Derivation-invariant not implies characteristic

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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)
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There can exist a Lie ring L with a subring S such that S is a derivation-invariant Lie subring of L, such that S is not a characteristic subring of L.

Related facts

Similar facts


Facts used

  1. Center is derivation-invariant


Suppose A is a non-abelian Lie ring, A_1, A_2 are isomorphic copies of A, and L is the direct sum A_1 \oplus A_2. Define S = A_1 +Z(L). Then, S is derivation-invariant but not characteristic.

Proof that the subring is not characteristic

Consider the coordinate exchange automorphism that interchanges A_1 and A_2. Under this automorphism, A_1 + Z(L) goes to A_2 + Z(L). Since A_2 is non-abelian, it is not contained in Z(L), so the image of A_1 + Z(L) is not equal to it.

Proof that the subring is derivation-invariant

Consider a derivation d:L \to L. There exist four abelian group endomorphisms d_{11}, d_{12}, d_{21}, d_{22} that describe d, namely:

d(x,0) = (d_{11}(x), d_{12}(x)), \qquad d(0,y) = (d_{21}(y),d_{22}(y)).

In other words:

\! d(x,y) = (d_{11}(x) + d_{21}(y), d_{12}(x) + d_{22}(y)).

The derivation condition states that:

\! d[(x,y),(x',y')] = [d(x,y),(x',y')] + [(x,y),d(x',y')].

This gives:

\! (d_{11}([x,x']) + d_{21}([y,y']), d_{12}([x,x']) + d_{22}([y,y'])) = ([d_{11}(x),x'] + [d_{21}(y),x']+ [x,d_{11}(x')] + [x,d_{12}(y')], [d_{12}(x),y'] + [d_{22}(y),y']) + [y,d_{21}(x')] + [y,d_{22}(y')]).

We thus get:

\! d_{11}([x,x']) + d_{21}([y,y'])  = [d_{11}(x),x'] + [d_{21}(y),x']+ [x,d_{11}(x')] + [x,d_{12}(y')]


\! d_{12}([x,x']) + d_{22}([y,y']) = [d_{12}(x),y'] + [d_{22}(y),y']) + [y,d_{21}(x')] + [y,d_{22}(y')].

Setting y = y' = 0 gives that d_{11} is a derivation. Setting x = x' = 0 gives that d_{22} is a derivation. Plugging these back in, we get:

\! d_{21}([y,y'])  = [d_{21}(y),x']+ [x,d_{12}(y')]


\! d_{12}([x,x']) = [d_{12}(x),y'] + [y,d_{21}(x')].

Setting y' = 0 in the first equation gives that [d_{21}(y),x'] = 0 for all x',y, implying that d_{21} takes values in the center of A_1. Similarly, setting x' = 0 in the second equation gives that d_{12}(x) is in the center of A_2. In particular, this implies that:

\! d(x,0) = (d_{11}(x), d_{12}(x))

takes values in A_1 \oplus Z(A_2) = A_1 + Z(L) = S.

Thus, d(A_1) \subseteq S. Since Z(L) is derivation-invariant by fact (1), d(Z(L)) \subseteq Z(L), so d(S) = d(A_1) + d(Z(L)) \subseteq S + Z(L) = S. Thus, S is a derivation-invariant subring of L.