Not all ideals are finitely generated

Question

Let $R=\{a_0+a_1X+...+a_nX^n \ | \ a_1,...,a_n \in \mathbb{Q}, a_o \in \mathbb{Z}, n\in \mathbb{Z}_{\geq 0} \}$ and $I=\{a_1X+...+a_nX^n \ | \ a_1,...,a_n \in \mathbb{Q}, n\in \mathbb{Z}^{+} \}.$ It is clear that $I$ is an ideal of $R.$

Show $I$ is not finitely generated as $R$-module.

Here is my attempt:

Suppose $I=Rf_1+...+Rf_m,$ for some $f_1,...,f_m \in I^{*}.$

Let $q_i (\in \mathbb{Q}) =$ coefficient of $X$ in $f_i \ (i=1,...,m).$

Given $aX \in I^{*}, $ by comparing coefficients, we obtain $\ a=a_1q_1 +...+ a_mq_m,$ for some $a_1,...,a_m \in \mathbb{Z}.$ Since $a \in \mathbb{Q}$ is arbitrary, this means $\mathbb{Q}$ is finitely generated by $q_1,...,q_m$ over $\ \mathbb{Z}.$ Contradiction.

Kindly advise, thank you.

Answer 1

It might be simpler to construct an $f \in I$ such that f is not in $Rf_1 + ... Rf_m$. Let $S$ be the finite set of all primes dividing the denominator of '$a_1$' of each $f_i$. (Hope this makes sense). Then for any prime $p$ not in $S$, $X/p$ cannot be in $Rf_1 + ... Rf_m$ but clearly in $I$.

Make this argument rigorous?

Answer 2

The approach you used looks good to me!

The elements of the form $qX$ could only result from integer combinations of the X monomials of the generators, and you're exactly right that finitely many rationals cannot generate all rationals using integer coefficients alone.