TOKYO J. MATH. VOL. 15, No. 1, 1992 Ring Derivations on Semi-Simple Commutative Banach Algebras Osamu HATORI and Junzo WADA Tokyo Medical College and Waseda University Introduction. Let $A$ be a commutative Banach algebra. An (resp. linear) operator $D$ on $A$ is called a ring (resp. linear) derivation on $A$ if equations $D(f+g)=D(f)+D(g)$ and $D(fg)=fD(g)+D(f)g$ are satisfied for every $f$ and $g$ in $A$ . The image of linear derivation was studied by Singer and Wermer [5] under the hypothesis ofcontinuity ofthe operator, and Thomas [6] has proved that every linear derivation on a commutative Banach algebra maps into the radical of the algebra. On the other hand there are ring derivations which do not map into the radical (cf. [1]). In this paper we characterize ring derivations on semi-simple commutative Banach algebras. A function algebra is semi-simple and so the results generalize our previous results in [3]. As a consequence of the results it is shown that only the zero operator is a ring derivation on a semi-simple commutative Banach algebra with the carrier space without an isolated point, which is a generaliza- tion of a theorem of Nandakumar [4]. 1. Lemmata. LEMMA 1. Let $A$ be a commutative Banach algebra with the carrier space $M_{A}$ . Suppose that $D$ is a ring derivation on A. Then $(D(\alpha f))^{\wedge}=\alpha(D(f))^{\wedge}for$ every $f$ in $A$ and for every rational number $\alpha$ in the complex number field $C$ , where denotes the Gel’fand representation. PROOF. If $\alpha$ is a rational real number, then $D(\alpha f)=\alpha D(f)$ by standard argument. So we only show that $(D(if))^{\wedge}=i(D(f))^{\wedge}$ , where $i$ is the imaginary unit. For every $f$ in $A$ , $2fD(f)=D(f^{2})=-D((if)^{2})=-2\iota fD(lf)$ , so we have $(D(f))^{\wedge}(x)=-i(D(if))(x)$ for every $x$ in $M_{A}$ with $\hat{f}(x)\neq 0$ . When $\hat{f}(x)=0$ , choose $g$ in $A$ with $\hat{g}(x)\neq 0$ . In the same way we have $(D(g))^{\wedge}(x)=-i(D(ig))^{\wedge}(x)$ and $(D(f+g))^{\wedge}(x)=-i(D(i(f+g)))^{\wedge}(x)$ since $(f+g)^{\wedge}(x)=f(x)+\hat{g}(x)\neq 0$ , so $(D(f))^{\wedge}(x)+(D(g))^{\wedge}(x)=-i(D(if))^{\wedge}(x)-i(D(ig))^{\wedge}(x)$ . Received May 8, 1991