Solvability, nilpotency and semisimplicity under extension of scalars #
Let L be a Lie algebra over a commutative ring R and let A be an R-algebra. Mathlib's
LieSubmodule.baseChange extends an ideal I of L to the ideal I.baseChange A of A ⊗[R] L,
and LieAlgebra.derivedSeriesOfIdeal_baseChange computes the derived series of an extended ideal
by extending the derived series term by term. This file records the same statement for the series
L ≥ ⁅I, L⁆ ≥ ⁅I, ⁅I, L⁆⁆ ≥ ⋯ that measures nilpotency of an ideal, and reads both series off as
transfer principles:
IsSolvable ↥(I.baseChange A) ↔ IsSolvable ↥I, IsNilpotent ↥(I.baseChange A) ↔ IsNilpotent ↥I
Ascent — transferring either property from I to I.baseChange A — asks nothing of A. Descent,
the → direction of the equivalences as displayed, is exactly where faithful flatness enters,
through Submodule.baseChange_inj: a term of either series can vanish after extending scalars
only if it vanished already. Applied to the two largest ideals, ascent gives
(radical R L).baseChange A ≤ radical A (A ⊗[R] L),
(nilradical R L).baseChange A ≤ nilradical A (A ⊗[R] L).
Neither containment is forced to be an equality by the transfer principles, because an ideal of
A ⊗[R] L need not be extended from L at all, and nothing above bounds the ideals that are
not. The case where both sides are ⊥ is the separate statement
LieAlgebra.hasTrivialRadical_baseChange_iff: over a field of characteristic zero a
finite-dimensional Lie algebra has trivial radical exactly when some field extension of it does,
so extending scalars can neither destroy nor create a solvable ideal of a semisimple algebra.
For a finite-dimensional Lie algebra over a field of characteristic zero the containment is an
equality: LieAlgebra.radical_baseChange says the radical commutes with a field extension, and
LieAlgebra.one_tmul_mem_radical_baseChange_iff reads that as a criterion, so membership in the
radical may be tested after extending scalars. The corresponding statement for the nilradical is
not proved here and does not follow, since L ⧸ nilradical K L need not have trivial
nilradical.
Main results #
LieIdeal.lcs_baseChange: the series⁅I, ⁅I, … ⁅I, M⁆…⁆⁆commutes with extension of scalars.LieIdeal.isSolvable_baseChange_iffandLieIdeal.isNilpotent_baseChange_iff: an ideal is solvable, respectively nilpotent, exactly when its faithfully flat extension of scalars is.LieAlgebra.baseChange_radical_leandLieAlgebra.baseChange_nilradical_le: the extension of the radical, respectively of the nilradical, lands in the radical, respectively the nilradical, of the extended algebra.LieAlgebra.hasTrivialRadical_baseChange_iff: in characteristic zero a finite-dimensional Lie algebra has trivial radical exactly when its extension to a field extension does.LieAlgebra.radical_baseChange: in characteristic zero the solvable radical of a finite-dimensional Lie algebra commutes with a field extension, withLieAlgebra.one_tmul_mem_radical_baseChange_iffreading it as a membership criterion that may be checked after extending scalars.
References #
- [N. Bourbaki, Lie Groups and Lie Algebras, Chapters 1--3][bourbaki1975], Chapter I, §5 and §6, for solvability, nilpotency and the radical under extension of scalars.
The series ⁅I, ⁅I, … ⁅I, M⁆…⁆⁆ commutes with extension of scalars. This is the
nilpotency counterpart of Mathlib's LieAlgebra.derivedSeriesOfIdeal_baseChange.
The extension of scalars of a solvable ideal is solvable. No hypothesis on the coefficient algebra is needed in this direction.
The extension of scalars of a nilpotent ideal is nilpotent. No hypothesis on the coefficient algebra is needed in this direction.
An ideal is solvable exactly when its faithfully flat extension of scalars is solvable.
An ideal is nilpotent exactly when its faithfully flat extension of scalars is nilpotent.
The extension of scalars of the solvable radical lands in the solvable radical. Nothing
formal makes the containment an equality, since an ideal of A ⊗[R] L need not be extended from
L; over a field of characteristic zero it is one, which is LieAlgebra.radical_baseChange.
The extension of scalars of the nilradical lands in the nilradical.
Triviality of the radical is insensitive to a field extension. In characteristic zero a finite-dimensional Lie algebra has trivial radical exactly when its extension of scalars to a field extension does.
The ← direction is the substantive one: it says that extending scalars cannot create a
solvable ideal. Neither direction follows from LieIdeal.isSolvable_baseChange_iff, which only
speaks of ideals extended from L. Characteristic zero is a genuine hypothesis here, not a
convenience.
In characteristic zero the solvable radical commutes with a field extension. For a
finite-dimensional Lie algebra L over a field K of characteristic zero and a field extension
A of K, the radical of A ⊗[K] L is the extension of scalars of the radical of L.
Characteristic zero is a genuine hypothesis, not a convenience. The equality determines the
radical of A ⊗[K] L completely, although it says nothing about individual solvable ideals of
A ⊗[K] L, which need not themselves be extended from L; as a test for membership in the
radical it is LieAlgebra.one_tmul_mem_radical_baseChange_iff.
Membership in the radical may be checked after a field extension. In characteristic zero a vector of a finite-dimensional Lie algebra lies in the solvable radical exactly when its canonical image in an extension of scalars does.