$R/\langle p \rangle$ is an integral domain, proof verification

Question

Let $R$ be a ring, and let $p$ be a prime element in $R$.

Then $R/\langle p \rangle$ is an integral domain.

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So first I like to try to decode a question.

$p$ being prime means $p|ab \implies p|a$ or $p|b$. $R/\langle p \rangle $ being an integral domain, means that if $ab \equiv 0 \pmod p$ then $a \equiv 0 \pmod p$ or $b\equiv 0 \pmod p$.

Then proof(?): Let $a,b\in R$ If $ab\equiv 0 \pmod p$ then $p|ab$ so either $p|a$ or $p|b$ so then $a\equiv 0 \pmod p$ or $b\equiv 0 \pmod p$, and we are done I believe.

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Is that all there is to it?

Answer 1

Yes, that's all there is to it (assuming that $R$ is commutative).

Your notation is a bit unusual though; the notation $x\equiv y\pmod{z}$ is used for $x,y,z\in\Bbb{Z}$, or slightly more generally when working in a number ring (like $\Bbb{Z}[i]$ or $\Bbb{Z}[\sqrt{-3}]$). In a general commutative ring one would denote the ideal generated by $p$ by $(p)$, and write:

The ring $R/(p)$ is an integral domain, if $ab\in(p)$ implies that $a\in(p)$ or $b\in(p)$.