Let $I, J$ be ideals of a ring $R$. Show that $I + J = \{a + b\vert a \in I, b \in J\}$ is an ideal of R.
Because $I,J$ are ideals of $R$, so $I,J$ both have $0$, thus $0+0=0\in I+J$. This shows that $I+J$ is not empty and the additive idendity.
Suppose $a_1+b_1$ and $a_2+b_2$ are in $I+J$ where $a_1,a_2\in I$, $b_1,b_2\in J$. Then $(a_1+b_1)+(a_2+b_2)=(a_1+a_2)+(b_1+b_2)$ where $a_1+a_2\in I$ and $b_1+b_2\in J$ since $I,J$ are ideals of $R$. Thus $(a_1+b_1)+(a_2+b_2)\in I+J$, so $I+J$ is closed under addition that tells $I+J$ is additive subgroup of $R$.
Since $I,J $are additive subgroups of $R$, let $a\in I$ and $b\in J$, then $\exists -a\in I$ an $\exists -b\in J$ such that $(a+b)+(-a-b)=(a-a)+(b-b)=0\in I+J$ ; therefore, $I+J$ is a additive subgroup of $R$.
Let $a+b\in I+J$ where $a\in I, b\in J$ and $r\in R$. Since $I,J$ are ideals, so $ar,ra$ are in $I$ and $br,rb$ are in $J$, thus $r(a+b)=ra+rb\in I+J$ and $(a+b)r=ar+br\in I+J$. Therefore, $I+J$ is an ideal of $R$.
Does my proof valid? If valid, is there any way to write a better proof? If not, can you tell me what I miss? Thanks