If $R$ is a ring (Integral Domain) then $R[x_1,x_2]$ is not a Principal Ideal Domain. Is it a unique factorization domain? Any hints on how to prove its not a Principal Ideal Domain?
$F[x_1,x_2]$ is not a Principal Ideal Domain
1
$\begingroup$
abstract-algebra
ring-theory
-
1Is there some reason for $F$ in title, $R$ in body? – 2012-11-18
1 Answers
3
Hint: Let $R$ have more than one element. To show $R[x_1,x_2]$ is not a Principal Ideal Domain, consider the polynomials with $0$ constant term. There are many other non-principal ideals.
-
1I expect you have no trouble showing the set of polynomials with $0$ constant term **is** an ideal. As to non-principal, suppose to the contrary $G$ is a generator. Then $x_1=GP$, $x_2=GQ$. Since $R$ is an integral domain, the coefficient of $x_1$ in $G$ is non-zero. But $Q$ has non-zero constant term, so the coefficient of $x_1$ in $GQ$ is non-zero, contradicting the fact that $GQ=x_2$. – 2012-11-18