From d4910357483a3cb3f27dedb425d437f370e64253 Mon Sep 17 00:00:00 2001
From: Luke Naylor <l.naylor@sms.ed.ac.uk>
Date: Sun, 14 May 2023 19:51:16 +0100
Subject: [PATCH] Correct recent work related to modular arithmetic
---
main.tex | 65 ++++++++++++++++++++++++++++++++------------------------
1 file changed, 37 insertions(+), 28 deletions(-)
diff --git a/main.tex b/main.tex
index a0c4921..ad2fced 100644
--- a/main.tex
+++ b/main.tex
@@ -951,7 +951,7 @@ this value of $\frac{1}{2n^2}\ZZ$ explicitly.
The expressions that will follow will be a bit more complicated and have more
parts which depend on the values of $q$ and $\beta$, even their numerators
$\aa,\bb$ specifically. The upcoming theorem (TODO ref) is less useful as a
-`nice' formula for a bound on the ranks of the pseudo-semistabilizers, but has a
+`clean' formula for a bound on the ranks of the pseudo-semistabilizers, but has a
purpose in the context of writing a computer program to find
pseudo-semistabilizers. Such a program would iterate through possible values of
$q$, then iterate through values of $r$ within the bounds (dependent on $q$),
@@ -959,7 +959,8 @@ which would then determine $c$, and then find the corresponding possible values
for $d$.
-Firstly, we only consider $r$-values for which $c:=\chern_1(E)$ is integral:
+Firstly, we only need to consider $r$-values for which $c:=\chern_1(E)$ is
+integral:
\begin{equation}
c =
@@ -982,24 +983,9 @@ rhs_numerator = (
\end{sagesilent}
\noindent
-Considering the numerator of the right-hand-side of
-(eqn \ref{eqn:positive_rad_condition_in_terms_of_q_beta}):
+Let $\aa^{'}$ be an integer representative of $\aa^{-1}$ in $\ZZ/n\ZZ$.
-\begin{align}
- \sage{rhs_numerator}
- &\equiv (\aa(-\aa^{-1}\bb)+2\bb)\aa &\mod n
-\\
- &\equiv \aa\bb &\mod n
-\end{align}
-
-\noindent
-And so, we also have $\aa(\aa r+2\bb) \equiv \aa\bb$ (mod $2n^2$).
-Now, suppose that $x/m$ is the smallest element of $\frac{1}{m}\ZZ$ strictly
-greater than the right-hand-side of
-(eqn \ref{eqn:positive_rad_condition_in_terms_of_q_beta}), and define $\epsilon$
-as the size of the gap.
-
-Using the following tautology:
+Considering the following tautology:
\begin{align}
&\frac{ x }{ m }
@@ -1009,25 +995,48 @@ Using the following tautology:
2n^2
}
= \frac{ k }{ 2mn^2 }
- \quad \text{for some } x \in \ZZ
+ \quad \text{for some } x \in \ZZ, k \in \ZZ_{>0}
+ \label{eqn:finding_better_eps_problem}
\\ &\iff
- (\aa r+2\bb)\aa m
\equiv k
\quad \mod 2n^2
+ \quad \text{for some } k \in \ZZ_{>0}
\\ &\iff
- - \aa\bb m
+ - \aa^2 m r - 2\aa\bb m
\equiv k
\quad \mod 2n^2
+ \quad \text{for some } k \in \ZZ_{>0}
+\\ &\Longrightarrow
+ \aa^2 \aa^{'}\bb m - 2\aa\bb m
+ \equiv k
+ \quad \mod \gcd(2n^2, \aa^2 mn)
+ \quad \text{for some } k \in \ZZ_{>0}
+ \label{eqn:better_eps_problem_k_mod_gcd2n2_a2mn}
+\\ &\Longrightarrow
+ -\aa\bb m
+ \equiv k
+ \quad \mod n
+ \quad \text{for some } k \in \ZZ_{>0}
+ \label{eqn:better_eps_problem_k_mod_n}
\end{align}
-We can recover how much greater $x/m$ is than the right-hand-side of
-(eqn \ref{eqn:positive_rad_condition_in_terms_of_q_beta}).
-First calculate the smallest $k_q \in \ZZ_{>0}$, such that
-$k_q \equiv -\aa\bb m \mod 2n^2$. Then we have
-$\epsilon = \epsilon_q := \frac{k_q}{2mn^2}$,
-an expression independent of $x$ and $r$, only depending on $q$.
+In our situation, we want to find the gap between the right-hand-side of eqn
+\ref{eqn:positive_rad_condition_in_terms_of_q_beta},
+and the least element of $\frac{1}{m}\ZZ$ which is strictly greater.
+This amounts to finding the least $k \in \ZZ_{>0}$ for which
+eqn \ref{eqn:finding_better_eps_problem} holds.
+Since such a $k$ satisfies eqn \ref{eqn:better_eps_problem_k_mod_n},
+we can pick the smallest $k \in \ZZ_{>0}$ which satisfies this new condition
+(a computation only depending on $q$ and $\beta$, but not $r$)
+and we are then guaranteed that the gap is at least $\frac{k}{2mn^2}$.
+
+A potentially larger gap can also be guaranteed if we choose the least
+$k \in \ZZ_{>0}$ satisfying eqn \ref{eqn:better_eps_problem_k_mod_gcd2n2_a2mn}
+instead, but at the cost of computing several $\gcd$'s and modulo reductions
+for each $q$ considered.
+
-%% TODO: check this^ result seems a bit strange
--
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