Section 18.3-19.1. Today we will discuss finite-dimensional associative algebras and their representations. Definition 1. Let A be a finite-dimensional associative algebra over a field F . An element a ∈ A is nilpotent if an = 0 for some positive integer n. An algebra is said to be nilpotent if all of its elements are. Exercise 2. Every subalgebra and factor algebra of a nilpotent algebra are nilpotent. Con- versely, if I ⊂ A is a nilpotent ideal, and the quotient algebra A/I is nilpotent, then so is A. Example. The algebra of all strictly lower- (or upper-) triangular n×n matrices is nilpotent. Proposition 3. If the algebra A is nilpotent, then An = 0 for some n ∈ Z+, that is the product of any n elements if the algebra A equals 0. Proof. Let B ⊂ A be the maximal subspace for which there exists n ∈ Z+ such that Bn = 0. Note that B is closed under multiplication, i.e. B ⊃ Bk for any k ∈ Z+. Assume B 6= A and choose an element a ∈ A\B. Since aBn = 0, there exists k ∈ Z+ such that aBk 6⊂ B. Replacing a with a non-zero element in aBk we obtain aB ⊂ B. Recall that there exists m ∈ Z+ so that am = 0. Now, let us set C = B ⊕〈a〉. Then we have Cmn = 0 which contradicts the definition of the subspace B. � Unless A is commutative, the set of all nilpotent elements of A does not have to be an ideal (in general, not even a subspace). On the other hand, if I,J are nilpotent ideals in A, then so is I + J = {x + y |x ∈ I,y ∈ J} . Therefore, there exists the maximal nilpotent ideal which contains every other nilpotent ideal of A. Definition 4. The radical of A is the maximal nilpotent ideal of A it is denoted rad(A). The algebra A is semisimple if rad(A) = 0. If char(F) = 0, there exists an alternative description of semisimple algebras. Consider the regular representation of the algebra A ρ: A → L(A), ρ(a)(b) = ab, and define a “scalar product” on A via (a,b) = tr(ρ(ab)) = tr(ρ(a)ρ(b)). One can see that (·, ·) is a symmetric bilinear function on A satisfying (ab,c) = (a,bc). Definition 5. For any ideal I ⊂ A, its orthogonal complement I⊥ is defined as I⊥ = {a ∈ A |(a,i) = 0 for all i ∈ I} . Proposition 6. For any ideal I ⊂ A, its orthogonal complement I⊥ ⊂ A is also an ideal. Proof. For any a ∈ A, x ∈ I⊥, and y ∈ I we have (xa,y) = (x,ay) = 0 and (ax,y) = (y,ax) = (ya,x) = 0. � 1 2 Proposition 7. If A is an algebra over a field F with char(F) = 0, then every element a ∈ A orthogonal to all of its powers is nilpotent. Proof. Let a ∈ A be such that (a,an) = tr ρ(a)n+1 = 0 for all n ∈ Z+. Let L ⊃ F be the splitting field of the characteristic polynomial f(x) of the operator ρ(a). Then, over L we have f(x) = tk0 s∏ i=1 (t−λi)ki, where λi are distinct for i = 1, . . . ,s, and tr ρ(a)n+1 = s∑ i=1 kiλ n+1 i = 0. Letting n run through the set 1, . . . ,s, the above equation yields a system of s homogeneous linear equations in variables k1, . . . ,ks. The determinant of this system equals λ21 . . .λ 2 nV (λ1, . . . ,λn), where V (λ1.