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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.

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3 Answers

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Suppose $I$ is generated by $g_1,\ldots,g_k$ over $R$. Observe that each $g_i$ has coefficient of degree $0$ equal to $0$, as $g_i=g_i\cdot 1\in I$. Since any $$a_1X+\cdots+a_nX^n=g_1(c_{10}+c_{11}X+\cdots+c_{1n}X^n)+\cdots+g_k(c_{k0}+c_{k1}X+\cdots+c_{kn}X^n)$$ if we let $g_{11},\ldots,g_{k1}$ be the coefficients of degree $1$ of $g_1,\ldots,g_k$ then we get $a_1=g_{11}c_{10}+\cdots+g_{k1}c_{k0}$ and since $a_1$ can be any element of $\mathbb Q$, we have that $g_{11},\ldots,g_{k1}$ generate $\mathbb Q$ over $\mathbb Z$. But $\mathbb Q$ is not finitely generated over $\mathbb Z$ (see my answer here for example), hence $g_1,\ldots,g_k$ do not generate $I$ over $R$.

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Hint: $\frac{1}{p}X \not\in P_1R + P_2R + \dots + P_kR$, where $P_1, P_2, \dots, P_k \in I$, with coefficients $\frac{a_1}{b_1}, \frac{a_2}{b_2}, \dots, \frac{a_k}{b_k}$ in degree one, and $p$ does not divide $b_1 b_2 \dots b_k$.

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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.)

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