## 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 $x \in \text{ann}_R(I) + \text{ann}_R(J). \ \Box$