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The classical Mazur–Gelfand theorem ([1]–[5]) implies that any subfield of a complex Banach algebra A is topologically isomorphic to C, provided it is a linear subspace of A. Here we present a somewhat surprising observation that if F is a subfield of A which is just a subring, and not a subalgebra, it need not be topologically isomorphic to a subfield of C. Let A be a complex Banach algebra and let F be a subfield of A. Denote by A0 the smallest closed subalgebra of A containing F . This is a commutative algebra with unit element equal to the unity of F . Thus A0 has a non-zero multiplicative-linear functional mapping isomorphically F into C. Therefore any subfield of A is isomorphic to a subfield of C under a continuous isomorphism. We shall show that in certain cases such an isomorphism cannot be a homeomorphic map. Denote by Q the set of all rational complex numbers, i.e. numbers of the form % = r1 + ir2 with rational r1 and r2. Denote by W the field of all rational functions in a variable t, with coefficients in Q; it contains the subfield of all constant functions, i.e. quotients of elements in Q. This subfield is clearly a dense subset of the complex plane C. Fixing a transcendental number c we obtain an isomorphic imbedding of W into C given by w → w(c), w ∈ W (a function w is uniquely determined by its value w(c) and this value is a well defined complex number, since c is transcendental). One can easily see that each isomorphism h of W into C is of the form w → w̃(d), where d is a transcendental number given by d = h(t), and w̃ is either w or w, depending on whether h(i) = i or h(i) = −i. Here w is an element of W obtained by replacing in w all coefficients by their complex conjugates. Take a complex Banach space X, dim X > 1, and take as A the algebra L(X) of all continuous endomorphisms of X. One can easily see that A contains a non-zero operator T satisfying

A subfield of a complex Banach algebra is not necessarily topologically isomorphic to a subfield of ℂ