$\mathrm{Ann}(I \cap J) = \mathrm{Ann}(I) +\mathrm{Ann}(J)$ if $R$ is self-injective

Question

If $R$ is injective as an $R$-module, then for every two ideals $I$ and $J$ we will have $\mathrm{Ann}(I \cap J) = \mathrm{Ann}(I) + \mathrm{Ann}(J)$.

I made an effort for an hour but it was not helpful.

Answer 1

Annihilator of an ideal $I$ in a module $M$ is just $\mathrm{Hom}_R(R/I, M)$. Now use the inclusion $R/I\cap J\to R/I\oplus R/J$ and injectivity of $M$.