Lemma 109.31.1. There exists a domain $A$ and a nonzero ideal $I \subset A$ such that $I_\mathfrak q \subset A_\mathfrak q$ is a principal ideal for all primes $\mathfrak q \subset A$ but $I$ is not an invertible $A$-module.

## 109.31 A noninvertible ideal invertible in stalks

Let $A$ be a domain and let $I \subset A$ be a nonzero ideal. Recall that when we say $I$ is invertible, we mean that $I$ is invertible as an $A$-module. We are going to make an example of this situation where $I$ is not invertible, yet $I_\mathfrak q = (f) \subset A_\mathfrak q$ is a (nonzero) principal ideal for every prime ideal $\mathfrak q \subset A$. In the literature the property that $I_\mathfrak q$ is principal for all primes $\mathfrak q$ is sometimes expressed by saying “$I$ is a locally principal ideal”. We can't use this terminology as our “local” always means “local in the Zariski topology” (or whatever topology we are currently working with).

Let $R = \mathbf{Q}[x]$ and let $M = \sum \frac{1}{x - n}R$ be the module constructed in Section 109.30. Consider the ring^{1}

and the ideal $I = M A = \bigoplus _{d \geq 1} \text{Sym}^ d_ R(M)$. Since $M$ is not finitely generated as an $R$-module we see that $I$ cannot be generated by finitely many elements as an ideal in $A$. Since an invertible module is finitely generated, this means that $I$ is not invertible. On the other hand, let $\mathfrak p \subset R$ be a prime ideal. By construction $M_\mathfrak p \cong R_\mathfrak p$. Hence

as a graded $R_\mathfrak p$-algebra. It follows that $I_\mathfrak p \subset A_\mathfrak p$ is generated by the nonzerodivisor $T$. Thus certainly for any prime ideal $\mathfrak q \subset A$ we see that $I_\mathfrak q$ is generated by a single element.

**Proof.**
See discussion above.
$\square$

## 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Your email address will not be published.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)