Derivation-invariant not implies characteristic: Difference between revisions
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===Proof that the subring is not characteristic=== | ===Proof that the subring is not characteristic=== | ||
Consider the ''coordinate exchange automorphism'' that interchanges <math>A_1</math> and <math>A_2</math>. Under this automorphism, <math>A_1 + Z(L)</math> goes to <math>A_2 + Z(L)</math>. Since <math>A_2</math> is non-abelian, it is not | Consider the ''coordinate exchange automorphism'' that interchanges <math>A_1</math> and <math>A_2</math>. Under this automorphism, <math>A_1 + Z(L)</math> goes to <math>A_2 + Z(L)</math>. Since <math>A_2</math> is non-abelian, it is not contained in <math>Z(L)</math>, so the image of <math>A_1 + Z(L)</math> is not equal to it. | ||
===Proof that the subring is derivation-invariant=== | ===Proof that the subring is derivation-invariant=== | ||
Latest revision as of 10:48, 19 August 2011
This article gives the statement and possibly, proof, of a non-implication relation between two Lie subring properties. That is, it states that every Lie subring satisfying the first Lie subring property (i.e., derivation-invariant Lie subring) need not satisfy the second Lie subring property (i.e., characteristic subring of a Lie ring)
View a complete list of Lie subring property non-implications | View a complete list of Lie subring property implications
Get more facts about derivation-invariant Lie subring|Get more facts about characteristic subring of a Lie ring
Definition
There can exist a Lie ring with a subring such that is a derivation-invariant Lie subring of , such that is not a characteristic subring of .
Related facts
Similar facts
- Perfect direct factor implies derivation-invariant
- Self-centralizing direct factor implies derivation-invariant
Converse
Facts used
Proof
Suppose is a non-abelian Lie ring, are isomorphic copies of , and is the direct sum . Define . Then, is derivation-invariant but not characteristic.
Proof that the subring is not characteristic
Consider the coordinate exchange automorphism that interchanges and . Under this automorphism, goes to . Since is non-abelian, it is not contained in , so the image of is not equal to it.
Proof that the subring is derivation-invariant
Consider a derivation . There exist four abelian group endomorphisms that describe , namely:
.
In other words:
.
The derivation condition states that:
.
This gives:
.
We thus get:
and:
.
Setting gives that is a derivation. Setting gives that is a derivation. Plugging these back in, we get:
and:
.
Setting in the first equation gives that for all , implying that takes values in the center of . Similarly, setting in the second equation gives that is in the center of . In particular, this implies that:
takes values in .
Thus, . Since is derivation-invariant by fact (1), , so . Thus, is a derivation-invariant subring of .