Condition for a quotient commutative ring to be an integral domain

Let $𝑅$ be a commutative ring and $𝐼 ◃ 𝑅$ be a proper, nontrivial (two-sided) ideal. Then the Quotient ring $𝑅 / 𝐼$ is an integral domain iff $𝐼$ is a prime ideal. #m/thm/ring

Proof

Assume $𝑅 / 𝐼$ is an integral domain and let $𝑎 𝑏 \in 𝐼$ . Then $(𝑎 + 𝐼) (𝑏 + 𝐼) = 𝑎 𝑏 + 𝐼 = 𝐼 \equiv 0$ , so either $𝑎 + 𝐼 = 𝐼 \equiv 0$ or $𝑏 + 𝐼 = 𝐼 \equiv 0$ .

For the converse, assume $𝐼 ⊴ 𝑅$ is prime. Since $𝑅 / 𝐼$ is automatically a commutative ring, it only remains to show that $𝑅 / 𝐼$ has no zero-divisors. To this end, assume $(𝑎 + 𝐼) (𝑏 + 𝐼) = 𝐼 \equiv 0$ . Then $𝑎 𝑏 \in 𝐼$ and hence $𝑎 \in 𝐼$ or $𝑏 \in 𝐼$ , whence $𝑎 + 𝐼 = 𝐼 \equiv 0$ or $𝑏 + 𝐼 = 𝐼 \equiv 0$ .

#state/tidy | #lang/en | #SemBr