Let $R$ be a commutative ring (not necessarily containing $1$). Say that $R$ is the trivial ring if it has trivial (zero) multiplication. If $R$ is the trivial ring, then $R$ has no prime ideals (as any ideal contains $0$, hence the square of every element). Is the converse true – i.e.,
if $R$ is not the trivial ring, then does $R$ necessarily have a prime ideal?
If $R$ has a unity 1, then $R$ has a maximal ideal, which is necessarily prime. However, if $R$ does not have a 1, then maximal ideals need not exist, and even if they do, they need not be prime (in fact, an ideal $I \subseteq R$ is a nonprime maximal ideal iff $R/I$ is the trivial ring of order $p$, for some prime $p \in \mathbb{Z}$). So at first glance, considering maximal ideals doesn't seem terribly helpful…
Best Answer
Take $R=2\mathbb Z/8\mathbb Z$. This ring has no prime ideals and the multiplication is clearly not trivial.
Since the OP took another way, I'd like to add a generalization of the example above: take $R=d\mathbb Z/n\mathbb Z$ with $d\mid n$, and $m=n/d$. The prime ideals of $R$ are of the form $pd\mathbb Z/n\mathbb Z$ with $p$ prime, $p\mid m$ and $p\not\mid d$. Now it's easy to give examples of such rings without prime ideals.