Lemma 110.31.1. 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.

## 110.31 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 110.30. 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.
$\square$

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