Maps Between Uniform Algebras Preserving Norms of Rational Functions

Let A, B be uniform algebras. Suppose that A ₀, B ₀ are subgroups of A ⁻¹, B ⁻¹ that contain exp A, exp B respectively. Let α be a non-zero complex number. Suppose that m, n are non-zero integers and d is the greatest common divisor of m and n. If T : A ₀ → B ₀ is a surjection with ||T(f)^mT(g)ⁿ - a||_￥ = ||f^mgⁿ - a||_￥{\|T(f)^{m}T(g)^{n} - \alpha\|_{\infty} = \|f^{m}g^{n} - \alpha\|_{\infty}} for all f,g ? A₀{f,g \in A_0}, then there exists a real-algebra isomorphism (T)\tilde] : A ? B{\tilde{T} : A \rightarrow B} such that (T)\tilde](f)^d = (T(f)/T(1))^d{\tilde{T}(f)^d = (T(f)/T(1))^d} for every f ? A₀{f \in A_0}. This result leads to the following assertion: Suppose that S _A , S _B are subsets of A, B that contain A ⁻¹, B ⁻¹ respectively. If m, n > 0 and a surjection T : S _A → S _B satisfies ||T(f)^mT(g)ⁿ - a||_￥ = ||f^mgⁿ - a||_￥{\|T(f)^{m}T(g)^{n} - \alpha\|_{\infty} = \|f^{m}g^{n} - \alpha\|_{\infty}} for all f, g ? S_A{f, g \in S_A}, then there exists a real-algebra isomorphism (T)\tilde] : A ? B{\tilde{T} : A \rightarrow B} such that (T)\tilde](f)^d = (T(f)/T(1))^d{\tilde{T}(f)^d = (T(f)/T(1))^d} for every f ? S_A{f \in S_A}. Note that in these results and elsewhere in this paper we do not assume that T(exp A) = exp B.