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Descended orbit idempotents are primitive and their blocks have one simple type

Statement

Let $A$ be a finite-dimensional semisimple unital $F$ -algebra and suppose $E / F$ is finite Galois with $E \otimes_{F} A$ split semisimple. The elements $e_{O}$ obtained by descending Galois orbit sums are precisely the primitive central idempotents of $A$ . Each $A e_{O}$ is a simple Artinian ring with unit $e_{O}$ and has exactly one simple left-module class. For $A = 0$ there are no such blocks. More precisely, if $A e_{O} ≅ M_{n} (D)$ and $S = D^{n}$ is its column simple, then $D ≅ {End}_{A} (S)^{o p}$ .

Facts & Assumptions

[F1]

Orbit sums descend uniquely to nonzero orthogonal central idempotents whose sum is $1$ , and central idempotents upstairs are unique subset sums of split factors: Orbit sums of primitive split central idempotents descend.

[F2]

Every nonzero semisimple ring is a finite product of full matrix rings over division rings: Wedderburn–Artin theorem for semisimple rings.

[F3]

Such a product has exactly one simple left-module class for each factor, its column module: Simple modules over a product of matrix rings over division rings.

[F4]

The matrix sizes and division rings in the decomposition are unique up to permutation and isomorphism: Uniqueness of the Wedderburn–Artin factors.

[F5]

A proper subspace of a finite-dimensional space has strictly smaller dimension: If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ .

[F6]

The canonical map $A \to E \otimes_{F} A$ , $a \mapsto 1 \otimes a$ , is injective: Galois fixed points recover finite-dimensional scalar extensions.

Proof

Given: $A, E / F$ , and the descended idempotents $e_{O}$ as stated.

1.1

Suppose $e_{O} = f + g$ with nonzero orthogonal central idempotents $f, g \in A$ . Their scalar extensions are nonzero: F6 makes the canonical scalar-extension map injective. Since $f e_{O} = f$ and $g e_{O} = g$ , the subset description in F1 writes $1 \otimes f$ and $1 \otimes g$ as disjoint nonempty subsets partitioning $O$ . Each subset is $Γ$ -stable because its sum is a tensor $1 \otimes a$ and the subset representation is unique. But a nonempty invariant subset of one orbit equals the entire orbit: for any of its points, every point of the orbit is a translate. Two such disjoint subsets cannot exist. Thus $e_{O}$ is primitive.

F1F6algebra

2.1

For any primitive central idempotent $e$ of $A$ , the products $e e_{O}$ are central orthogonal idempotents summing to $e$ . At least one is nonzero since $e \neq 0$ ; exactly one is nonzero by primitivity of $e$ . For that orbit, $e = e e_{O}$ . Now $e_{O} = e + (e_{O} - e)$ is a sum of orthogonal central idempotents: $(e_{O} - e)^{2} = e_{O} - e$ and $e (e_{O} - e) = 0$ . Primitivity from step 1.1 forces $e_{O} - e = 0$ . Thus every primitive central idempotent is exactly one $e_{O}$ .

F1step 1.1algebra

3.1

For $A \neq 0$ , choose $A ≅ \prod_{j} M_{n_{j}} (D_{j})$ by F2. A matrix commuting with all $E_{k k}$ has zero off-diagonal entries, and commuting with all $E_{k l}$ makes its diagonal entries equal. Commuting also with matrices having any $d \in D_{j}$ in position $(1, 1)$ forces that common entry into $Z (D_{j})$ . Conversely such scalar matrices commute with everything. Since a division ring has no idempotents except $0, 1$ , the primitive central idempotents in this product are exactly its factor identities. By step 2.1, each $A e_{O}$ is therefore one factor $M_{n} (D)$ .

F2step 2.1algebra

4.1

A nonzero two-sided ideal $J \subset M_{n} (D)$ contains a matrix $X$ with entry $x_{p q} \neq 0$ . For any $a, b$ , $E_{a p} X E_{q b} = x_{p q} E_{a b} \in J$ . Left multiplying by the diagonal matrix having $x_{p q}^{- 1}$ in position $(a, a)$ gives $E_{a b} \in J$ . Hence $I = \sum_{a} E_{a a} \in J$ and $J = M_{n} (D)$ . Also every left or right ideal is an $F$ -subspace of the finite-dimensional algebra $A e_{O}$ ; a strict descending chain strictly decreases dimension and so terminates. Thus this ring is simple and Artinian, without asserting that its regular left module is simple when $n > 1$ .

F5step 3.1algebra

5.1

F3 gives its unique simple left-module class $S = D^{n}$ . A commuting endomorphism of this column module preserves the first coordinate line, and its value there determines all other coordinates by the matrix units; commuting with left multiplication by $D$ forces it to be right multiplication by a fixed $d \in D$ . Composition reverses the order of these right multiplications, so ${End}_{A} (S) ≅ D^{o p}$ , or $D ≅ {End}_{A} (S)^{o p}$ . F4 ensures that using another product description does not change the factor data up to the stated equivalences. For $A = 0$ , F1 supplies empty families and there is no primitive idempotent or simple module; for $n = 1$ the same calculations give the division ring itself. [F1, F3, F4, step 3.1, step 4.1, algebra] QED

Remarks

Zheng, Theorem 3.2.1 and Remark 3.2.3, pp.117–118, supply the semisimple decomposition; §3.3, especially Lemma 3.3.2 and Warning 3.3.3(1), explains the simple-ring versus simple-module distinction. The matrix-ideal computation above proves that distinction's needed positive assertion locally.

Depends on

Used by

Galois orbits classify simple modules after splitting base change Theorem

Dependency tree · two levels

41 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

Weizhe Zheng, Lectures on Algebra (10 January 2025) (standard reference, not scraped)