Lemma 3

Let A be a K-algebra and L a subalgebra of A.

a) The algebra L is a maximal commutative subalgebra of A if and only if L is equal to its commutant $L'$ in A.

b) Let $K'$ be a nonzero commutative K-algebra. Then L is a maximal commutative subalgebra of A if and only if $L_{(K')}$ is a maximal commutative subalgebra of $A_{(K')}$.

Let us prove a). First suppose that L is equal to $L'$. Then L is commutative; if M is a commutative subalgebra of A containing L, then we have $xy=yx$ for $x$ in L and $y$ in M, so $M\subset L'$ and therefore M = L. Consequently, L is a maximal commutative subalgebra of A.

Conversely, suppose that L is a maximal commutative subalgebra of A, and let $x$ be an element of $L'$. The subalgebra M of A generated by $L\cup \{x\}$ is then commutative and contains L. Because L is maximal, we have M = L, and so $x\in L$ and finally $L = L'$, which gives a).

By Proposition 6 of III, §4, No. 4, p. 468, the commutant of $L_{(K')}$ in $A_{(K')}$ is $L'_{(K')}$. Since the equalities $L = L'$ and $L_{(K')}= L'_{(K')}$ are equivalent (II, §7, No. 9, p. 311, Proposition 19), assertion b) follows from a).