Starting from a nonminimal resolution F of the carpet over a larger finite prime field, we lift the complex to the integers, and compute the diagonal entries of the Smith normal form. The critical constrand strand for a carpet of type (a,b) with a>=b is the a+1-st strand. Green's conjecture for carpet says that the map has maximal rank over QQ.
i1 : a=5,b=5
o1 = (5, 5)
o1 : Sequence
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i2 : I = carpet(a,b);
ZZ
o2 : Ideal of -----[x ..x , y ..y ]
32003 0 5 0 5
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i3 : F = res(I, FastNonminimal => true)
ZZ 1 ZZ 36 ZZ 187 ZZ 491 ZZ 793 ZZ 833 ZZ 573 ZZ 250 ZZ 63 ZZ 7
o3 = (-----[x ..x , y ..y ]) <-- (-----[x ..x , y ..y ]) <-- (-----[x ..x , y ..y ]) <-- (-----[x ..x , y ..y ]) <-- (-----[x ..x , y ..y ]) <-- (-----[x ..x , y ..y ]) <-- (-----[x ..x , y ..y ]) <-- (-----[x ..x , y ..y ]) <-- (-----[x ..x , y ..y ]) <-- (-----[x ..x , y ..y ]) <-- 0
32003 0 5 0 5 32003 0 5 0 5 32003 0 5 0 5 32003 0 5 0 5 32003 0 5 0 5 32003 0 5 0 5 32003 0 5 0 5 32003 0 5 0 5 32003 0 5 0 5 32003 0 5 0 5
10
0 1 2 3 4 5 6 7 8 9
o3 : ChainComplex
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i4 : L = analyzeStrand(F,a); #L
-- 0.02973 seconds elapsed
o5 = 350
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i6 : betti F_a, betti F
0 0 1 2 3 4 5 6 7 8 9
o6 = (total: 833, total: 1 36 187 491 793 833 573 250 63 7)
6: 350 0: 1 . . . . . . . . .
7: 468 1: . 36 160 342 436 350 174 49 6 .
8: 15 2: . . 27 148 351 468 379 186 51 6
3: . . . 1 6 15 20 15 6 1
o6 : Sequence
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i7 : factor product L
266 15
o7 = 2 3
o7 : Expression of class Product
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i8 : L3 = select(L,c->c%3==0); #L3
o9 = 14
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i10 : carpetBettiTable(a,b,3)
-- 0.00159207 seconds elapsed
-- 0.00469944 seconds elapsed
-- 0.0207872 seconds elapsed
-- 0.027244 seconds elapsed
-- 0.00251372 seconds elapsed
0 1 2 3 4 5 6 7 8 9
o10 = total: 1 36 160 315 302 302 315 160 36 1
0: 1 . . . . . . . . .
1: . 36 160 315 288 14 . . . .
2: . . . . 14 288 315 160 36 .
3: . . . . . . . . . 1
o10 : BettiTally
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