Realization Calculus

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1 \section{Rational spectral projectors and resultant bounds}\label{sec:resultant}
2 
3 Let $\OO$ be a discrete valuation ring with fraction field $K$ and uniformizer $\pi$.  Let $\Lambda$ be a finite free $\OO$ module and put
4 \[
5 V=K\otimes_{\OO}\Lambda.
6 \]
7 Suppose $A\in\End_{\OO}(\Lambda)$.
8 
9 \begin{theorem}[Bézout spectral projector]\label{thm:bezout_projector}
10 Assume that
11 \[
12 \chi_A(x)=f(x)g(x)
13 \]
14 with monic polynomials $f,g\in\OO[x]$ that are coprime in $K[x]$.  Choose $u,v\in\OO[x]$ and a nonzero $r\in\OO$ such that
15 \[
16 u(x)f(x)+v(x)g(x)=r.
17 \]
18 Then
19 \[
20 E_f=r^{-1}v(A)g(A)
21 \]
22 is the projector of $V$ onto $\Ker f(A)$ along $\Ker g(A)$.  In particular,
23 \[
24 E_f^2=E_f,
25 \qquad
26 rE_f\in\End_{\OO}(\Lambda).
27 \]
28 \end{theorem}
29 
30 \begin{proof}
31 Cayley Hamilton gives
32 \[
33 f(A)g(A)=0.
34 \]
35 If $x\in\Ker f(A)$, then the Bézout identity gives
36 \[
37 v(A)g(A)x=rx,
38 \]
39 so $E_fx=x$.  If $x\in\Ker g(A)$, then $E_fx=0$.  The same identity gives the direct sum decomposition
40 \[
41 V=\Ker f(A)\oplus\Ker g(A).
42 \]
43 Indeed, every $x\in V$ is the sum of
44 \[
45 r^{-1}v(A)g(A)x\in\Ker f(A)
46 \]
47 and
48 \[
49 r^{-1}u(A)f(A)x\in\Ker g(A).
50 \]
51 Thus $E_f$ is the stated projector and is idempotent.  The formula for $rE_f$ has coefficients in $\OO$, so it preserves $\Lambda$.
52 \end{proof}
53 
54 \begin{theorem}[Several coprime spectral packets]\label{thm:several_projectors}
55 Assume
56 \[
57 \chi_A(x)=\prod_{i=1}^rf_i(x),
58 \]
59 where the monic factors $f_i\in\OO[x]$ are pairwise coprime in $K[x]$.  Put
60 \[
61 g_i=\prod_{j\ne i}f_j.
62 \]
63 There are complementary projectors $E_i$ with
64 \[
65 \im E_i=\Ker f_i(A),
66 \qquad
67 \sum_iE_i=I,
68 \qquad
69 E_iE_j=0
70 \quad(i\ne j).
71 \]
72 One may choose $E_i$ so that
73 \[
74 \Res(f_i,g_i)E_i\in\End_{\OO}(\Lambda)
75 \]
76 up to multiplication of the resultant by a unit.
77 \end{theorem}
78 
79 \begin{proof}
80 Apply \cref{thm:bezout_projector} to the factorization $\chi_A=f_ig_i$.  The resulting projector is one on $\Ker f_i(A)$ and zero on every $\Ker f_j(A)$ with $j\ne i$.  The direct sum decomposition follows from the polynomial Chinese remainder theorem.  The orthogonality and sum identities follow by evaluating the corresponding idempotents in $K[x]/(\chi_A)$.  The Sylvester Bézout relation for $(f_i,g_i)$ gives the stated denominator bound.
81 \end{proof}
82 
83 \begin{corollary}[Resultant denominator bound]\label{cor:resultant_bound}
84 Under the hypotheses of \cref{thm:bezout_projector}, one may take
85 \[
86 r=\Res(f,g)
87 \]
88 up to multiplication by a unit.  Therefore
89 \[
90 \pi^{\val_\pi(\Res(f,g))}E_f\Lambda\subseteq\Lambda.
91 \]
92 \end{corollary}
93 
94 \begin{proof}
95 The Sylvester matrix construction gives a Bézout relation with right side equal to the resultant.  Apply \cref{thm:bezout_projector}.  Standard facts about resultants may be found in \cite{Lang2002,GohbergLancasterRodman2009}.
96 \end{proof}
97 
98 \begin{example}[A resultant bound that is not exact]\label{ex:loose_resultant}
99 Let
100 \[
101 A=\diag(0,\pi^s)
102 \]
103 on the standard lattice $\Lambda=\OO^2$.  The factors $x$ and $x-\pi^s$ have resultant $-\pi^s$, up to the harmless choice of resultant order, but the two spectral projectors are the integral coordinate projectors.  Their exact separation weight is zero.  The resultant sees the closeness of the two eigenvalues.  It does not see that the chosen lattice already splits along their eigenspaces.
104 \end{example}
105 
106 \begin{interpretation}
107 A resultant bounds the denominator through the characteristic factors. The exact separation weight also depends on how the lattice sits inside the rational packet spaces. These are two stages of the calculation. First construct a rational projector with a known clearing denominator. Then measure the gluing of the chosen lattice to find the least denominator that actually works.
108 \end{interpretation}
109 
110 \subsection{Evaluated certificates and explicit lattice action}
111 
112 \subsubsection{Minimal evaluated B\'ezout identities}
113 
114 Let $\OO$ be a DVR, $K=\Frac(\OO)$, and $A\in\End_\OO(\Lambda)$ for a finite free lattice $\Lambda$. Suppose the characteristic polynomial factors over $\OO$ as
115 \[
116 \chi_A(x)=f(x)g(x),
117 \]
118 where $f$ and $g$ are coprime over $K[x]$.
119 
120 \begin{theorem}[Evaluated B\'ezout spectral projector]\label{thm:bezout-projector}
121 Let $F,G,U,V\in\End_K(\Lambda\otimes_\OO K)$ and let $r\in K^\times$.
122 Suppose
123 \[
124 UF+VG=rI,
125 \qquad FG=0,
126 \qquad FG=GF,
127 \qquad GU=UG.
128 \]
129 Put $N=VG$. Then
130 \[
131 \boxed{N^2=rN.}
132 \]
133 Consequently the operator
134 \[
135 \boxed{
136 E=r^{-1}VG
137 }
138 \]
139 is idempotent, acts as the identity on $\ker F$, and acts as zero on
140 $\ker G$. If $V$ and $G$ preserve $\Lambda$, then
141 \[
142 rE=VG\in\End_\OO(\Lambda).
143 \]
144 \end{theorem}
145 
146 \begin{proof}
147 The B\'ezout identity gives $N=rI-UF$. The two stated commutation relations
148 and $FG=0$ give
149 \[
150 N(UF)=VGUF=VUGF=VUFG=0.
151 \]
152 Therefore
153 \[
154 N^2=N(rI-UF)=rN.
155 \]
156 Since $r\ne0$, $E^2=r^{-2}N^2=E$. If $Fx=0$, the evaluated B\'ezout
157 identity gives $VGx=rx$, hence $Ex=x$. If $Gx=0$, then $Ex=0$.
158 The lattice assertion follows directly from $rE=VG$.
159 \end{proof}
160 
161 The equality $N^2=rN$ before division uses neither $r\ne0$ nor the other
162 pairwise commutation relations that hold in the polynomial application.
163 Normalizing to $E$ uses $r\ne0$. Its action on $\ker F$ uses only the B\'ezout
164 identity and $r\ne0$, while vanishing on $\ker G$ follows immediately from $N=VG$.
165 The bundled Lean certificate records all pairwise commutations as convenient
166 evaluated polynomial data, but the minimal theorems expose only the hypotheses
167 they consume.
168 
169 For the polynomial application, one supplies $F=f(A)$, $G=g(A)$,
170 $U=u(A)$, and $V=v(A)$. A B\'ezout or resultant identity
171 $uf+vg=r$ evaluates to the first certificate, while
172 $(fg)(A)=0$ supplies the second. When $fg=\chi_A$, this last equality is
173 Cayley Hamilton. Standard B\'ezout, resultant, and DVR facts are recalled
174 from \cite{Lang2002,Serre1979}. The inherited certificate theorem begins with the evaluated
175 certificate. It does not infer exact action merely from nilpotence of
176 $f(A)$ or $g(A)$, which would be insufficient.
177 
178 \begin{definition}[Separation weight]\label{def:separation-weight}
179 For a rational projector $E$ and a specified $\OO$ lattice $\Lambda$, provided
180 at least one clearing exponent exists, define
181 \[
182 s_{\pi,\Lambda}(E)
183 =
184 \min\{s\ge0:\pi^sE(\Lambda)\subseteq\Lambda\}.
185 \]
186 If the evaluated certificate arises from integral $v,g$ and
187 $r=u\pi^s$ with $u^{-1}$ preserving the lattice, then
188 \cref{thm:bezout-projector} gives the upper bound
189 \[
190 s_{\pi,\Lambda}(E)\le s=\val_\pi(r).
191 \]
192 Indeed,
193 \[
194 \boxed{\pi^sE=u^{-1}VG}
195 \]
196 preserves $\Lambda$. The minimal Lean theorem needs only the lattice action of
197 $u^{-1}$. The bundled wrapper for lattice units records both $u$ and $u^{-1}$.
198 A nonzero scalar in the fraction field need not be an integral unit.
199 Existence of a clearing exponent is an explicit hypothesis. Once one exists,
200 the least one is obtained from the least element property of the natural numbers.
201 Clearing exponents are upward closed provided multiplication by $\pi$
202 preserves the lattice.
203 \end{definition}
204 
205 \subsubsection{Certificate-level precision delay}
206 
207 \begin{theorem}[Projector precision delay]\label{thm:precision-delay}
208 Let $E$ be a rational projector with $\pi^sE(\Lambda)\subseteq\Lambda$. If
209 \[
210 x\equiv y\pmod{\pi^{m+s}\Lambda},
211 \]
212 then
213 \[
214 Ex\equiv Ey\pmod{\pi^m\Lambda}.
215 \]
216 Thus $m+s$ input digits suffice to determine $m$ output digits of the projected packet.
217 \end{theorem}
218 
219 \begin{proof}
220 Write $x-y=\pi^{m+s}z$. Since $\pi^sE(z)\in\Lambda$,
221 \[
222 E(x-y)=\pi^m\bigl(\pi^sE(z)\bigr)\in\pi^m\Lambda.
223 \]
224 \end{proof}
225 
226 The theorem gives a precision budget for delayed separation. A packet with weight $s$ can be reconstructed at the requested output precision by retaining $s$ additional input digits.
227 
228 
229 \section{Exact lattice separation}\label{sec:lattice}
230 
231 The exact denominator belongs to the pair consisting of the rational packet decomposition and the integral lattice.
232 
233 \Needspace{8\baselineskip}
234 \begin{definition}[Lattice packet data]\label{def:lattice_packet}
235 Let
236 \[
237 V=\bigoplus_{\alpha\in I}V_\alpha
238 \]
239 be a finite direct sum over $K$, with packet projectors $E_\alpha$.  Let $\Lambda\subseteq V$ be an $\OO$ lattice.  Put
240 \[
241 \Lambda_\alpha=\Lambda\cap V_\alpha,
242 \qquad
243 \Lambda_{\mathrm{split}}=\bigoplus_\alpha\Lambda_\alpha,
244 \]
245 and define the gluing module
246 \[
247 G_{\Lambda,\Pcal}
248 =
249 \Lambda/\Lambda_{\mathrm{split}}.
250 \]
251 The separation weight of $E_\alpha$ is
252 \[
253 s_\Lambda(E_\alpha)
254 =
255 \min\{s\geq0:\pi^sE_\alpha\Lambda\subseteq\Lambda\}.
256 \]
257 The global separation weight is
258 \[
259 s_\Lambda(\Pcal)=\max\bigl(\{0\}\cup\{s_\Lambda(E_\alpha):\alpha\in I\}\bigr).
260 \]
261 For a finitely generated torsion $\OO$ module $G$, write $\expn_\pi(G)$ for the least $s$ such that $\pi^sG=0$.
262 \end{definition}
263 
264 The gluing module $G_{\Lambda,\Pcal}$ is a finitely generated torsion $\OO$ module, hence has finite length, because $\Lambda_{\mathrm{split}}$ and $\Lambda$ are full lattices in the same vector space. To see that the split lattice is full, clear denominators of bases of the rational summands. Each $\Lambda\cap V_\alpha$ then spans $V_\alpha$, so their direct sum spans $V$.
265 
266 \begin{theorem}[Gluing module and exact separation weight]\label{thm:gluing_weight}
267 With the notation of \cref{def:lattice_packet}, the map
268 \[
269 \Phi:G_{\Lambda,\Pcal}
270 \longrightarrow
271 \bigoplus_{\alpha\in I}E_\alpha\Lambda/\Lambda_\alpha,
272 \qquad
273 [x]\longmapsto(E_\alpha x+\Lambda_\alpha)_\alpha
274 \]
275 is injective, and every coordinate projection of its image is surjective.  We also have
276 \[
277 \boxed{
278 \expn_\pi(G_{\Lambda,\Pcal})
279 =
280 s_\Lambda(\Pcal).
281 }
282 \]
283 If
284 \[
285 G_{\Lambda,\Pcal}
286 \cong
287 \bigoplus_{j=1}^r\OO/\pi^{a_j}\OO
288 \]
289 with $a_1\leq\cdots\leq a_r$, then
290 \[
291 s_\Lambda(\Pcal)=\max\bigl(\{0\}\cup\{a_j:1\leq j\leq r\}\bigr).
292 \]
293 \end{theorem}
294 
295 \begin{proof}
296 If $x\in\Lambda$ maps to zero, then $E_\alpha x\in\Lambda_\alpha$ for every $\alpha$.  Since
297 \[
298 x=\sum_\alpha E_\alpha x,
299 \]
300 we have $x\in\Lambda_{\mathrm{split}}$.  Thus $\Phi$ is injective.  The coordinate map onto $E_\alpha\Lambda/\Lambda_\alpha$ is surjective by the definition of $E_\alpha\Lambda$.
301 
302 Let $s=s_\Lambda(\Pcal)$.  For every $x\in\Lambda$ and every $\alpha$,
303 \[
304 \pi^sE_\alpha x\in\Lambda_\alpha.
305 \]
306 Hence
307 \[
308 \pi^sx=\sum_\alpha\pi^sE_\alpha x\in\Lambda_{\mathrm{split}},
309 \]
310 so $\pi^sG_{\Lambda,\Pcal}=0$.  This proves
311 \[
312 \expn_\pi(G_{\Lambda,\Pcal})\leq s.
313 \]
314 If $d<s$, then for some $\alpha$ there is an $x\in\Lambda$ with
315 \[
316 \pi^dE_\alpha x\notin\Lambda_\alpha.
317 \]
318 The class $\pi^d[x]$ is therefore nonzero in the gluing module.  Thus $\pi^d$ does not annihilate the gluing module, and equality follows.  The final formula is the usual description of the exponent through Smith normal form.
319 \end{proof}
320 
321 \begin{corollary}[Total gluing and worst precision]\label{cor:length_and_weight}
322 If
323 \[
324 G_{\Lambda,\Pcal}
325 \cong
326 \bigoplus_{j=1}^r\OO/\pi^{a_j}\OO,
327 \]
328 then
329 \[
330 \length_{\OO}(G_{\Lambda,\Pcal})
331 =
332 \sum_{j=1}^ra_j,
333 \qquad
334 s_\Lambda(\Pcal)=\max\bigl(\{0\}\cup\{a_j:1\leq j\leq r\}\bigr).
335 \]
336 Thus the lattice index measures the total amount of gluing, while the separation weight measures the worst precision loss in one direction.
337 \end{corollary}
338 
339 \begin{proof}
340 The length and exponent of a finitely generated torsion module are read directly from its Smith decomposition.  Apply \cref{thm:gluing_weight}.
341 \end{proof}
342 
343 \begin{corollary}[Two complementary packets]\label{cor:binary_gluing}
344 Suppose
345 \[
346 V=V_1\oplus V_2
347 \]
348 with projector $E$ onto $V_1$ and complementary projector $I-E$.  Then
349 \[
350 \Lambda/(\Lambda_1\oplus\Lambda_2)
351 \cong
352 E\Lambda/\Lambda_1
353 \cong
354 (I-E)\Lambda/\Lambda_2.
355 \]
356 Consequently
357 \[
358 s_\Lambda(E)=s_\Lambda(I-E)
359 =
360 \expn_\pi\bigl(\Lambda/(\Lambda_1\oplus\Lambda_2)\bigr).
361 \]
362 \end{corollary}
363 
364 \begin{proof}
365 Map $x\in\Lambda$ to $Ex+\Lambda_1$.  Its kernel consists of the $x$ for which $Ex\in\Lambda_1$.  In that case
366 \[
367 (I-E)x=x-Ex\in\Lambda\cap V_2=\Lambda_2,
368 \]
369 so the kernel is $\Lambda_1\oplus\Lambda_2$.  The map is surjective by definition.  The complementary map is handled in the same way.  The equality of weights follows from \cref{thm:gluing_weight}.
370 \end{proof}
371 
372 \begin{corollary}[Integral splitting criterion]\label{cor:integral_split}
373 Every packet projector preserves $\Lambda$ if and only if
374 \[
375 \Lambda=\bigoplus_\alpha(\Lambda\cap V_\alpha).
376 \]
377 Equivalently,
378 \[
379 s_\Lambda(\Pcal)=0
380 \quad\Longleftrightarrow\quad
381 G_{\Lambda,\Pcal}=0.
382 \]
383 \end{corollary}
384 
385 \begin{proof}
386 If every projector preserves $\Lambda$, then
387 \[
388 x=\sum_\alpha E_\alpha x
389 \]
390 expresses every $x\in\Lambda$ as a sum of vectors in $\Lambda_\alpha$.  The converse is immediate because each projector is the identity on one direct summand and zero on the others.  The final equivalence follows from \cref{thm:gluing_weight}.
391 \end{proof}
392 
393 \subsection{Refinement and separation weight}
394 
395 A finer rational packet system imposes more integral splitting conditions.
396 
397 \begin{theorem}[Gluing exact sequence under refinement]\label{thm:gluing_refinement}
398 Let $\Qcal=(Q_\beta)$ refine $\Pcal=(P_\alpha)$ on $V$, and let $\Lambda$ be a lattice.  Define
399 \[
400 \Lambda_{\Pcal}
401 =
402 \bigoplus_\alpha(\Lambda\cap P_\alpha V),
403 \qquad
404 \Lambda_{\Qcal}
405 =
406 \bigoplus_\beta(\Lambda\cap Q_\beta V).
407 \]
408 Then
409 \[
410 \Lambda_{\Qcal}\subseteq\Lambda_{\Pcal}\subseteq\Lambda
411 \]
412 and there is a short exact sequence
413 \[
414 0
415 \longrightarrow
416 \Lambda_{\Pcal}/\Lambda_{\Qcal}
417 \longrightarrow
418 \Lambda/\Lambda_{\Qcal}
419 \longrightarrow
420 \Lambda/\Lambda_{\Pcal}
421 \longrightarrow0.
422 \]
423 In particular,
424 \[
425 \boxed{
426 s_\Lambda(\Pcal)\leq s_\Lambda(\Qcal).
427 }
428 \]
429 \end{theorem}
430 
431 \begin{proof}
432 Every fine packet space lies in its parent packet space, so every element of $\Lambda_{\Qcal}$ belongs to $\Lambda_{\Pcal}$.  The displayed exact sequence is the standard quotient sequence for nested submodules.  The exponent of a quotient cannot exceed the exponent of the module from which it is obtained.  Apply \cref{thm:gluing_weight} to the two gluing modules.
433 \end{proof}
434 
435 \begin{example}[Strict growth under refinement]\label{ex:strict_weight_growth}
436 Take the standard lattice $\mathbb Z_2^2$ and the involution from \cref{prop:sign_projectors}.  The one packet system $(I)$ has weight zero.  Refining it into the two rational sign packets produces weight one.  The coarse lattice is already split because there is only one summand.  The fine eigenspaces are glued together by one dyadic digit.
437 \end{example}
438 
439 \begin{remark}[Strict growth can occur]\label{rem:strict_growth}
440 A coarse packet family may preserve the lattice even though a fine spectral split does not.  Refinement can reveal rational directions that are glued together integrally.  The weight records the extra depth needed to separate them.
441 \end{remark}
442 
443 \subsection{Exact precision delay}
444 
445 The separation weight has an operational meaning.  It is the exact number of extra input digits needed to recover a projected packet uniformly.
446 
447 % Keep the final proof marker with its displayed conclusion.
448 \enlargethispage{\baselineskip}
449 \Needspace{8\baselineskip}
450 \begin{theorem}[Necessary and sufficient precision delay]\label{thm:precision_delay}
451 Let $E$ be a rational projector on $V$, let $\Lambda$ be a lattice, and let
452 \[
453 s=s_\Lambda(E).
454 \]
455 For every $m\geq0$, the rule
456 \[
457 [x]_{m+s}\longmapsto[Ex]_m
458 \]
459 from $\Lambda/\pi^{m+s}\Lambda$ to $V/\pi^m\Lambda$ is well defined.  Here $m$ is an integer.  If $d$ is an integer with $0\leq d<s$, then for every integer $m\geq0$ there are $x,y\in\Lambda$ such that
460 \[
461 x\equiv y\pmod{\pi^{m+d}\Lambda}
462 \]
463 but
464 \[
465 Ex\not\equiv Ey\pmod{\pi^m\Lambda}.
466 \]
467 Thus $s$ extra digits are both sufficient and necessary for a uniform precision guarantee.
468 \end{theorem}
469 
470 \begin{proof}
471 If
472 \[
473 x-y=\pi^{m+s}z,
474 \]
475 then
476 \[
477 E(x-y)=\pi^m(\pi^sEz)\in\pi^m\Lambda.
478 \]
479 This proves that the rule is well defined.  Now take $d<s$.  By the minimality of $s$, there is a $z\in\Lambda$ with
480 \[
481 \pi^dEz\notin\Lambda.
482 \]
483 Set
484 \[
485 x=\pi^{m+d}z,
486 \qquad
487 y=0.
488 \]
489 Then $x\equiv y$ modulo $\pi^{m+d}\Lambda$, while
490 \[
491 E(x-y)=\pi^m(\pi^dEz)\notin\pi^m\Lambda.
492 \]
493 \end{proof}
494 
495 \begin{keyidea}
496 The necessity statement is uniform over the lattice. A particular input may need fewer extra digits, but a smaller delay cannot recover every projected packet at the prescribed precision. This is why the least clearing exponent gives an exact precision guarantee rather than only a sufficient budget.
497 \end{keyidea}
498 
499 \begin{example}[The same rational split with two lattices]\label{ex:lattice_dependence}
500 Let
501 \[
502 V=Ke_1\oplus Ke_2
503 \]
504 and let $E$ project onto $Ke_1$.  On the split lattice
505 \[
506 \Lambda_0=\OO e_1\oplus\OO e_2,
507 \]
508 the weight is zero.  Fix $s\geq1$ and consider
509 \[
510 \Lambda_s
511 =
512 \OO e_1+
513 \OO(\pi^{-s}e_1+e_2).
514 \]
515 Then
516 \[
517 \Lambda_s\cap Ke_1=\OO e_1,
518 \qquad
519 \Lambda_s\cap Ke_2=\pi^s\OO e_2.
520 \]
521 The gluing module is isomorphic to $\OO/\pi^s\OO$, so
522 \[
523 s_{\Lambda_s}(E)=s.
524 \]
525 The rational decomposition has not changed.  Only the lattice has changed.
526 \end{example}
527 
528 The example can be realized by an integral operator.  Define
529 \[
530 Ae_1=0,
531 \qquad
532 Ae_2=\pi^se_2.
533 \]
534 In the basis
535 \[
536 e_1,
537 \quad
538 \pi^{-s}e_1+e_2,
539 \]
540 the matrix of $A$ is
541 \[
542 \begin{pmatrix}
543 0&-1\\
544 0&\pi^s
545 \end{pmatrix}.
546 \]
547 Its eigenvalue factors have resultant equal to $\pi^s$ up to a unit, and the exact projector weight on $\Lambda_s$ is also $s$.
548 
549 \section{The dyadic sign split}\label{sec:dyadic}
550 
551 The smallest example already shows why coefficient selectors and spectral averages must be kept distinct.
552 
553 \begin{proposition}[Sign projectors over a lattice]\label{prop:sign_projectors}
554 Let
555 \[
556 A=
557 \begin{pmatrix}
558 1&1\\
559 0&-1
560 \end{pmatrix}
561 \]
562 act on $\mathbb Z_p^2$.  Over $\mathbb Q_p$, the sign projectors are
563 \[
564 E_+=\frac{I+A}{2}
565 =
566 \begin{pmatrix}
567 1&1/2\\
568 0&0
569 \end{pmatrix},
570 \]
571 \[
572 E_-=\frac{I-A}{2}
573 =
574 \begin{pmatrix}
575 0&-1/2\\
576 0&1
577 \end{pmatrix}.
578 \]
579 For every odd prime $p$, both projectors preserve $\mathbb Z_p^2$.  At $p=2$,
580 \[
581 s_2(E_+)=s_2(E_-)=1.
582 \]
583 The factors $x-1$ and $x+1$ have resultant $2$.
584 \end{proposition}
585 
586 \begin{proof}
587 The displayed matrices are complementary idempotents.  The scalar $2$ is a unit in $\mathbb Z_p$ for odd $p$, so both matrices are integral there.  Over $\mathbb Z_2$, multiplying by $2$ clears the denominators, while neither projector preserves the standard lattice.  Thus both weights are one.  The resultant of the two linear factors is their root difference, which is $2$.
588 \end{proof}
589 
590 Modulo $2$, the two eigenvalues become equal.  The rational sign packets remain distinct over $\mathbb Q_2$, but the standard lattice does not split into their intersections until one more dyadic digit is retained.
591 
592 \Needspace{4\baselineskip}
593 \begin{proposition}[Integral character averages away from the group order]\label{prop:character_integrality}
594 Let $\OO$ be a discrete valuation ring, let $N$ be a unit in $\OO$, and suppose $\OO$ contains a primitive $N$th root of unity $\omega$.  If an integral operator $U$ satisfies $U^N=I$, then
595 \[
596 E_j=\frac1N\sum_{k=0}^{N-1}\omega^{-jk}U^k
597 \]
598 preserves every $U$ stable $\OO$ lattice.  At primes dividing $N$, this formula alone gives no integral projector.
599 \end{proposition}
600 
601 \begin{proof}
602 Every scalar coefficient in the displayed sum belongs to $\OO$ because $N$ is a unit and $\omega$ is integral.  Each power of $U$ preserves the lattice, so their $\OO$ linear combination does as well.  When $N$ is not a unit, division by $N$ is not an operation in $\OO$, so the same argument is unavailable.
603 \end{proof}
604 
605 At a prime dividing $N$, the factor $1/N$ no longer guarantees that the character average preserves the lattice. Cancellation may still make a particular projector integral. Coefficient residue selectors remain exact because they never divide by the group order.
606 
607 \subsection{The concrete rational sign matrix}
608 
609 Consider the integral involution
610 \[
611 A=
612 \begin{pmatrix}
613 1&1\\
614 0&-1
615 \end{pmatrix},
616 \qquad
617 A^2=I.
618 \]
619 Over a field of characteristic different from two, its sign projectors are
620 \[
621 E_+=\frac{I+A}{2}
622 =
623 \begin{pmatrix}
624 1&1/2\\
625 0&0
626 \end{pmatrix},
627 \]
628 \[
629 E_-=\frac{I-A}{2}
630 =
631 \begin{pmatrix}
632 0&-1/2\\
633 0&1
634 \end{pmatrix}.
635 \]
636 They are orthogonal idempotents and sum to $I$.
637 
638 \begin{figure}[htbp]
639 \centering
640 \includegraphics[width=0.96\textwidth]{07_dyadic_delayed_separation.pdf}
641 \caption{The rational sign projectors for an integral involution. Their denominator has $2$-adic weight one.}
642 \label{fig:dyadic-projectors}
643 \end{figure}
644 
645 \begin{proposition}[Odd-adic versus dyadic behavior]\label{prop:dyadic-sign}
646 For every prime $p\ne2$, every entry of $E_+$ and $E_-$ is $p$-integral,
647 and each projector has exact local denominator weight zero. At $p=2$, each
648 projector has exact local denominator weight one. Multiplication by $2$ makes
649 it integral, while the unscaled matrix is not integral:
650 \[
651 s_{2,\Z_2^2}(E_+)=s_{2,\Z_2^2}(E_-)=1.
652 \]
653 The characteristic factors $x-1$ and $x+1$ have resultant $2$. Its $2$-adic
654 valuation is one, matching both local projector weights above, and its
655 $p$-adic valuation is zero at every prime $p\ne2$.
656 \end{proposition}
657 
658 \begin{proof}
659 The reduced denominators of the displayed entries are $1$ or $2$, which proves
660 the local integrality and minimality statements directly. The resultant is
661 \[
662 \Res(x-1,x+1)=2.
663 \]
664 Thus $v_2(2)=1$ and $v_p(2)=0$ for primes $p\ne2$.
665 \end{proof}
666 
667 The two characteristic factors $x-1$ and $x+1$ have the same reduction in
668 $\F_2[x]$. The rational matrices $E_+$ and $E_-$ themselves cannot be reduced
669 modulo $2$, because their entries contain $1/2$. Consequently the integral
670 special fiber does not inherit their separate idempotent splitting, even
671 though the two projectors remain distinct over $\Q_2$. This is the simplest
672 model of delayed dyadic separation.
673 
674 \begin{figure}[htbp]
675 \centering
676 \includegraphics[width=0.96\textwidth]{08_odd_prime_separation.pdf}
677 \caption{The denominator weight of the sign projector is nonzero only at the prime two.}
678 \label{fig:odd-dyadic}
679 \end{figure}
680 
681 \Needspace{7\baselineskip}
682 \subsubsection{Why coefficient packets remain available}
683 
684 The failure of the character formula
685 \[
686 \frac12(I\pm A)
687 \]
688 to be integral does not invalidate coefficient selectors such as even and odd residue extraction. It shows that two distinct constructions must be distinguished:
689 \[
690 \boxed{
691 \text{integral address selector}
692 \quad\neq\quad
693 \text{character-average spectral projector}
694 }
695 \]
696 when the group order is not a unit. The first can remain exact over $\Z_2$. The second can acquire a denominator weight.
697 
698 
699 \section{Smith normal form and congruence lifting}\label{sec:smith}
700 
701 The gluing module gives the exact common denominator of packet projectors.  Smith normal form gives the corresponding coordinate by coordinate solvability and branching rule.
702 
703 \begin{standardfact}[Smith normal form over a discrete valuation ring]\label{fact:smith}
704 Let $M\in M_{m,n}(\OO)$.  There are invertible matrices
705 \[
706 U\in\operatorname{GL}_m(\OO),
707 \qquad
708 V\in\operatorname{GL}_n(\OO)
709 \]
710 such that
711 \[
712 UMV
713 =
714 \diag(\pi^{s_1},\ldots,\pi^{s_r},0,\ldots,0),
715 \]
716 where
717 \[
718 0\leq s_1\leq\cdots\leq s_r
719 \]
720 and $r$ is the rank over $K$.  The exponents are uniquely determined.  See \cite{Newman1972}.
721 \end{standardfact}
722 
723 \begin{proposition}[Computing a packet weight from an inclusion matrix]\label{prop:weight_from_inclusion}
724 Choose an $\OO$ basis of $\Lambda$ and an $\OO$ basis of $\Lambda_{\mathrm{split}}$.  Let $B$ be the matrix of the inclusion
725 \[
726 \Lambda_{\mathrm{split}}\hookrightarrow\Lambda.
727 \]
728 If the Smith form of $B$ is
729 \[
730 \diag(\pi^{a_1},\ldots,\pi^{a_n}),
731 \qquad
732 0\leq a_1\leq\cdots\leq a_n,
733 \]
734 then
735 \[
736 G_{\Lambda,\Pcal}
737 \cong
738 \bigoplus_{j=1}^n\OO/\pi^{a_j}\OO,
739 \qquad
740 s_\Lambda(\Pcal)=\max\bigl(\{0\}\cup\{a_j:1\leq j\leq n\}\bigr).
741 \]
742 \end{proposition}
743 
744 \begin{proof}
745 The gluing module is the cokernel of the inclusion matrix $B$.  Smith normal form gives its invariant factors.  The weight formula follows from \cref{thm:gluing_weight}.
746 \end{proof}
747 
748 \begin{theorem}[Smith weighted congruence solve]\label{thm:smith_congruence}
749 Let $M\in M_{m,n}(\OO)$ have Smith exponents $s_1,\ldots,s_r$.  Assume the residue field is finite of cardinality $q$, fix $L\geq1$, and consider
750 \[
751 Mx\equiv b\pmod{\pi^L}.
752 \]
753 Put
754 \[
755 \widetilde b=Ub
756 \]
757 for a Smith matrix $U$ as in \cref{fact:smith}.  The congruence is solvable if and only if
758 \[
759 \widetilde b_i\equiv0
760 \pmod{\pi^{\min(s_i,L)}}
761 \qquad(1\leq i\leq r)
762 \]
763 and
764 \[
765 \widetilde b_i\equiv0
766 \pmod{\pi^L}
767 \qquad(r<i\leq m).
768 \]
769 When it is solvable, the number of solutions in $(\OO/\pi^L\OO)^n$ is
770 \[
771 \boxed{
772 q^{(n-r)L+\sum_{i=1}^r\min(s_i,L)}.
773 }
774 \]
775 \end{theorem}
776 
777 \begin{proof}
778 Set $x=Vy$.  Multiplying the congruence by $U$ reduces it to independent equations
779 \[
780 \pi^{s_i}y_i\equiv\widetilde b_i\pmod{\pi^L}
781 \qquad(1\leq i\leq r)
782 \]
783 and zero equations in the remaining rows.  Multiplication by $\pi^{s_i}$ on $\OO/\pi^L\OO$ has image equal to the elements divisible by $\pi^{\min(s_i,L)}$ and kernel of size
784 \[
785 q^{\min(s_i,L)}.
786 \]
787 Each of the $n-r$ unused variables is free and contributes $q^L$ choices.  Multiplying the independent counts gives the formula.
788 \end{proof}
789 
790 \begin{corollary}[Obstruction and branching data]\label{cor:smith_obstruction}
791 The transformed entries of $\widetilde b$ that fail the stated divisibility conditions are the complete finite congruence obstructions.  When all obstructions vanish, the Smith exponents determine the number of solution branches at depth $L$.
792 \end{corollary}
793 
794 \begin{proof}
795 This is exactly the solvability and counting statement of \cref{thm:smith_congruence} written in invariant coordinates.
796 \end{proof}
797 
798 \begin{example}[Three weighted directions]\label{ex:smith_three}
799 Take $p=2$, $L=6$, and the diagonal weights
800 \[
801 (0,1,3).
802 \]
803 The right side
804 \[
805 (1,2,8)
806 \]
807 is admissible.  The three coordinate kernels have sizes
808 \[
809 1,
810 \quad2,
811 \quad8,
812 \]
813 so there are $16$ solutions.  Replacing the final entry $8$ by $4$ violates the required divisibility by $2^3$ and makes the system unsolvable.
814 \end{example}
815 
816 \subsection{Concrete diagonal fibers over integer quotients}
817 
818 Let $M$ be a matrix over a DVR. Smith normal form gives invertible matrices $U,V$ such that
819 \[
820 UMV
821 =
822 \diag(\pi^{s_1}u_1,\ldots,\pi^{s_r}u_r,0,\ldots,0),
823 \]
824 where each $u_i$ is a unit and
825 \[
826 0\le s_1\le\cdots\le s_r.
827 \]
828 The exponents $s_i$ are directional lifting weights.
829 
830 \begin{theorem}[Concrete Smith-weighted congruence solve]\label{thm:smith-solve}
831 Let $p>1$ and consider the diagonal system over
832 $(\Z/(p^L)\Z)^d$
833 \[
834 \diag(p^{s_1},\ldots,p^{s_d})x=b
835 \pmod{p^L}.
836 \]
837 The system is solvable if and only if
838 \[
839 p^{\min(s_i,L)}\mid (b_i)_{\mathrm{val}}
840 \]
841 for every $i$, where $(b_i)_{\mathrm{val}}$ is the canonical natural
842 representative. Conditional on this admissibility condition, the number of
843 solutions is
844 \[
845 \boxed{
846 p^{\sum_{i=1}^d\min(s_i,L)}.
847 }
848 \]
849 If any coordinate is inadmissible, the solution set is empty and its
850 cardinality is zero.
851 \end{theorem}
852 
853 \begin{proof}
854 Lean proves the one-coordinate arithmetic directly over
855 $\mathbb Z/p^L\mathbb Z$ for every $p>1$:
856 \[
857 \bigl(\exists x\in\mathbb Z/p^L\mathbb Z:\quad p^s x=b\pmod{p^L}\bigr)
858 \quad\Longleftrightarrow\quad
859 p^{\min(s,L)}\mid b_{\mathrm{val}},
860 \]
861 where $b_{\mathrm{val}}$ is the canonical natural representative, and every
862 admissible fiber has exactly $p^{\min(s,L)}$ elements. The coordinates are
863 independent, so the product equivalence gives the displayed global count. If
864 one coordinate fiber is empty, the global fiber is empty. Lean proves both the
865 global solvability equivalence and the conditional count, including
866 cardinality zero on the inadmissible branch.
867 \end{proof}
868 
869 For an arbitrary DVR $\OO$ with residue field cardinality $q$, the analogous
870 formula
871 \[
872 q^{\sum_i\min(s_i,L)}
873 \]
874 follows from the solvability and fiber count for each coordinate. The formal
875 correspondence records direct proofs of this DVR coordinate arithmetic and
876 the rectangular transport as separate targets.
877 The inherited abstract product interface remains available when the coordinate
878 facts are supplied as certificates. A Smith form and its changes of basis are
879 inputs to the transport statement. It does not provide an extracted
880 diagonalization routine. See \cite{Serre1979,Lang2002}.
881 
882 If a Smith form is rank deficient, a literal zero diagonal factor requires
883 the corresponding target coordinate to vanish modulo $p^L$ and then
884 contributes all $p^L$ possible coordinate values. In the concrete quotient it
885 can equivalently be represented by any weight $s_i\ge L$, since
886 $p^{s_i}=0$ in $\Z/(p^L)\Z$. The theorem then gives exactly that condition and
887 factor. The numbered theorem takes the diagonal data as its input.
888 
889 \begin{example}
890 At $p=2$, depth $L=6$, and weights $(0,1,3)$, the right side $(1,2,8)$ is admissible. The coordinate kernels have sizes
891 \[
892 1,2,8,
893 \]
894 so the system has $16$ solutions. Replacing $8$ by $4$ in the final coordinate violates the divisibility condition and makes the system unsolvable.
895 \end{example}
896 
897 \begin{interpretation}
898 The gluing calculation takes the largest Smith weight because every packet must be separated to the requested precision. The congruence calculation retains the individual directional weights. A coordinate with weight $s_i$ contributes a solvability threshold and $\min(s_i,L)$ to the branching exponent. The independent coordinate counts multiply, so these truncated weights add in the exponent.
899 \end{interpretation}
900 
901 

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