The polynomial ring over an integral domain is an integral domain

Let $𝐷$ be a ring and $𝐷 [𝑥]$ be the polynomial ring in indeterminate $𝑥$ . Then $𝐷 [𝑥]$ is an integral domain iff $𝐷$ is an integral domain. #m/thm/ring

Proof

Assume $𝐷$ is an integral domain. Clearly $𝐷 [𝑥]$ is commutative since $𝐷$ is. Let $𝑓 (𝑥), 𝑔 (𝑥) \in 𝐷 [𝑥]$ be nonzero with leading terms $𝑓 𝑛 𝑥 𝑛$ and $𝑔 𝑚 𝑥 𝑚$ respectively. Then the leading term of $𝑓 (𝑥) 𝑔 (𝑥)$ is $𝑓 𝑛 𝑔 𝑚 𝑥 𝑛 + 𝑚$ so $𝑓 (𝑥) 𝑔 (𝑥) \neq 0$ .

Note if $𝐷 [𝑥]$ is an integral domain, then so are its subrings, including $𝐷$ .

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