Ideal not finitely generated Let $R=\{a_0+a_1 X+a_2 X^2 +\cdots + a_n X^n\}$, where $a_0$ is an integer and the rest of the coefficients are rational numbers.
Let $I=\{a_1 X+a_2 X^2+\cdots +a_n X^n\}$ where all of the coefficients are rational numbers.
Prove that I is an ideal of R.
Show further that I is not finitely generated as an R-module.
I have managed to prove that I is an ideal of R, by showing that I is the kernel of the evaluation map that maps a polynomial in Q[x] to its constant term. Hence I is an ideal of R.
However, I am stuck at showing I is not finitely generated as an R-module.
Sincere thanks for any help.

This ring is an example of a Bézout domain that is not a unique factorization domain (since not all nonzero noninvertible elements decompose into irreducibles in the first place; for instance $X$ does not). The wikipedia page gives a proof of the Bézout property, namely that any finitely generated ideal is in fact a principal ideal. So if $I$ were finitely generated, it would have a single generator. But it cannot, since an element without constant term and with coefficient $c$ of $X$ only generates elements whose coefficient of $X$ is an integer multiple of $c$. (One also sees directly that a finite set of elements of $I$ only generate elements whose coefficient of $X$ is an integer multiple of the gcd of their coefficients of $X$ which cannot be all of $\mathbb Q$, as is indicated in the anwers by Alexander Thumm and Alex Becker.)
