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TestIdeals :: QGorensteinGenerator

QGorensteinGenerator -- finds an element representing the Frobenius trace map of a Q-Gorenstein ring

Synopsis

Description

Suppose that R is a ring such that (pe-1)KR is linearly equivalent to zero (for example, if R is Q-Gorenstein with index not divisible by p). Then if we write R = S/I where S is a polynomial ring, we have that I[pe] : I = u S + I[pe] for some u ∈S. By Fedder’s criterion, this element u represents the generator of the R1/pe-module Hom(R1/pe, R). For example if I = (f) is principal, then u = fpe-1 works.

This function produces the element f described above. If do not specify an integer e, it assumes e = 1.

i1 : S = ZZ/3[x,y,z];
i2 : f = x^2*y - z^2;
i3 : I = ideal(f);

o3 : Ideal of S
i4 : R = S/I;
i5 : u = QGorensteinGenerator(1, R)

      4 2    2   2    4
o5 = x y  + x y*z  + z

o5 : S
i6 : u%I^3 == f^2%I^3

o6 = true

If Macaulay2 does not recognize that I[pe] : I / I[pe] is principal, an error is thrown. Note in the nongraded case, Macaulay2 is not guaranteed to do this simplification.

Ways to use QGorensteinGenerator :