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The identity of a finite-dimensional algebra decomposes into primitive central idempotents

Statement

Let $F$ be a field and $A$ a finite-dimensional unital $F$ -algebra with centre $Z (A) : = {z \in A : z a = a z for all a \in A}$ (Algebras over a commutative ring, central structure maps, and algebra homomorphisms). Then $Z (A)$ is a finite-dimensional commutative $F$ -algebra, and the identity $1 \in Z (A)$ admits a decomposition $1 = e_{1} + \dots + e_{k}$ into pairwise orthogonal primitive central idempotents of $A$ : each $e_{i}$ is a nonzero idempotent of $Z (A)$ that is not the sum of two nonzero orthogonal idempotents of $Z (A)$ . No semisimplicity of $A$ is assumed and no choice principle is used.

Facts & Assumptions

Given: A field $F$ , a finite-dimensional unital $F$ -algebra $A$ , and the centre $C : = Z (A) = {z \in A : z a = a z for all a \in A}$ .

[F1]

$F$ -algebra conventions: a unital $F$ -algebra $A$ is a unital ring together with a unital ring homomorphism $η : F \to A$ whose image is central, and the induced scalar action $r a : = η (r) a$ makes $A$ an $F$ -module with multiplication $A \times A \to A$ bilinear (Algebras over a commutative ring, central structure maps, and algebra homomorphisms).

[F2]

Subspace dimension: if $V$ is a finite-dimensional vector space over $F$ with $\dim_{F} V = n$ and $U$ a linear subspace of $V$ , then $U$ is finite-dimensional with $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ (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 $U = V$ ).

[F3]

Well-ordering principle: every nonempty subset of $N$ has a least element (The well-ordering principle).

Proof

technique · direct

1.1givenF1

If $A = 0$ , then $C = Z (A) = {0}$ and the empty family ( $k = 0$ ) satisfies $1 = e_{1} + \dots + e_{k}$ with the stated conditions vacuously, so assume $A \neq 0$ ; then in $A$ one has $1 \neq 0$ and hence $1 \neq 0$ in $C$ .

1.2givenF1algebra

The centre $C$ is a commutative unital $F$ -subalgebra of $A$ and, in particular, a linear subspace of the $F$ -vector space $A$ : $1 \in C$ ; if $x, y \in C$ , $α \in F$ and $a \in A$ , then $(x + y) a = x a + y a = a x + a y = a (x + y)$ , $(α x) a = α (x a) = α (a x) = a (α x)$ and $(x y) a = x (y a) = x (a y) = (x a) y = (a x) y = a (x y)$ , while $0 \in C$ ; and distinct elements of $C$ commute with each other because each commutes with all of $A$ .

1.3givenF3

Suppose, for contradiction, that the following general claim fails: every finite-dimensional commutative unital $F$ -algebra $D$ with $1_{D} \neq 0$ and $\dim_{F} D \geq 1$ admits a decomposition of $1_{D}$ into pairwise orthogonal idempotents of $D$ that are primitive in $D$ . Let $S \subseteq N$ be the nonempty set of $F$ -dimensions of such counterexamples; by [F3] let $d \in S$ be least and fix a counterexample $D$ with $\dim_{F} D = d$ .

2.1step 1.2F2

The centre $C$ is finite-dimensional over $F$ with $\dim_{F} C \leq \dim_{F} A$ , by [F2] applied to the linear subspace $C$ of the finite-dimensional space $A$ .

2.2step 1.3given

Since $D$ is a counterexample, $1_{D}$ is not primitive in $D$ : otherwise $k = 1$ , $e_{1} = 1_{D}$ would be the required decomposition. Hence there are nonzero orthogonal idempotents $u, v \in D$ with $1_{D} = u + v$ .

3.1step 2.2F1algebra

The set $D_{u} : = u D = {u x : x \in D}$ is a linear subspace of $D$ and, with the inherited operations, a commutative unital $F$ -algebra with identity $u$ : it is closed since $u x + u y = u (x + y)$ , $α (u x) = u (α x)$ and $(u x) (u y) = u^{2} x y = u (x y)$ for $x, y \in D$ , $α \in F$ , while $u \cdot u x = u x$ ; and $u \neq 0$ . The same holds for $D_{v} : = v D$ with identity $v$ .

4.1step 3.1step 1.3F2algebra

One has $D_{u} \neq D$ because $v \notin D_{u}$ : if $v = u x$ then $0 = u v = u^{2} x = u x = v \neq 0$ , a contradiction using $u v = 0$ and $u^{2} = u$ . Hence by [F2], applied to the subspace $D_{u}$ of $D$ with $\dim_{F} D = d$ , one has $\dim_{F} D_{u} < d$ ; the same argument gives $\dim_{F} D_{v} < d$ .

5.1step 4.1step 1.3given

By the minimality of $d$ in $S$ , no counterexample has dimension $< d$ , and since $\dim_{F} D_{u} < d$ and $\dim_{F} D_{v} < d$ with $1_{D_{u}} = u \neq 0$ , $1_{D_{v}} = v \neq 0$ , the general claim holds for $D_{u}$ and for $D_{v}$ : there are pairwise orthogonal idempotents $f_{1}, \dots, f_{a}$ of $D_{u}$ primitive in $D_{u}$ with $u = f_{1} + \dots + f_{a}$ , and pairwise orthogonal idempotents $g_{1}, \dots, g_{b}$ of $D_{v}$ primitive in $D_{v}$ with $v = g_{1} + \dots + g_{b}$ .

6.1step 5.1step 3.1algebra

The combined family $f_{1}, \dots, f_{a}, g_{1}, \dots, g_{b}$ is a decomposition of $1_{D}$ into pairwise orthogonal idempotents of $D$ that are primitive in $D$ : the elements lie in $D_{u} \cup D_{v} \subseteq D$ ; distinct $f$ 's are orthogonal and distinct $g$ 's are orthogonal by step 5.1; $f_{i} g_{j} = (u f_{i}) (v g_{j}) = u v f_{i} g_{j} = 0$ ; and if $f_{i} = p + q$ with $p, q$ nonzero orthogonal idempotents of $D$ , then $p = f_{i} p = u f_{i} p = u p \in D_{u}$ and likewise $q = u q \in D_{u}$ , contradicting primitivity of $f_{i}$ in $D_{u}$ (and symmetrically for $g_{j}$ in $D_{v}$ ).

7.1step 6.1step 1.3

Step 6.1 contradicts the choice of $D$ as a counterexample, so no counterexample exists and the general claim of step 1.3 holds: every finite-dimensional commutative unital $F$ -algebra $D$ with $1_{D} \neq 0$ and $\dim_{F} D \geq 1$ decomposes its identity into pairwise orthogonal primitive idempotents.

8.1step 7.1step 2.1step 1.1∎

Apply step 7.1 to $D = C = Z (A)$ : by steps 1.1 and 1.2, $C$ is a commutative unital $F$ -algebra with $1 \neq 0$ and $\dim_{F} C \geq 1$ ; by step 2.1 it is finite-dimensional. This produces $1 = e_{1} + \dots + e_{k}$ with the $e_{i}$ pairwise orthogonal idempotents of $C = Z (A)$ primitive in $C$ ; idempotents of $Z (A)$ are exactly the central idempotents of $A$ , and primitive in $C$ says precisely that $e_{i}$ is not the sum of two nonzero orthogonal idempotents of $Z (A)$ . Since no semisimplicity was used and the only invoked principles are the choice-free subspace theorem [F2], the choice-free well-ordering principle [F3] and finite selections, the proof is complete.

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Sources

Andrew Mathas, Hecke algebras and Schur algebras of the symmetric group, complete prepublication manuscript (1998; expanded published edition: AMS University Lecture Series 15, 1999); complete 79-page manuscript, SHA-256 8cfeb611a95a3ff25cb95ad22bfd3a68475818301c34b04d131a8480a1b997e0 (standard reference, not scraped)