind unless there's some reason to expect otherwise 1d comment Are Z and Z* (defined below) isomorphic as rings? @J.Medley $x =1-(1-x)=f(1-x)$ you know that invertible functions are both 1-1 and onto, right? 2d comment Isomorphism of a ring of matrices @IDontKnowMath there are a lot of ways to tell if it isn't a subring. I can't think of a broad way to tell if one is a subring, though 2d comment Isomorphism of a ring of matrices @IDontKnowMath I switched it to a different description for you. 2d revised Isomorphism of a ring of matrices deleted 24 characters in body 2d comment Examples of applications of the Theorems of Pappus and Ménélaüs. The most striking thing about Pappus' theorem for me is that for a division ring $D$, $D$ is commutative (a field) iff Pappus holds in the plane $D^2$. That doesn't seem to satisfy your simplicity requirement though 2d comment Out of curiosity, which numbers are necessary? This is off topic because it is too hypothetical/speculative/opinion-based. Might be good in the chat, though, if not on this site then something like the xmcd forums or quora 2d comment Prove that a semigroup which satisfies a certain conditions is a group Not only is it considered inappropriate to post questions without effort, but it is also bad to post questions whose answers are already easily found here and elsewhere in the Internet. Try the search features first. If nobody uses the search feature, then why would we store all of this stuff in the first place? 2d answered Are Z and Z* (defined below) isomorphic as rings? 2d comment Are Z and Z* (defined below) isomorphic as rings? @learner But $(\Bbb Z, \circ)$ is cyclic, generated by the multiplicative identity $0$. Circle adding $0$ to itself gets you the negative integers, and circle adding the additive inverse of $0$ ( which is $2$) gets you the positive integers. 2d answered Isomorphism of a ring of matrices 2d revised Sum of nil right ideals as an ideal added 2 characters in body Apr 28 comment Let R* be the set of units of R and S* be the set of units of S. Prove that f(R*) = S*. @j. Medley it is a terrible idea to split your question between the title and the body of the post. Always put a complete question in the body Apr 28 comment Let R* be the set of units of R and S* be the set of units of S. Prove that f(R*) = S*. @s.harp well, ring isomorphisms of rings with identity preserve identity whether or not that is among your axioms :) Apr 28 revised Commutativity of a ring from idempotents. added 117 characters in body Apr 28 comment Sum of nil right ideals @karparvar there are obviously elements of arbitrarily high index of nilpotency. Apr 28 comment Sum of nil right ideals @karparvar there are obviously elements of arbitrarily high index of nilpotency. Apr 28 answered Commutativity of a ring from idempotents. Apr 27 comment Rings in which $ab=0$ implies $axb=0$ @Gro-Tsen well, not many people have these definitions in their heads at all times. This is a great example of a bunch of closely knit conditions that "tease apart" particular aspects of something that happens in a commutative ring. (The nilpotent elements forming an ideal) it's also nice to see the variety of examples it produces Apr 27 comment Rings in which $ab=0$ implies $axb=0$ @Gro-Tsen I'm glad you like it! I really need to get back to work on that website.