commutative rings with prime ideals are fields

Let $R$ be a commutative ring. If every ideal of $R$ is prime, then $R$ is a field.

A commutative ring with identity is an integral domain if it has no zero divisors. We proceed by establishing that $R$ is an integral domain and then showing that inverses of elements exist, this $R$ is a field. Recall that an ideal $P$ is a prime ideal if $P \neq R$ and $ab \in{P} \Longrightarrow a \in{P}$ or $b \in{P}$.

$\textit{Proof}$ : Consider the ideal generated by the additive identity $(0)=0 \subseteq R$. $ab \in{(0)} \Longleftrightarrow ab=0 \Longrightarrow a=0$ or $b=0$. This is exactly the definition of not having zero divisors. So, $R$ is an integral domain.
Now let $x \neq 0$. $x^2 \in (x^2) \Longrightarrow x \cdot x \in{(x^2)}$. So, if $x \in{(x^2)} \Longrightarrow \exists a\in{R}$ s.t. $x=ax^2$. Since we have cancellation laws in an integral domain, then $xx^{-1}=ax^2x^{-1}$ so $1=ax$. So, $x$ has an inverse. $\therefore R$ is a field. $\blacksquare$