The Stacks project

Example 110.22.1. Let $k$ be a field, and let $A = k[x_1, x_2, x_3, \dots ] / (x_1^2, x_2^2, x_3^2, \dots )$. Any prime ideal of $A$ contains the nilpotents $x_1, x_2, x_3, \dots$, so $\mathfrak p = (x_1, x_2, x_3, \dots )$ is the only prime ideal of $A$. Therefore the underlying topological space of $\operatorname {Spec} A$ is a single point and in particular is Noetherian. However $\mathfrak p$ is clearly not finitely generated.

Example 110.22.2. Let $k$ be a field, and let $A \subseteq k[x, y]$ be the subring generated by $k$ and the monomials $\{ xy^ i\} _{i \ge 0}$. The prime ideals of $A$ that do not contain $x$ are in one-to-one correspondence with the prime ideals of $A_ x \cong k[x, x^{-1}, y]$. If $\mathfrak p$ is a prime ideal that does contain $x$, then it contains every $xy^ i$, $i \ge 0$, because $(xy^ i)^2 = x(xy^{2i}) \in \mathfrak p$ and $\mathfrak p$ is radical. Consequently $\mathfrak p = (\{ xy^ i\} _{i \ge 0})$. Therefore the underlying topological space of $\operatorname {Spec} A$ is Noetherian, since it consists of the points of the Noetherian scheme $\mathop{\mathrm{Spec}}(A[x, x^{-1}, y])$ and the prime ideal $\mathfrak p$. But the ring $A$ is non-Noetherian because $\mathfrak p$ is not finitely generated. Note that in this example, $A$ also has the property of being a domain.