Let $I,J$ be ideals in a ring $R$. Then the following are ideals in $R$:
(1) $I\cap J$
(2) $I+J=$ $\{$ $i+j$ $:$ $i\in I$ and $j\in J$ $\}$
My attempt: (1) Since the intersection of rings is a ring, $I\cap J$ is a ring. Let $a\in I\cap J$ and $r\in R$. Since $I$, $J$ are ideals, $ar,ra\in I\cap J$. Hence the intersection is an ideal.
(2) Since $0\in I\cap J$, it follows that $0=0+0$, hence $0\in I+J$. Let $x,y \in I+J$. So $x=a_1+a_2$ and $y=a_1'+a_2'$ . Then, $-y=-a_1'-a_2'$. Hence, $x-y=$ $(a_1+a_2)+(-a_1'-a_2')$ $=$ $(a_1+a_2)+((-a_1')+(-a_2'))$. Since $R$ is an abelian group, $x-y=(a_1+(-a_1'))+(a_2+(-a_2'))\in I+J$. Let $a\in R$ and $x=i+j \in I+J$. Then $ax=a(i+j)=ai+aj \in I+J$, since $I,J$ are ideals. Furthermore, since $R$ is a ring, the distributive properties hold, so $xa\in I$ as well.
Note: An ideal is a subring of a ring, $R$, with the absorption properties.
Is my attempt correct?