Exercise A0.V.1.16

Proposition

Let

$$R = \frac{\mathbb{Z}[x_1,x_2,x_3,\ldots]}{(x_1 - x_2^2, x_2 - x_3^2, \ldots)}.$$

The ascending chain condition for principal ideals does not hold in $R.$

Proof

Consider the chain of principal ideals

$$(x_1) \subseteq (x_2) \subseteq \ldots \subseteq(x_n)\subseteq \ldots.$$

The containment holds since $x_i = x_{i + 1}^2 \in (x_{i + 1}).$ For containment to be proper, we need to verify that $x_{i+1} \not\in (x_i).$

Suppose, for the sake of contradiction, that $x_{k + 1} \in (x_k),$ and that $k$ is the minimal index at which this happens.

Suppose $k > 1.$

Let $c \in R$ be such that $cx_{k} = x_{k + 1}.$ Then

$$x_k = x_{k + 1}^2 = c^2x_k^2 = c^2 x_{k - 1}.$$

So, $x_k \in (x_{k - 1}),$ which contradicts minimality. So we’re done.

Suppose instead that $k = 1.$ Consider the homomorphism $\tau : R \to \mathbb{R}$ given by $\tau(x_1) = 2$ and $\tau(x_{n+1}) = \sqrt{\tau(x_n)}.$

Since $x_2 \in (x_1)$ there exists $c \in R$ such that $cx_1 = x_2.$ In $\mathbb{R}$ this corresponds to $2\tau(c) = \sqrt{2}$ and hence $\tau(c) = \sqrt{2}/2.$

Observe that $\tau(x_n)$ is an algebraic integer, since it is a root of the polynomial $x^{2^{n}} - 2.$ Hence, the image of $\tau$ is a subring of the ring of algebraic integers.

Since $(\sqrt{2}/2)^2 = 1/2,$ it cannot be an algebraic integer. Thus, there exists no such $c$ for which $\tau(c) = \sqrt{2}/2.$

$\blacksquare$

NOTE

I rely on some basic properties of algebraic integers. This is a bit of a “cheat,” since Aluffi doesn’t cover them at this stage, but I was unable to solve this problem otherwise.