Lesson plan

Idealizer

In abstract algebra, the idealizer of a subsemigroup T of a semigroup S is the largest subsemigroup of S in which T is an ideal. Such an idealizer is given by : \mathbb{I}_S(T)=\{s\in S \mid sT\subseteq T \text{ and } Ts\subseteq T\}. In ring theory, if A is an additive subgroup of a ring R, then \mathbb{I}_R(A) (defined in the multiplicative semigroup of R) is the largest subring of R in which A is a two-sided ideal. In Lie algebra, if L is a Lie ring (or Lie algebra) with Lie product [x,y], and S is an additive subgroup of L, then the set : \{r\in L\mid [r,S]\subseteq S\} is classically called the normalizer of S, however it is apparent that this set is actually the Lie ring equivalent of the idealizer. It is not necessary to specify that [S,r] ⊆ S, because anticommutativity of the Lie product causes [s,r] = −[r,s] ∈ S. The Lie "normalizer" of S is the largest subring of L in which S is a Lie ideal.

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Article summary

In abstract algebra, the idealizer of a subsemigroup T of a semigroup S is the largest subsemigroup of S in which T is an ideal. Such an idealizer is given by : \mathbb{I}_S(T)=\{s\in S \mid sT\subseteq T \text{ and } Ts\subseteq T\}. In ring theory, if A is an additive subgroup of a ring R, then \mathbb{I}_R(A) (defined in the multiplicative semigroup of R) is the largest subring of R in which A is a two-sided ideal. In Lie algebra, if L is a Lie ring (or Lie algebra) with Lie product [x,y], and S is an additive subgroup of L, then the set : \{r\in L\mid [r,S]\subseteq S\} is classically called the normalizer of S, however it is apparent that this set is actually the Lie ring equivalent of the idealizer. It is not necessary to specify that [S,r] ⊆ S, because anticommutativity of the Lie product causes [s,r] = −[r,s] ∈ S. The Lie "normalizer" of S is the largest subring of L in which S is a Lie ideal.

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