Sum and intersection of ideals Let $R$ be a commutative ring and let $I,J⊂R$ be ideals. Define: 
$I∩J=\{a∈R : a∈I \mbox{ and }a ∈J\}$
$I+J=\{a+b∈R : a∈I, b∈J\}$. 
(a) Prove that $I∩J$ and $I+J$ are ideals. 
(b) Suppose $R=ℤ$ or $F[x]$, $I=\langle a\rangle$, and $J=\langle b\rangle$. Identify $I∩J$ and $I+J$. 
For part (a), a∈I and b∈J. For r∈R, ra∈I and rb∈J. Therefore, ra+rb=r(a+b)∈I+J, so I+J is an ideal. Is this correct for I+J?
Note: For part (b), suppose I= and J=. Now, I∩J and I+J are both principal in ℤ or in F[x].
I need to feed a formula for the generators of these ideals in terms of a and b?

You just have to check that the two properties defining ideals holds. 
So if $a, b \in I \cap J$, then $a$ and $b$ are both in $I$ and $J$, so, because $I$ and $J$ are ideals, $a +b \in I$ and  $a +b \in J$. Moreover, for $r \in R$, $ra \in I$ and $ra \in J$ because closure under multiplication by elements of $R$ holds in the two ideals, so $ra$ lies in their intersection. Try the same strategy for $I + J$. 
I think something is missing in your point (b) in the definition of $I$ and $J$.
