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I am just wondering if that is the case in the ring $\mathbb{Q}[x]$ because in class we showed that it is maximal in $\mathbb{Z}[x]$

I believe that (x,2) is also not principal in $\mathbb{Q}[x]$

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Hint $\ 2$ is a unit (invertible) in $\Bbb Q,$ and any ideal containing a unit $u$ contains $\,u u^{-1}\! = 1$.

Note also that $\Bbb Q[x]$ is a PID, because it is Euclidean, using the polynomial division algorithm.

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    Yes, to both. $\,r\in R,\ 1\in I\:\Rightarrow\: r\cdot 1 = r\in I\,$ so $I$ contains every $\,r\in R,\,$ i.e. if an ideal contains $u$ then it contains all multiples of $u$. When $u=1,$ every element is a multiple of $1$.2012-09-29