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idealizers

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Plural form of idealizer.

Idealizers might refer to
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* * * * * * I * * * S * * * ( * T * ) * = * { * s * ∈ * S * ∣ * s * T * ⊆ * T * * and * * T * s * ⊆ * T * } * * * {\displaystyle \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 * * * * * * I * * * R * * * ( * A * ) * * * {\displaystyle \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 * ∈ * L * ∣ * [ * r * , * S * ] * ⊆ * S * } * * * {\displaystyle \{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 mention 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 S in which S is a Lie ideal.

Idealizers might refer to

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*
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* {\displaystyle \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
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* {\displaystyle \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
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* {\displaystyle \{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 mention 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 S in which S is a Lie ideal.

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