Every ideal is prime implies that $R$ is a field

Question

Let $R$ be a commutative ring with $1$ such that every proper ideal is a prime ideal. Prove that $R$ is a field. [Hint: for $a \neq 0$, consider the ideals $(a)$ and $(a^2)$].

Attempt.

Let $a \in R$ with $a \neq 0$. Now, $(a^2)$ is prime. Then $ra^2 \in (a^2)$ for $r \in R$. But I can write $ra^2 = ra \cdot a$. As $(a^2)$ is prime, then either $ra \in (a^2)$ or $a \in a^2$. Either way, $a \in (a^2)$ and so $(a) \subseteq (a^2)$. Trivially, we always have $(a^2) \subseteq (a)$. Thus, $(a)=(a^2)$.

Observe as well that since every ideal is prime, then $\langle 0 \rangle$ is prime. Thus, when $xy \in \langle 0 \rangle$ i.e. when $xy = 0$, then $x \in \langle 0 \rangle$ or $y \in \langle 0 \rangle$, i.e. $x=0$ or $y=0$. So $R$ is an integral domain.

Then as $(a) = (a^2)$, there is $r \in R$ so that $ra^2 = a$. As $R$ is an integral domain, we can cancel $a$ from both sides to get $ra=1$. And so $a$ is a unit, but then $R$ is a field.

Answer 1

The question here needs some qualification: $R$ should be required to be non-trivial and the assumption should be that every proper ideal is prime.

You introduced $r$, so that you can talk about a general member of $(a^2)$, but it has led to confusion. If you look at the simplest particular example of a member of $(a^2)$, then the fact that $(a^2)$ is prime should tell us something useful.

So, let $a$ be a non-zero element of $R$. Clearly $a^2 \in (a^2)$ which is a prime ideal by assumption. So by the definition of a prime ideal $a \in (a^2)$. So $a = xa^2$ for some $x \in R$. As the $0$ ideal is prime, $R$ is an integral domain (if $bc = 0$ then at least one of $b$ or $c$ is $0$, so $R$ has no non-trivial zero-divisors). Hence, as we have $1 \cdot a = x \cdot a \cdot a$, we can cancel $a$ on the right, to get $1 = x \cdot a$, giving us that $x$ is an inverse for $a$. So any non-zero element of $R$ has an inverse, i.e. $R$ is a field.