In the ring $R=\mathbb Z[X],$ is $(X)+(X^2)=(X)$?
It is known that if $R$ is a UFD, then $R[X]$ is a UFD. Is the converse true?
Two ring questions
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0Thanks a lot! (No this isn't homework; I was trying to find an example of a non-PID, where the sum of two principal ideals is also a principal ideal) The second question was a query that popped out during my revision. – 2012-04-03
2 Answers
For $\rm(2)$ the key observation is that $\rm R$ is inertly embedded in $\rm R[X],$ i.e. factorizations in $\rm R[X]$ of $\rm r\in R^*$ already lie in $\rm R,$ i.e $\rm\: 0\ne r = fg,\ f,g\in R[X]\:$ $\rm\Rightarrow$ $\rm\:f,g\in R,\:$ by comparing degrees (and employing $\rm R$ is a domain). This implies the factorization theory of $\rm R[X]$ restricts faithfully to $\rm R$.
Thus, since $\rm R[X]$ is a UFD, any nonunit $\rm\:r\in R^*\:$ is a product of atoms in $\rm R[X],$ hence in $\rm R,$ by inertness. Further, such atoms $\rm\:p\:$ are prime in $\rm R[X]$ so also in $\rm R$ since $\rm p\ |\ a,b\:$ $\rm\Rightarrow$ $\rm\:p\ |\ a\:$ or $\rm\:p\ |\ b\:$ in $\rm R[X]$ so in $\rm R$, i.e. $\rm\:p\ |\ a\:$ in $\rm\:R[x]\:$ $\Rightarrow$ $\rm\:a = p\:\!f,\ f\in R[X]\:$ $\rm\Rightarrow$ $\rm\:f\in R\:$ $\Rightarrow$ $\rm\:p\ |\ a\:$ in $\rm R,\:$ by inertness. Hence $\rm R$ is a UFD, since prime factorizations of $\rm\:r\in R[X]\:$ pull back to prime factorizations in $\rm R.$
Similarly one can show that any inertly embedded subdomain of a GCD domain is also a GCD domain, and gcds remain the same in the subdomain. In a classic paper, Paul Cohn showed that any GCD domain can be inertly embedded in a Bezout domain, i.e. a domain $\rm D$ where gcds are linearly representable $\rm\gcd(a,b) = ac + bd,\ c,d\in D,\:$ see Cohn: Bezout rings and their subrings.
Hints:
As was pointed out in the comments, if $I$ and $J$ are ideals of $R$ with $I\subseteq J$, then show that $I+J=J$.
Suppose that $R[x]$ is a UFD. If we have an irreducible element $\pi\in R$, then is $\pi$ irreducible in $R[x]$? If so, is $\pi$ then a prime element of $R$? Why?