Example 52.16.4. Let $k$ be a field. Let $A = k[x, y][[t]]$ with $I = (t)$ and $\mathfrak a = (x, y, t)$. Let us use notation as in Situation 52.16.1. Observe that $U \cap Y = (D(x) \cap Y) \cup (D(y) \cap Y)$ is an affine open covering. For $n \geq 1$ consider the invertible module $\mathcal{L}_ n$ of $\mathcal{O}_ U/t^ n\mathcal{O}_ U$ given by glueing $A_ x/t^ nA_ x$ and $A_ y/t^ nA_ y$ via the invertible element of $A_{xy}/t^ nA_{xy}$ which is the image of any power series of the form

$u = 1 + \frac{t}{xy} + \sum _{n \geq 2} a_ n \frac{t^ n}{(xy)^{\varphi (n)}}$

with $a_ n \in k[x, y]$ and $\varphi (n) \in \mathbf{N}$. Then $(\mathcal{L}_ n)$ is an invertible object of $\textit{Coh}(U, I\mathcal{O}_ U)$ which is not the completion of a coherent $\mathcal{O}_ U$-module $\mathcal{L}$. We only sketch the argument and we omit most of the details. Let $y \in U \cap Y$. Then the completion of the stalk $\mathcal{L}_ y$ would be an invertible module hence $\mathcal{L}_ y$ is invertible. Thus there would exist an open $V \subset U$ containing $U \cap Y$ such that $\mathcal{L}|_ V$ is invertible. By Divisors, Lemma 31.28.3 we find an invertible $A$-module $M$ with $\widetilde{M}|_ V \cong \mathcal{L}|_ V$. However the ring $A$ is a UFD hence we see $M \cong A$ which would imply $\mathcal{L}_ n \cong \mathcal{O}_ U/I^ n\mathcal{O}_ U$. Since $\mathcal{L}_2 \not\cong \mathcal{O}_ U/I^2\mathcal{O}_ U$ by construction we get a contradiction as desired.

Note that if we take $a_ n = 0$ for $n \geq 2$, then we see that $\mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{L}_ n)$ is nonzero: in this case we the function $x$ on $D(x)$ and the function $x + t/y$ on $D(y)$ glue. On the other hand, if we take $a_ n = 1$ and $\varphi (n) = 2^ n$ or even $\varphi (n) = n^2$ then the reader can show that $\mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{L}_ n)$ is zero; this gives another proof that $(\mathcal{L}_ n)$ is not algebraizable in this case.

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