Fixed bug in A1BrouwerDegrees

[wgabrielong]

A bug was identified and fixed by Claude Code in A1BrouwerDegrees.

Consider the quadratic form
beta = makeDiagonalForm(QQ, (-1,-1,-1,-1,-1,1))

Since beta ≅ <-1,-1,-1,-1> + H, its anisotropic part is <-1,-1,-1,-1>, of rank 4.
However, getAnisotropicPart beta silently returns the rank-2 form <-1,1> — the
hyperbolic plane, which is isotropic and violates the Witt decomposition identity:

i6 : a = getAnisotropicPart beta
o6 = | -1 0 |
     | 0  1 |
i8 : isAnisotropic a
o8 = false
i9 : isIsomorphicForm(beta, addGW(a, makeHyperbolicForm(QQ, 2*getWittIndex beta)))
o9 = false

Cause. reduceAnisotropicPartQQDimension3 (Code/Decomposition.m2) must return
<alpha> such that beta + <-alpha> has anisotropic dimension 2. It chooses alpha
from congruence conditions at the finite primes only, ignoring the real place, where the
anisotropic dimension equals |signature|. Here the running form <-1,-1,-1,-1,-1,1,1>
has signature -3 and an empty relevant-prime list, so the solver returns alpha = 1;
the real anisotropic part is negative definite, so alpha must be negative. The
anisotropic dimension rises from 3 to 4 instead of dropping to 2, the == 2/== 1
branches of getAnisotropicPartQQ never fire, and a truncated form is returned with no
error.

Fix (by Claude Code). The congruences determine alpha only modulo the product M
of the moduli, so the real place is accommodated by shifting to a representative of the
correct sign:

M := product(L1 | L2);
if getSignature(beta) == -3 and alpha > 0 then alpha = (alpha % M) - M;
if getSignature(beta) == 3 and alpha < 0 then alpha = (alpha % M) + M;

getAnisotropicPart beta now returns <-1,-1,-1,-1>. Tests 48–49 add regression
coverage (the existing Test 47 never reached the reduction chain); the full package
suite passes under Macaulay2 1.25.11.