> [!NOTE] Lemma > Let $(R,+,\times)$ be a [[Rings|ring]] and let $J$ be an [[Ideal of Ring|ideal]] of $R.$ Then $J$ is a [[Subring|subring]] of $R.$ **Proof**: By definition, $(J,+)$ is a subgroup of $(R,+).$ Also by definition, every product involving a member of $J$ is in $J$ and $J$ is closed is closed under multiplication. Thus $J$ passes the [[Three Step Subring Test|subring test]] and is a subring of $R.$