Let $R$ be a commutative ring. If $R[X]$ is a principal ideal domain, then $R$ is a field.
Solution 1:
If $f$ is a unit then there exists $g \in R[x]$ such that $1=gf \in (f)$. So $1 \in (f)$ implies that $(f)=(1)$.
Let $R$ be a commutative ring. If $R[X]$ is a principal ideal domain, then $R$ is a field.
Solution 1:
If $f$ is a unit then there exists $g \in R[x]$ such that $1=gf \in (f)$. So $1 \in (f)$ implies that $(f)=(1)$.