Let $W$ be a unital algebra and let $V$ be its maximal abelian subalgebra. Must the commutant $V^\prime$ of $V$ be commutative?
Commutants of commutative algebras
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$\begingroup$
abstract-algebra
ring-theory
ideals
noncommutative-algebra
1 Answers
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If $V$ is abelian then $V \subseteq V'$ by definition of the commutant.
If $V$ is maximal abelian, then I claim that this is an equality: $V = V'$. Indeed, suppose that $v \in V'$. Then the set $V \cup \{v\}$ consists of pairwise commuting elements and therefore generates an abelian subalgebra of $W$. This abelian subalgebra contains $V$, so it must be equal to $V$ by maximality. So it must be the case that $v \in V$.
In particular, yes, $V'$ is commutative.
(As another exercise, you may wish to prove that $V = V'$ if and only if $V$ is a maximal abelian subalgebra, without even assuming that the subset $V \subseteq W$ is a subalgebra!)
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0@Lubin, you're absolutely right, and I've included more explanation. (It was late when I first wrote my answer, so I was feeling short on words.) – 2012-09-06