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I am trying to loosely follow Casselman's "The Bruhat-Tits Trees of SL(2)" instead using the field $F=\mathbb R_\rho$, a quotient of a subring of the hyperreals. It has a non-archimedean valuation with value group $\mathbb R$ and valuation ring $\mathcal O$. I am trying to show every bounded open subgroup of $\mathrm{GL}_2(F)$ stabilizes a lattice over $F^2$, that is a finitely generated $\mathcal O$-module which spans $F^2$ as an $F$-vector space.

To this end, I am trying to prove the following lemma: If $M$ is a lattice, $M\subseteq X\subseteq F^2$ with $X$ bounded, then the intersection of all lattices containing $X$ is finitely generated (and therefore a lattice.) Here $X$ is an $\mathcal O$-module which is not necessarily finitely generated. To this end I am trying to make a Zorn's lemma argument on the set of lattices containing $X$: If $C$ is a descending chain of lattices $$L_1\supseteq L_2 \supseteq \cdots$$ then, $\bigcap L_i$ is finitely generated, which is where I am stuck. My advisor told me to scrape the literature for lemmas about nesting intersections of lattices over Bézout domains, or more generally Prüfer domains. Can anyone recommend any sources? Thank you.

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    $\begingroup$ I'm a bit confused about where the objects in $M \subseteq X \subseteq \operatorname{GL}(F)$ live. I guess $\operatorname{GL}(F)$ means $\operatorname{GL}_2(F)$, but then I'm not sure what it means to speak of $M$ as a lattice in it. Is it supposed to be $M \subseteq X \subseteq F^2$? $\endgroup$ Commented May 30 at 23:53
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    $\begingroup$ @LSpice oops! I just fixed that. I have become far too comfortable associating $\mathcal O$-lattices in $F^2$ with $2\times 2$ invertible matrices whose columns are thought of as basis elements. I also slightly strengthened the assumptions although I doubt it makes a difference. $\endgroup$ Commented May 31 at 3:00
  • $\begingroup$ Does your field contain elements like $x+ x^{1/2} + x^{1/4} + x^{1/8} + \dots$ where $x$ is an element of valuation $-1$? $\endgroup$ Commented May 31 at 15:58
  • $\begingroup$ @WillSawin If by that you mean the limit to such a series, then no; any such series will be divergent. $\endgroup$ Commented Jun 1 at 1:13

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I don't know what the field $\mathbb R_\rho$ is exactly but I think this is not true. I guess because the value group is $\mathbb R$ we can find an element $x$ with valuation $-1$ that admits roots $x^{1/2}, x^{1/4},\dots$ of valuation progressively closer to $0$.

Now let $L_n$ be the lattice generated by the vectors $\begin{pmatrix} 0 \\ x^{1/2^n} \end{pmatrix}$ and $\begin{pmatrix} 1 \\ \sum_{i=0}^{n-1} x^{1/2^i} \end{pmatrix}$. We have $L_{n+1} \subset L_n$ since it is easy to check both generators lie in $L_n$.

Let's show the intersection of the $L_n$ is not finitely generated. Observe that the intersection contains the vector

$$\begin{pmatrix} x^{-1/2^m} \\ \sum_{i=0}^{m-1} x^{1/2^i -1/2^m} \end{pmatrix}$$

as it differs from $1/2^m$ times the second generator of any $L_n$ by a vector whose first coordinate is $0$ and whose second coordinate has valuation $\geq 0$ and thus is a multiple of the first generator of any $L_n$.

Thus the intersection contains a vector whose first coordinate has valuation $1/2^m$ for each $m$. If the intersection is finitely generated, this can only happen if the intersection contains a vector whose first coordinate has valuation $0$. It follows that by multiplication by a unit in $\mathcal O$ that the intersection contains a vector whose first coordinate has valuation $0$. Since this vector lies in $L_n$, its second coordinate must be congruent modulo $x^{1/2^n}$ to $\sum_{i=0}^{n-1} x^{1/2^i} $.

However, the field $\mathbb R^\rho$ does not contain, I think, an element whose second coordinate is congruent modulo $x^{1/2^n}$ to $\sum_{i=0}^{n-1} x^{1/2^i} $. (This is not quite the claim you answered in your comment, since such a vector would not be the limit of the series, so I am not completely sure.) This gives a contradiction, so the lattice is not finitely generated.

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  • $\begingroup$ Thank you for your reply. I am trying to understand your reasoning: what makes you say that you don't think $\mathbb R_\rho$ has an element congruent to $\sum_{i=0}^{n-1}x^{1/2^i}$ modulo $x^{1/2^n}$. I am trying to work out the details, but I am coming up short on intuition. $\endgroup$ Commented Jun 1 at 4:30
  • $\begingroup$ @4u9ust It's entirely a guess because you have not stated the definition of this field - it would be helpful to do this. I thought that the field does not contain such elements because if they did there would be a notation like $x + x^{1/2} + x^{1/4} + \dots$ to describe them and you were not aware of such notation so I assumed it doesn't exist. $\endgroup$ Commented Jun 1 at 10:36

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