In this section, we give examples of common operations involving modules. Throughout this section, we suppose that the base ring
R is graded, with each variable having degree one, and that
M is a graded
R-module. If the ring is not graded, or is multi-graded, or if
M is not graded, some of these functions still work, but care must be taken in interpreting the output. Here, we just consider the standard grading case.
checking homogeneity
Let's start by making a module over a ring with 18 variables
i1 : R = ZZ/32003[vars(0..17)];
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i2 : M = coker genericMatrix(R,a,3,6)
o2 = cokernel | a d g j m p |
| b e h k n q |
| c f i l o r |
3
o2 : R-module, quotient of R
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Use
isHomogeneous to check whether a given module is graded.
i3 : isHomogeneous M
o3 = true
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codimension, degree, and sectional arithmetic genera
Use
codim(Module),
degree(Module), and
genera(Module) for some basic numeric information about a module.
i4 : codim M
o4 = 4
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i5 : degree M
o5 = 15
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i6 : genera M
o6 = {-2, 2, -2, 2, -2, 2, -2, 2, -2, 2, -2, 2, 4, 14}
o6 : List
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The last number in the list of genera is the degree minus one. The second to last number is the genus of the generic linear section curve, ..., and the first number is the arithmetic genus.
the Hilbert series
The Hilbert series (
hilbertSeries(Module)) of
M is by definition the formal power series
H(t) = sum(d in ZZ) dim(M_d) t^d. This is a rational function with denominator
(1-t)^n, where
n is the number of variables in the polynomial ring. The numerator of this rational function is called the poincare polynomial, and is obtained by the
poincare(Module) function.
i7 : poincare M
4 5 6
o7 = 3 - 6T + 15T - 18T + 6T
o7 : ZZ[T]
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i8 : hf = hilbertSeries M
4 5 6
3 - 6T + 15T - 18T + 6T
o8 = --------------------------
18
(1 - T)
o8 : Expression of class Divide
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It is often useful to divide the poincare polynomial by
(1-t) as many times as possible. This can be done by using
reduceHilbert:
i9 : reduceHilbert hf
2
3 + 6T + 6T
o9 = ------------
14
(1 - T)
o9 : Expression of class Divide
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i10 : poincare' = (M) -> (
H := poincare M;
t := (ring H)_0; -- The variable t above
while H % (1-t) == 0 do H = H // (1-t);
H);
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i11 : poincare' M
2
o11 = 3 + 6T + 6T
o11 : ZZ[T]
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free resolutions
The minimal free resolution
C is computed using
resolution(Module). The specific matrices are obtained by indexing
C.dd.
i12 : C = resolution M
3 6 15 18 6
o12 = R <-- R <-- R <-- R <-- R <-- 0
0 1 2 3 4 5
o12 : ChainComplex
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i13 : C.dd_3
o13 = {4} | m -n o p -q r 0 0 0 0 0 0 0 0 0 0 0 0 |
{4} | -j k -l 0 0 0 p 0 0 0 -q r 0 0 0 0 0 0 |
{4} | g -h i 0 0 0 0 p 0 0 0 0 -q 0 0 r 0 0 |
{4} | -d e -f 0 0 0 0 0 p 0 0 0 0 -q 0 0 r 0 |
{4} | a -b c 0 0 0 0 0 0 p 0 0 0 0 -q 0 0 r |
{4} | 0 0 0 -j k -l -m 0 0 0 n -o 0 0 0 0 0 0 |
{4} | 0 0 0 g -h i 0 -m 0 0 0 0 n 0 0 -o 0 0 |
{4} | 0 0 0 -d e -f 0 0 -m 0 0 0 0 n 0 0 -o 0 |
{4} | 0 0 0 a -b c 0 0 0 -m 0 0 0 0 n 0 0 -o |
{4} | 0 0 0 0 0 0 g j 0 0 -h i -k 0 0 l 0 0 |
{4} | 0 0 0 0 0 0 -d 0 j 0 e -f 0 -k 0 0 l 0 |
{4} | 0 0 0 0 0 0 a 0 0 j -b c 0 0 -k 0 0 l |
{4} | 0 0 0 0 0 0 0 -d -g 0 0 0 e h 0 -f -i 0 |
{4} | 0 0 0 0 0 0 0 a 0 -g 0 0 -b 0 h c 0 -i |
{4} | 0 0 0 0 0 0 0 0 a d 0 0 0 -b -e 0 c f |
15 18
o13 : Matrix R <-- R
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For more information about chain complexes and resolutions, see
chain complexes and
computing resolutions.
betti numbers
Use
betti to display the graded betti numbers of
M.
i14 : betti C
0 1 2 3 4
o14 = total: 3 6 15 18 6
0: 3 6 . . .
1: . . . . .
2: . . 15 18 6
o14 : BettiTally
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This table should be interpreted as follows: the number in the
i-th row and
j-th column (indices starting at 0), is the number of
j-th syzygies in degree
i+j. In the above example, there are 15 second syzygies of degree 4, and the entries of the maps
CC.d_1, CC.d_3, CC.d_4 are all linear.