## In a commutative unital ring, if a nontrivial prime ideal contains no zero divisors, then the ring is an integral domain

Let $R$ be a commutative ring with $1 \neq 0$. Prove that if $P \subseteq R$ is a prime ideal containing no zero divisors (in $R$), then $R$ is an integral domain.

Let $a,b \in R$ such that $ab = 0$. Since $P$ is a prime ideal and $ab \in P$, without loss of generality $a \in P$. If $a \neq 0$, then since $P$ contains no zero divisors in $R$, $b = 0$.

• ai  On July 3, 2011 at 11:23 am

How do we know that R has a prime ideal P to begin with? I have a longer proof that doesn’t assume R has a prime ideal, but this proof is nicer, assuming of course that we can always assume an arbitrary, commutative ring with a 1 has prime ideals.
Thanks.

• ai  On July 3, 2011 at 11:26 am

How do we know that R has a prime ideal P to begin with? I have a longer proof that doesn’t assume R has a prime ideal, but this proof is nicer, assuming of course that we can always assume an arbitrary, commutative ring with a 1 has prime ideals.
Thanks.

Oh, never mind, I got it:

If R has a 1, then every proper ideal is contained in a maximal ideal, and if R also commutative, then every maximal ideal is a prime ideal.