Warning: the resolution can have free modules with unexpected ranks when the module
is not homogeneous. Here is an example where even the lengths of the resolutions differ. We compute a resolution of the kernel of a ring map in two ways. The ring
will turn out to be homogeneous.
i1 : k = ZZ/101; T = k[v..z];
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i3 : m = matrix {{x,y,z,x^2*v,x*y*v,y^2*v,z*v,x*w,y^3*w,z*w}}
o3 = | x y z vx2 vxy vy2 vz wx wy3 wz |
1 10
o3 : Matrix T <--- T
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i4 : n = rank source m
o4 = 10
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i5 : R = k[u_1 .. u_n]
o5 = R
o5 : PolynomialRing
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i6 : S = k[u_1 .. u_n,Degrees => degrees source m]
o6 = S
o6 : PolynomialRing
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i7 : f = map(T,R,m)
2 2 3
o7 = map(T,R,{x, y, z, v*x , v*x*y, v*y , v*z, w*x, w*y , w*z})
o7 : RingMap T <--- R
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i8 : g = map(T,S,m)
2 2 3
o8 = map(T,S,{x, y, z, v*x , v*x*y, v*y , v*z, w*x, w*y , w*z})
o8 : RingMap T <--- S
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i9 : res ker f
1 17 57 76 46 12 1
o9 = R <-- R <-- R <-- R <-- R <-- R <-- R <-- 0
0 1 2 3 4 5 6 7
o9 : ChainComplex
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i10 : res ker g
1 14 35 35 15 2
o10 = S <-- S <-- S <-- S <-- S <-- S <-- 0
0 1 2 3 4 5 6
o10 : ChainComplex
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i11 : isHomogeneous f
o11 = false
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i12 : isHomogeneous g
o12 = true
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