Assume $R$ is a commutative ring and $I$ is a nonzero proper ideal of $R$ satisfying:
$(1)$ If $I_1$ and $I_2$ are ideals such that $I = I_1 \cap I_2$, then $I = I_1$ or $I = I_2$;
$(2)$ If $a^n \in I$, then $a \in I$.
Prove that $I$ is prime.
Here is my strategy thusfar. Suppose $xy \in I$. Ultimately, it seems the goal is to find ideals $I_x \ni x$ and $I_y \ni y$ such that $I = I_x \cap I_y$. By $(2)$, either $x \in I$ or $y \in I$.
I thought the natural thing to check was $I_x = I + (x)$ and $I_y = I + (y)$. For then $I \subset I_x \cap I_y$ and $I_x I_y \subset I$. Generally, $I_x I_y \subset I_x \cap I_y$, but the two are equal if $I_x + I_y = R$.
So I thought I might be able to use $(2)$ somehow to prove $I + (x) + (y) = R$. At the same time, I wonder whether this is simply not true and I am going about this the wrong way.
Am I on the right track? Can someone give me a hint that will help me get unstuck?
Thanks.
I'll just add a solution that continues your approach: Try to show that $I_x\cap I_y=I$, I'm not sure what you are trying to do with the product $I_xI_y$. So suppose we have an element in the intersection $$ax+i = a'y+i'\in I_x\cap I_y.$$ Now we just need a little trick: Since we know that $xy\in I$, multiply by $y$ so you get $$ a'y^2+i'y= axy+iy. $$ From this you should be able to conclude that $a'y^2\in I$ and then use (2) to get $a'y\in I$ and hence $a'y+i'\in I$. This shows that $I_x\cap I_y=I$. Now apply (1) to conclude.