84 reputation
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location Sweden
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visits member for 3 years, 5 months
seen Oct 10 '12 at 8:20

Jul
2
awarded  Curious
May
5
comment Is a ring closed under both operations?
You wrote "the group" instead of "the set" in the formulation of your question, so let me just remind you that $(R,\cdot)$ is in general not a group.
Feb
13
accepted Do non-commutative algebras with dense commutative subalgebras exist?
Feb
13
awarded  Benefactor
Feb
13
accepted Proper ideals generated by central ideals
Feb
13
comment Proper ideals generated by central ideals
Parsa, thanks! This is a nice counterexample! Ewan, from the relations it is clear that $k[Z] \subseteq Z(R)$. Again, using the relations, it is not difficult to verify the other inclusion.
Feb
12
awarded  Editor
Feb
12
awarded  Promoter
Feb
12
revised Proper ideals generated by central ideals
Clarified the notation RI.
Nov
19
accepted Conditions on $X$ ensuring that a non-constant continuous function $f: X\to \mathbb{C}$ exists
Nov
19
asked Conditions on $X$ ensuring that a non-constant continuous function $f: X\to \mathbb{C}$ exists
Nov
9
asked Proper ideals generated by central ideals
Oct
17
accepted Dense *-subalgebras of C*-algebras and their intersections with sub-C*-algebras
Oct
16
answered Find $a$ and $b$ with whom this expression $x\bullet y=(a+x)(b+y)$ is associative
Oct
16
asked Dense *-subalgebras of C*-algebras and their intersections with sub-C*-algebras
Oct
14
comment Do non-commutative algebras with dense commutative subalgebras exist?
Thanks! I tried to do epsilon-stuff with $||xy-yx||$ and got lost. Your method was very quick and nice!
Oct
14
asked Do non-commutative algebras with dense commutative subalgebras exist?
Sep
10
comment Proof that a bijection has unique two-sided inverse
Thomas, $\beta=\alpha^{-1}$. I think that this is the main goal of the exercise. The unique map that they look for is nothing but the inverse.
Sep
10
comment Proof that a bijection has unique two-sided inverse
robjohn, this is the whole point - there is only ONE such bijection, and usually this is called the 'inverse' of $\alpha$. What one needs to do is suppose that there is another map $\beta'$ with the same properties and conclude that $\beta=\beta'$.
Aug
23
awarded  Supporter