(1) Let $A$ be a UFD, $a \in A$ irreducible(=prime). It is well-known (and easily shown) that $Aa$, the ideal of $A$ generated by $a$, is a prime ideal.
If $a_1,a_2 \in A$ are irreducible, what can be said about $Aa_1+Aa_2$? Namely, is it a prime ideal? I guess there are counterexamples.
(2) Let $R$ be a (noetherian) UFD, $T$ an indeterminate, and $h \in R[T]$ irreducible of degree $\geq 2$.
- When $h'$ is also irreducible?
- Is the ideal of $R[T]$ generated by $h$ and $h'$ a prime ideal?
Any comment is welcome. (I apologize if my questions are somewhat trivial).
Edit: A similar question can be found here.
In general, I would suggest that you work some of these out instead of being too optimistic.
For 1), take for example, $A=k[x,y]$, $a_1=x-y^2, a_2=x$, both are prime, but $Aa_1+Aa_2$ is not.
For 2), again take $R=k[x]$ so that $R[y]=A$ as above. Consider for the first bullet, $h(y)=y^3-x^2$. Then $h$ is irreducible, but $h'(y)=3y^2$ is not. For the last one, take $h(y)=y^2-x^3$, again irreducible, but so is $h'(y)=2y$ (characteristic not 2). But the ideal generated by $h,h'$ is not prime.