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A nondegenerate regular-trace pairing forces semisimplicity

Statement

Let $K$ be a field and $H = H_{ϕ}$ a finite-dimensional field specialization from Finite Hecke specialization and its regular-trace determinant. For $x \in H$ , let $L_{x} : H \to H$ be left multiplication and define $B (x, y) : = {Tr}_{K} (L_{x y})$ . The pairing $B$ is nondegenerate if $B (x, y) = 0$ for every $y \in H$ implies $x = 0$ . If $B$ is nondegenerate, then the left regular $H$ -module is semisimple, meaning an internal direct sum of simple left submodules (Semisimple modules as direct sums of simple modules). This implication holds over every field; no converse is asserted.

Facts & Assumptions

Given: A field $K$ , a finite-dimensional unital $K$ -algebra $H = H_{ϕ}$ , and the regular-trace pairing $B (x, y) = {Tr}_{K} (L_{x y})$ .

[F1]

$H_{ϕ}$ is a finite-dimensional algebra with regular-trace pairing $B (x, y) = {Tr}_{K} (L_{x y})$ (Finite Hecke specialization and its regular-trace determinant).

[F2]

A finite-dimensional module has a maximal proper submodule when nonzero, and iterating gives a finite composition series (Finite matrix and module preliminaries: right inverses, rank invariance, finite length and nilpotent trace).

[F3]

A submodule of a finite direct sum of simple modules is isomorphic to a direct sum of a subfamily and has a complement (Finite matrix and module preliminaries: right inverses, rank invariance, finite length and nilpotent trace).

[F4]

A nilpotent endomorphism of a finite-dimensional vector space has trace zero (Finite matrix and module preliminaries: right inverses, rank invariance, finite length and nilpotent trace).

[F5]

A module is semisimple when it is an internal direct sum of simple submodules (Semisimple modules as direct sums of simple modules).

[F6]

A simple left module is nonzero and has no proper nonzero submodule (Simple module: a nonzero module with no proper nonzero submodule).

Proof

technique · direct

1.1givenF1F6algebra

By [F1], $H$ is finite-dimensional and $B$ is its regular-trace pairing. Define $J : = ⋂_{S} {Ann}_{H} (S)$ , intersecting over all simple left $H$ -modules as in [F6]. This equals the intersection of all maximal left ideals: if $a$ lies in every maximal left ideal and $s \neq 0$ lies in a simple module $S$ , the map $H \to S$ , $x \mapsto x s$ , is onto and its kernel is maximal, so $a s = 0$ ; conversely, if $a$ annihilates every simple module then it kills $1 + M$ in each simple quotient $H / M$ , so $a \in M$ for every maximal left ideal $M$ . In particular $J$ is a two-sided ideal.

1.2givenF2algebra

By [F2], choose a finite composition series $0 = M_{0} ⊊ M_{1} ⊊ \dots ⊊ M_{r} = H$ of the left regular module. Since $J$ annihilates each simple factor $M_{i} / M_{i - 1}$ , $J M_{i} \subseteq M_{i - 1}$ for every $i$ . Repeating this inclusion gives $J^{r} H = 0$ , and then $J^{r} = 0$ because $1 \in H$ .

2.1givenstep 1.1algebrachoose

Starting with $K_{0} = H$ , if $K_{i} \neq J$ choose $x_{i} \in K_{i} ∖ J$ . The equality in step 1.1 gives a maximal left ideal $N_{i}$ not containing $x_{i}$ ; then $K_{i + 1} : = K_{i} \cap N_{i}$ contains $J$ and has strictly smaller $K$ -dimension than $K_{i}$ . Thus after at most $\dim_{K} H$ steps the intersection is $J$ . This finite descent selects finitely many maximal left ideals and uses no arbitrary Choice.

2.2step 1.2F1F4algebra

For $j \in J$ and $b \in H$ , the two-sided-ideal property gives $j b \in J$ ; by step 1.2, $(j b)^{r} = 0$ . Hence $L_{j b}^{r} = L_{(j b)^{r}} = 0$ , so [F4] gives $B (j, b) = {Tr}_{K} (L_{j b}) = 0$ for every $b \in H$ .

3.1step 2.1F3F6algebra

The map $H / J \to ⨁_{i} H / N_{i}$ sending $x + J$ to $(x + N_{i})_{i}$ is an injective $H$ -module homomorphism: its kernel is $(⋂_{i} N_{i}) / J = 0$ , and each quotient $H / N_{i}$ is simple by [F6]. By [F3] its image is semisimple, so $H / J$ is semisimple.

4.1givenstep 3.1step 2.2F5algebra∎

Nondegeneracy of $B$ and step 2.2 imply $j = 0$ for every $j \in J$ , hence $J = 0$ . Therefore $H ≅ H / J$ is semisimple by step 3.1 and [F5]. No separable-division-algebra trace converse is used.

Depends on

Used by

Finite Coxeter Hecke algebras are semisimple over the generic characteristic-zero field Theorem

Dependency tree · two levels

42 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Arun Ram, Lectures in Representation Theory, Lecture 5 (standard reference, not scraped)
Maarten Solleveld, Affine Hecke Algebras and Their Representations (standard reference, not scraped)