If P is a prime ideal of a ring R and P contains no non-zero zero divisors, then R is an integral domain

Question

Here, $R$ is a commutative ring with unity. I realize there are many proofs of this online, but all of them seem to assume a definition of zero divisors that Dummit and Foote does not. Here is what the definition is in Dummit and Foote:

If $a \in P$ and $b \in P$, then $ab = 0$ implies either $a = 0$ or $b = 0$. The crucial assumption here is that both $a,b \in P$. So the solution for this problem typically goes something like this:

Let $P$ be a prime ideal, and suppose we have $a,b \in R$ such that $ab = 0$. But since $P$ is an ideal, $ab \in P$, and since $P$ is prime, this implies that either $a \in P$ or $b \in P$. Without loss of generality, assume $a \in P$. If $a = 0$, we are done, so suppose $a \neq 0$. Then since $P$ has no zero divisors, this implies $b = 0$.

This last sentence is my main issue. It seems to imply that $b \in P$, which we are not given at all. So how would one prove this? I also tried using the fact that $R/P$ is an integral domain, but I got stuck at the same place.

Answer 1

You must mean "$P$ contains no (nonzero) zero divisors" or else the hypothesis would never hold: $0$ is always in $P$.

So you are starting with "suppose $a,b\in R$, both are nonzero, and $ab=0$." This begins a proof by contradiction.

So here is where things can be improved:

Without loss of generality, assume $a \in P$. If $a = 0$, we are done, so suppose $a \neq 0$. Then since $P$ has no zero divisors, this implies $b = 0$.

can be more like either

Without loss of generality, assume $a \in P$. By Assumption $P$ contains no zero divisors, so $ab=0$ implies $b=0$. But this contradicts the earlier assumption that $b\neq 0$.

or

Without loss of generality, assume $a \in P$. But $b\neq 0$ witnesses that $a$ is a zero divisor of $R$. But this contradicts the earlier assumption that $P$ contains no zero divisors.