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 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