> [!NOTE] Theorem (Sum of ideals is ideal) > Let $R$ be a [[Rings|ring]]. Let $I,J$ be [[Ideal of Ring|ideals]] of $R.$ Then their sum given by $I+J = \{ i+j \mid i\in I, j \in J \}$is an ideal of $R.$ *Proof*. (Identity) $0 \in I+J$ since $0=0+0$ and $0\in I$ and $0\in I$ as $I,J$ are subgroups. (Closure) Let $x,y\in I+J.$ Then there exist $i_{1},i_{2}\in I$ and $j_{1},j_{2}\in J$ such that $x=i_{1}+j_{1}.$ Then $x+y=(i_{1}+i_{2})+(j_{1}+j_{2})\in I+J$ since $i_{1}+i_{2} \in I$ and $j_{1}+j_{2}\in J.$ (Inverses) $-x=(-i_{1})+(-j_{1}) \in I+J$ since $-i_{1} \in I$ and $-j_{1} \in J.$ Thus by [[Two-Step Subgroup Test]], $I+J$ is also a subgroup of $R.$ Now let $r\in R.$ Then $r\times x=r\times(i_{1}+j_{1})=r\times i_{1}+r\times j_{1} \in I+J$ since $r\times i_{1}\in I$ and $r\times j_{1} \in R$ as $I,J$ are ideals. Similarly, $x\times r \in I+J.$ Thus $I+J$ is an ideal of $R.$