Let $D$ be an integral domain and suppose the polynomial $D[x]$ is an unique factorization domain (UFD). Does $D$ have to be an unique factorization domain? In other words, does the equivalence $$D\text{ is an }UFD\Longleftrightarrow D[x]\text{ is an }UFD$$ holds?
If $D[x]$ is unique factorization domain (UFD) then $D$ is an UFD?
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abstract-algebra
polynomials
ring-theory
1 Answers
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Every element $x \in D$ has a factorization in $D[X]$. Now use $\deg(pq) = \deg(p) + \deg(q)$ for polynomials in an integral domain to conclude.
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0Out of curiosity, is there a non-UFD $D$ such that $D[X]$ is a UFD once we drop the restriction that $D$ should be an integral domain? – 2017-02-19
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1If $D[X]$ is a domain, $D$ is also a domain. – 2017-02-19
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0Oh, duh. Thanks! – 2017-02-19