Example 108.20.1. 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.

## 108.20 Underlying space Noetherian not Noetherian

We give two examples to show that a scheme whose underlying topological space is Noetherian may not be a Noetherian scheme.

Example 108.20.2. 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.

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)