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  1. In the ring $R=\mathbb Z[X],$ is $(X)+(X^2)=(X)$?

  2. It is known that if $R$ is a UFD, then $R[X]$ is a UFD. Is the converse true?

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  • $\begingroup$ Are these questions homework? Anyway, two hints: Q1) Note that for two ideals $I_1, I_2$, if $I_1 \supseteq I_2$, then $I_1 + I_2 = I_1$ (why?). Q2) The converse is true. Proof sketch: Take any non-unit element in $R$, consider prime factorization over $R[X]$. Show that this prime factorization uses elements from $R$ only. Furthermore, if $a \in R$ is prime in $R[X]$, it is prime in $R$ (why?), which proves the claim. $\endgroup$ – Johannes Kloos Apr 2 '12 at 14:11
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    $\begingroup$ Finally, aside from possibly appearing on the same homework assignment, these two questions are not obviously related and so should be posted as separate questions. (Since it would be completely reasonable to answer just one of the two) $\endgroup$ – Daniel Martin Apr 2 '12 at 14:13
  • $\begingroup$ Thanks 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. $\endgroup$ – yoyostein Apr 3 '12 at 4:13
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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.

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Hints:

  1. As was pointed out in the comments, if $I$ and $J$ are ideals of $R$ with $I\subseteq J$, then show that $I+J=J$.

  2. 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?

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