In my textbook (Abstract Algebra by Saracino), it gives $(\mathbb{Z}_2,\oplus,\odot)$ as an example of a commutative ring that has a maximal ideal that is not prime. It claims that the trivial ideal $\{0\}$ (which is maximal because it's the only ideal) is not prime. But the only way to express $0$ as the product of elements in $\mathbb{Z}_2$ is $1 \cdot 0 = 0 \cdot 1 = 0 \cdot 0$. In each case, $0$ appears as a factor, and $0 \in \{0\}$. So wouldn't this imply that $\{0\}$ is a prime ideal of $(\mathbb{Z}_2,\oplus,\odot)$? I just want to make sure I'm not crazy here...
This is the textbook's definition of prime:
Let $R$ be a ring, $I$ an ideal in $R$. Then $I$ is prime if whenever $a,b \in R$ and $ab \in I$, then at least one of $a$ or $b$ is in $I$.
Usually, we consider $\{0\}$ to be a prime ideal in any (commutative, unital) ring without zero divisors. The alternative is just cumbersome.
A ring is an integral domain iff $\{0\}$ is prime. $I\subseteq R$ is prime iff $R/I$ is an integral domain. Formulating these and similar statements while trying to take into account that $\{0\}$ can't be prime just isn't worth the hassle.
This also makes every maximal ideal into a prime ideal.