Description
If
M or
N is a sheaf of rings, it is regarded as a sheaf of modules in the evident way.
M and
N must be coherent sheaves on the same projective variety or scheme
X.
As an example, we compute Hom_X(I_X,OO_X), and Ext^1_X(I_X,OO_X), for the rational quartic curve in $P^3$.
i1 : S = QQ[a..d];
|
i2 : I = monomialCurveIdeal(S,{1,3,4})
3 2 2 2 3 2
o2 = ideal (b*c - a*d, c - b*d , a*c - b d, b - a c)
o2 : Ideal of S
|
i3 : R = S/I
o3 = R
o3 : QuotientRing
|
i4 : X = Proj R
o4 = X
o4 : ProjectiveVariety
|
i5 : IX = sheaf (module I ** R)
o5 = cokernel {2} | c2 bd ac b2 |
{3} | -b -a 0 0 |
{3} | d c -b -a |
{3} | 0 0 -d -c |
1 3
o5 : coherent sheaf on X, quotient of OO (-2) ++ OO (-3)
X X
|
i6 : Ext^1(IX,OO_X)
o6 = 0
o6 : QQ-module
|
i7 : Hom(IX,OO_X)
16
o7 = QQ
o7 : QQ-module, free
|
The Ext^1 being zero says that the point corresponding to I on the Hilbert scheme is smooth (unobstructed), and vector space dimension of Hom tells us that the dimension of the component at the point I is 16.
The method used may be found in: Smith, G.,
Computing global extension modules, J. Symbolic Comp (2000) 29, 729-746
If the module $\oplus_{d\geq 0} Ext^i(M,N(d))$ is desired, see
Ext^ZZ(CoherentSheaf,SumOfTwists).