Annihilators in self-injective rings

Posted: February 19, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: ,

Problem. Let R be a commutative ring and suppose that R is injective as R-module. Prove that for any ideals I,J of R  we have \text{ann}_R(I \cap J)=\text{ann}_R(I) + \text{ann}_R(J).

Solution. The inclusion \text{ann}_R(I) + \text{ann}_R(J) \subseteq \text{ann}_R(I \cap J) is trivial. Now let x \in \text{ann}_R(I \cap J). See that the map f: I + J \longrightarrow R defined by f(a + b) = ax, \ \ a \in I, b \in J, is a well-defined R-homomorphism. Hence, since R is self-injective, there exists some r \in R such that ax = f(a + b) = (a + b)r, for all a \in I, b \in J. Therefore if we let a=0, we will get br=0 for all b \in J, and if we let b=0, we will get ax = ar for all a \in I. Thus x - r \in \text{ann}_R(I) and r \in \text{ann}_R(J). So x \in \text{ann}_R(I) + \text{ann}_R(J). \ \Box

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