normed algebras , 5 Stephen William Semmes Rice University Houston , Texas

LetA be a finite-dimensional commutative algebra over the complex numbers with nonzero multiplicative identity element e. Let M denote the set of homomorphisms from A onto the complex numbers. Thus we get a homomorphism from A into the algebra of complex-valued functions on M, with respect to pointwise addition and multiplication of functions, by sending an element x of A to the function on M whose value at a homomorphism φ : A → C is φ(x). Suppose that φ, ψ are distinct elements of M. This means that there is an x ∈ A such that φ(x) 6= ψ(x). By choosing y to be a linear combination of x and e we may arrange it so that φ(y) = 1 and ψ(y) = 0. If E is a finite subset of M, then we can use this construction to show that for any fixed φ ∈ E there is a z ∈ A such that φ(z) = 1 and ψ(z) = 0 for all ψ ∈ E such that ψ 6= φ. Thus, for any finite subset E of M and any complex-valued function f on E there is a w ∈ A such that φ(w) = f(φ) for all φ ∈ E. In particular the number of elements of E is at most the dimension of A. It follows that the number of elements of M is at most the dimension of A, and that our homomorphism from A into the algebra of functions on M has image equal to the whole algebra of functions on M. Suppose that x ∈ A is in the kernel of the canonical homomorphism from A onto the algebra of functions onM. This is the same as saying that φ(x) = 0 for all homomorphisms φ from A onto the complex numbers. Equivalently, the spectrum of x consists of exactly 0. By the Cayley–Hamilton theorem, this is equivalent to saying that x is nilpotent, which is to say that xn = 0 for some positive integer n.