Exercise A0.III.6.14

Proposition

Let $R$ be the ring $\mathbb{Z}[x_1,x_2,\ldots]$ and let $I = (x_1, x_2, \ldots)$ be the ideal of $R$ generated by all indeterminates. Then $I$ is not finitely generated as an $R$-module.

Proof

Suppose $I$ is indeed finitely generated, that is, there exists a surjective $R$-module homomorphism $\pi : R^{\oplus n} \to I$ for some $n \in \mathbb{N}.$

Since $\pi$ has a finitely generated free module as its domain, we can describe it fully by specifying where it sends each basis element. This gives us a finite list of polynomials. Let $x_{\xi}$ be the lowest index indeterminate not present in any of these polynomials.

Since $\pi$ is surjective, it must send some $R$-linear combination of basis elements $r_1j_1 + r_2j_2 + \dots + r_n j_n$ to $x_\xi.$ This gives us a relation that holds in $R$:

$$x_\xi = r_1 \pi(j_1) + r_2 \pi(j_2) + \dots + r_n \pi(j_n).$$

Recall that $R$ is itself a free object on the set $\{x_1, x_2, \ldots\}$ in $\mathbf{Ring}.$ Consider the morphism $\varphi : R \to \mathbb{Z}$ sending $x_\xi$ to $1$ and all other indeterminates to $0.$ Note that $\varphi(\pi(j_k)) = 0$ for all $k.$ This gives us $1 = 0$ (as integers), a contradiction.

$\blacksquare$