Prove that $p$ is prime if and only if there are no zero divisors in the quotient ring.

Question

$R$ is a factorial ring, and $p\in R$ is a non-zero non-invertible element. I need to prove that $p$ is prime if and only if there are no zero divisors in the quotient ring $R/(p)$.
Here is what I've done:
Since $p\in R$ where $R$ is a factorial ring, and $p$ is a non-zero non-invertible element, it is true that $p=p_1\times\dots\times p_n$ where $p_i$ is prime for $1\leq i\leq n$. Also, I know that $a$ is called a zero divisor of the ring $R/(p)$ if $\exists b\ne0: ab=ba=0$. The problem is I don't know what to do with the so-called quotient ring.

Answer 1

Suppose $p$ is prime, and let's assume $(a+(p))(b+(p))=(p)$. By the definition of multiplication in the quotient ring it means that $ab+(p)=(p)$, and hence $p|ab$. Since $p$ is prime we conclude that $p|a$ or $p|b$ which implies $a+(p)=(p)$ or $b+(p)=(p)$. So $R/(p)$ has no zero divisors.

For the other direction suppose $R/(p)$ has no zero divisors and assume $p|ab$. Then $(p)=ab+(p)=(a+(p))(b+(p))$. Since there are no zero divisors we conclude that $a+(p)=(p)$ or $b+(p)=(p)$ which means that $p|a$ or $p|b$.

Answer 2

Hint:

$p\mid x\iff x\equiv 0\mod p\iff \bar x=\bar 0 \:\text{ in }\: R/(p)$. You can use Euclid's lemma.

Answer 3

$$\begin{align} p\ \ \lnot \rm prime \iff&\ \ p\ \mid\ ab,\ \ p\ \nmid\ a,b,\ \ {\rm some}\ a,b\in R\\[.2em] \iff&\ \ 0 = \bar a\bar b,\ \ 0\neq \bar a,\bar b,\ \ {\rm some}\ \bar a,\bar b\in R/p\\[.2em] \iff&\ R/p\ \ \text{has a zero divisor}\end{align}\qquad\qquad$$