Here we give an example of a spectral sequence that takes n+2 steps to degenerate, where n is the embedding dimesion of the ring. We present this when n = 2 but the user with computational power can easily do a bigger case.

i1 : S = ZZ/101[x,y]; |

i2 : I = ideal(x^2,x*y,y^2); o2 : Ideal of S |

i3 : R = S/I; |

i4 : kR = coker vars R; |

i5 : kS = coker vars S; |

i6 : CS = res kS; |

i7 : CR = res(kR,LengthLimit=>6); |

i8 : CS' = CS**R; |

i9 : E = prune spectralSequence (CS' ** filteredComplex CR); |

Since this spectral sequence only consists of $k$ vector spaces, and all are generated in a single degree, for ease of presentation we may as well just look at the rank and degree which we can easily encode in a matrix with $rt^d$ encoding the rank $r$ and degree $d$ of each vector space $E_{i,j}$.

i10 : use ZZ[t] o10 = ZZ[t] o10 : PolynomialRing |

i11 : easyPresentation = (P,n,m) -> ( transpose matrix apply(n, i-> apply(m, j-> (length (P_{i,j}))*t^( if (L = unique flatten degrees P_{i,j})!= {} then first L else 0) ) )); |

To see what we're going for, we compute the E_{infinity} page and also some earlier pages. Notice that it's clear that all terms except those in the top row of the matrix must eventually disappear, but for this to happen, there must a map of the right degree mapping to them.

i12 : easyPresentation(E_infinity,6,3) o12 = {-2} | 1 2t t2 0 0 0 | {-7} | 0 0 0 0 0 96t7 | {-8} | 0 0 0 0 32t7 64t8 | 3 6 o12 : Matrix (ZZ[t]) <--- (ZZ[t]) |

i13 : easyPresentation(E_1,6,3) o13 = {-5} | 1 2t 4t2 8t3 16t4 32t5 | {-7} | 3t2 6t3 12t4 24t5 48t6 96t7 | {-8} | 2t3 4t4 8t5 16t6 32t7 64t8 | 3 6 o13 : Matrix (ZZ[t]) <--- (ZZ[t]) |

i14 : easyPresentation(E_2,6,3) o14 = {-5} | 1 2t 4t2 8t3 16t4 32t5 | {-7} | 3t2 6t3 12t4 24t5 48t6 96t7 | {-8} | 2t3 4t4 8t5 16t6 32t7 64t8 | 3 6 o14 : Matrix (ZZ[t]) <--- (ZZ[t]) |

i15 : easyPresentation(E_3,6,3) o15 = {-5} | 1 2t t2 2t3 4t4 8t5 | {-7} | 0 0 0 0 0 96t7 | {-8} | 2t3 4t4 8t5 16t6 32t7 64t8 | 3 6 o15 : Matrix (ZZ[t]) <--- (ZZ[t]) |

i16 : length image ((E_2).dd_{3,0}) o16 = 6 |

i17 : length image (E_3).dd_{3,0} o17 = 2 |

The final two computations are meant to explain that the copy of $k^8$ in degree 3 that appears on the $E_1$ cancels in two steps via an $E_2$ map with $k^6$ and via an $E_3$ map with a $k^2$.