Find an Ideal of $\mathbb{Z}+x \mathbb{Q}[ x ]$ that is NOT principal

Question

The ring $\mathbb{Z}+x \mathbb{Q}[ x ]$ cannot be a principal ideal domain since it is not a unique factorization domain. Find an ideal of $\mathbb{Z}+x \mathbb{Q}[ x ]$ that is not principal.

My book gives no examples of how to show an ideal is not principal. I'm pretty sure if I let $I=(2,1/2 x)$ then I can show it's not principal.

I'm pretty confident that the best route to go is to do a proof by contradiction. But how do I start? What is my initial assumption?

Thanks!

Answer 1

Call $R= \Bbb{Z}+ x \Bbb{Q}[x]$. I highly suspect that $R$ is a Bezout domain, (i.e. every finitely generated ideal is principal), so I give you a non finitely generated ideal. Consider the ideal $$I=(x, x/2 , x/4 , x/8 , \dots) = \bigcup_{k \ge 1} \left( \frac{1}{2^k}x \right)$$ Clearly, for all $k \ge 1$ we have $$\left( \frac{1}{2^k}x \right) \subsetneq \left( \frac{1}{2^{k-1}}x \right)$$ because $$\frac{1}{2^{k-1}}x = 2 \frac{1}{2^k}x$$ and $2$ is not a unit of $R$ : this shows that $R$ is not Noetherian, and that the union of this chain of ideals is not finitely generated.

Answer 2

Another ideal which is not principal is $X\mathbb Q[X]$. (In fact, it is not even finitely generated.) Suppose the contrary, and let $f\in X\mathbb Q[X]$, say $f=Xg$ with $g\in\mathbb Q[X]$, such that $X\mathbb Q[X]=(f)$. Then since $X(\frac12g)\in X\mathbb Q[X]$ there is $h\in R$ such that $X(\frac12g)=fh$, that is, $X(\frac12g)=Xgh$. It follows $h=\frac12\notin\mathbb Z$, a contradiction.