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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?

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This is a standard fact in a moderately advanced algebra course. $R[x_1,x_2,x_3,\cdots]$ is UFD if and only if $R$ is UFD. –  Lubin Nov 18 '12 at 7:01
    
Is there some reason for $F$ in title, $R$ in body? –  Gerry Myerson Nov 18 '12 at 11:38
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1 Answer 1

up vote 2 down vote accepted

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.

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Yes, I am trying to show that Ideal generated by $(x_1,x_2)$ is not a Principal Ideal, but I am not able to move forward. –  rTeja Nov 18 '12 at 11:53
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I 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 $GQ=x_2$. –  André Nicolas Nov 18 '12 at 16:29
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