Let $A$ be an integrity domain with unity and $a \in A$. Show that an ideal generated by $a^{2}$ is a prime ideal if and only if $a = 0$.
Denote by $<a^{2}>$ or ideal generated by $a^{2}$. The "only if" is trivial. Since $A$ is the integrity domain and $a \in A$, if $a = 0$ then $a^{2} = 0$. Then $<a^{2}> = <0> = \lbrace 0 \rbrace$. Hence, if $bc \in <a^{2}>$ then $bc = 0$ and therefore $b = 0$ or $c = 0$, since $A$ is integrity domain.
I'm having a hard time proving the "if". Let $<a^{2}>$ be prime ideal, so if $bc \in <a^{2}>$ then $b \in <a^{2}>$ or $c \in <a^{2}>$. What would be the best way forward? I thank you for your help.
You know that $a^2\in <a^{2}>$, and $a^2$ is the product of $a$ with itself. If this ideal were prime, you would know that $a\in <a^{2}>$, which means that $a^2$ divides $a$. There must be some element $b\in A$ such that $a=a^2b$, hence $a(1-ab)=0$. So either $a=0$, either $1=ab$. The second case however is impossible, because it would imply that $a$ is invertible (assuming the ring is commutative), hence $<a>=<a^{2}>=A$ is not prime.
NB: It's the first time I read about "integrity" domain. Generally, it is called "integral domain" if I'm not mistaken.