How to prove that a subring R of the PID R[x] is an integral domain

The reader should know that only proving that the subring has no zero divisors is not enough. A subring should also contain the identity element.

Prove that a subring R of the PID R[x] is an integral domain

Proof: we take the principal ideal domain $R[x]$ and let $R$ be its subring. Let’s assume that we have the non-zero elements $a,b \in R$ such that $ab = 0$, i.e., $a$ and $b$ are zero divisors. Since $R \subset R[x]$, we have that $ab = 0$, but that is a contradiction since $R[x]$ has no zero divisors since $R[x]$ is an integral domain.

Now we need to figure out if $R$ contains the identity element. But, $R[x]$ contains the identity element since it is a principal ideal domain, and therefore, $R$ too (so $1_R \neq 0_R$).