Let $R$ be the ring given by $R=\mathbb Z+x\mathbb Q[x]$. Then show that:
1) $R$ is an integral domain and its units are $+1$ and $-1$.
2) $x$ is not prime in $R$ and describe the quotient ring $R/(x)$.
3) Compute an ideal of $R$ which is not principal.
$R$ being a subset of $\mathbb Q[x]$ inherits non-zero divisor, unity, and commutativity. So $1$st question is done. For the third one I think the ideal $(2,{1/2}x)$ may be but can't give proper proof. Please help me for $2$nd and $3rd$ ones. Thanks in advance.
This is going to sound a little strange at first, but it ends up at your solution. There is quite a bit of checking for you to do.
The set $S=\mathbb Z\times \mathbb Q/\mathbb Z$ is an Abelian group under coordinatewise addition, and it becomes a ring if you use the multiplication $(y, q+\mathbb Z)(z, p+\mathbb Z)=(yz, yp+\mathbb Z+zq+\mathbb Z)$. It's called the trivial extension of $\mathbb Q/\mathbb Z$ by $\mathbb Z$.
Now, it is not hard to prove that $S\cong R/(x)$ for the $R$ in your problem. Just look at $(x)=xR=x\mathbb Z+x^2\mathbb Q[x]$ and consider the quotient carefully.
You can check that $I=\{0\}\times \mathbb Q/\mathbb Z$ is an ideal of $S$, and furthermore $ab=0$ for any $a,b\in I$. Therefore $S$ isn't an integral domain... and that should be enough information for you to conclude that $(x)$ is not prime in $R$.
Finally, $x\mathbb Q[x]$ is your candidate for a non-finitely-generated ideal. You'll see that you have to generate everything of the form $x\mathbb Q$ using integer multiples of your generators (which won't be possible with finitely many generators.)