Elementary uses of Groebner bases
\def\P{\bf P}
In this tutorial we introduce a number
of basic operations using Groebner bases, and
at the same time become familiar with a range
of useful Macaulay2 constructs. The sections are:
A. First steps; example with a monomial curve
B. Random regular sequences
C. Division with remainder
D. Elimination theory
E. Quotients and saturation
\beginsection{ A. First Steps; example with a monomial curve}\par
To compute the Groebner basis of an ideal
$(x^2y,xy^2+x^3)$ in the polynomial ring in
four variables we proceed as follows:
Our favorite field
i1 : KK = ZZ/31991
o1 = KK
o1 : QuotientRing |
The polynomial ring
i2 : R = KK[x,y,z,w]
o2 = R
o2 : PolynomialRing |
and the ideal
i3 : I = ideal(x^2*y,x*y^2+x^3)
2 3 2
o3 = ideal (x y, x + x*y )
o3 : Ideal of R |
now the punch line:
i4 : J = gens gb I
o4 = {0} | x2y x3+xy2 xy3 |
1 3
o4 : Matrix R <--- R |
From this we can for example compute the
codimension, dimension,
degree, and the whole Hilbert
function and polynomial.
This will be more fun if we work with an
example having some meaning. We choose
to work with the ideal defining the
rational quartic curve in $\P^3$ given
parametrically in an affine representation
by
$$t \mapsto (t,t^3,t^4).$$
(The reader who doesn't understand this
terminology may ignore it for the moment,
and treat the ideal given below as a
gift from the gods... .)
We obtain the ideal by first making the
polynomial ring in 4 variables (the
homogeneous coordinate ring of $\P^3$)
i5 : R = KK[a..d]
o5 = R
o5 : PolynomialRing |
and then using a function {\tt monomialCurve}, which we shall
treat for now as a black box
i6 : I = monomialCurve(R,{1,3,4})
3 2 2 2 3 2
o6 = ideal (b*c - a*d, c - b*d , a*c - b d, b - a c)
o6 : Ideal of R |
From Macaulay's point of view, $I$ is an
ideal, and the codimension of its support
is 2, while its dimension is 2:
This is the codimension of $R/I$ in $R$
and the dimension of $R/I$. We could work with
the module $R/I$ as well.
Precision requires writing $R^1$ instead
of $R$ ($R$ is a ring, and $R^1$ is
the free module of rank 1 over it)
i9 : codim (R^1/(I*R^1))
o9 = 2 |
We could also extract the generators of
$I$ (as a matrix) and take the cokernel to
get the same thing:
i10 : M = coker gens I
o10 = cokernel {0} | bc-ad c3-bd2 ac2-b2d b3-a2c |
1
o10 : R - module, quotient of R |
And similarly for the degree:
As one might expect, the degree of the quartic
is 4 !
The Hilbert polynomial is obtained by
i15 : hilbertPolynomial M
o15 = - 3*P + 4*P
0 1
o15 : ProjectiveHilbertPolynomial |
The term $\P^i$ represents the Hilbert polynomial of
projective $i$-space. This formula tells
us that the Hilbert polynomial of $M$ is
$H(m) = 4(m+1) - 3 = 4m + 1$. Thus the degree
is four, the dimension of the projective variety
which is the support of $M$ is 1 (and so the affine
dimension is 2),
and that the (arithmetic) genus is 0 (1 minus the
constant term, for curves).
The Hilbert series of $M$ (the generating function
for the dimensions of the graded pieces of $M$) is
i16 : hilbertSeries M
3 2
- $T + 2$T + 2$T + 1
o16 = ----------------------
2
(- $T + 1)
o16 : Divide |
The indeterminate in this expression is \$T.
Another way to get information about
the module $M$ is to see its free resolution
i17 : Mres = res M
1 4 4 1
o17 = R <-- R <-- R <-- R
0 1 2 3
o17 : ChainComplex |
To get more precise information about Mres,
we could do
i18 : betti Mres
o18 = total: 1 4 4 1
0: 1 . . .
1: . 1 . .
2: . 3 4 1
o18 : Net |
The display is chosen for compactness:
the first line gives the total betti
numbers, the same information given when
we type the resolution. The remaining
lines express the degrees of each of the
generators of the free modules in the
resolution. The $j$th column after the colons
gives the degrees of generators of the
$j$th module(counting from $0$);
an $n$ in the $j$th column in the
row headed by ``$d$:'' means that the $j$th
free module has $n$ generators of degree
$n+j$. Thus for example in our case, the
generator of the third (last) free module in the
resolution has degree $3+2=5$.
\beginsection{ B. Random regular sequences}\par
An interesting and illustrative open problem
is to understand the initial ideal (and
the Groebner basis) of a ``generic''
regular sequence. To study a very simple case
we take a matrix of 2 random forms
in a polynomial ring in
3 variables:
i19 : R = KK[x,y,z]
o19 = R
o19 : PolynomialRing |
i20 : F = random(R^1, R^{-2,-3})
o20 = {0} | -8514x2+5827xy-9613y2+7886xz+5702yz+11835z2 -15791x3+12712x2y-11866xy2-11819y3-2169x2z+1548xyz+10647y2z+11189xz2+6279yz2+8360z3 |
1 2
o20 : Matrix R <--- R |
makes $F$ into a $1 \times 2$ matrix whose elements
have degrees $2,3$ (that is, $F$ is a random map
to the free module $R^1$, which has its one
generator in the (default) degree, $0$, from
the free module with generators in the listed
degrees, $\{2,3\}$). We now can compute
i21 : GB = gens gb F
o21 = {0} | x2-14140xy+5100y2+12053xz+14443yz+7878z2 xy2-6589y3-2097xyz-8415y2z-5942xz2+4719yz2-9862z3 y4+2957y3z-7768xyz2+13545y2z2+7229xz3+5489yz3-4313z4 |
1 3
o21 : Matrix R <--- R |
i22 : LT = leadTerm gens gb F
o22 = {0} | x2 xy2 y4 |
1 3
o22 : Matrix R <--- R |
i23 : betti LT
o23 = total: 1 3
0: 1 .
1: . 1
2: . 1
3: . 1
o23 : Net |
shows that there are Groebner basis elements
of degrees 2,3, and 4. This result is
dependent on the monomial order in the ring $R$;
for example we could take the lexicographic
order
i24 : R = KK[x,y,z, MonomialOrder => Lex]
o24 = R
o24 : PolynomialRing |
(see {\tt help MonomialOrder} for other possibilities).
We get
i25 : F = random(R^1, R^{-2,-3})
o25 = {0} | -9394x2+8312xy+12075xz-3147y2-11542yz+3907z2 -13293x3-1763x2y+10247x2z-8437xy2+2153xyz-4835xz2+2447y3-1786y2z+12876yz2+12592z3 |
1 2
o25 : Matrix R <--- R |
i26 : GB = gens gb F
o26 = {0} | x2+2417xy+2168xz+13360y2-639yz+14449z2 xy2+198xyz-4686xz2+1906y3-585y2z+12211yz2+3780z3 xyz2-5641xz3+3811y4-12973y3z-2454y2z2+11675yz3+3823z4 xz4-3470y5-8776y4z-8700y3z2+1864y2z3-14123yz4-5748z5 y6-8183y5z+7264y4z2-9842y3z3-14142y2z4-12444yz5-15013z6 |
1 5
o26 : Matrix R <--- R |
i27 : LT = leadTerm gens gb F
o27 = {0} | x2 xy2 xyz2 xz4 y6 |
1 5
o27 : Matrix R <--- R |
i28 : betti LT
o28 = total: 1 5
0: 1 .
1: . 1
2: . 1
3: . 1
4: . 1
5: . 1
o28 : Net |
and there are Groebner basis elements of degrees
$2,3,4,5,6.$
\beginsection{ C. Division With Remainder}\par
A major application of Groebner bases is
to decide whether an element is in a given
ideal, and whether two elements reduce to
the same thing modulo an ideal. For
example, everyone knows that the trace
of a nilpotent matrix is 0. We can produce
an ideal $I$ that defines the variety $X$ of
nilpotent $3 \times 3$ matrices by taking the cube
of a generic matrix and setting the entries
equal to zero. Here's how:
i29 : R = KK[a..i]
o29 = R
o29 : PolynomialRing |
i30 : M = genericMatrix(R,a,3,3)
o30 = {0} | a d g |
{0} | b e h |
{0} | c f i |
3 3
o30 : Matrix R <--- R |
i31 : N = M^3
o31 = {0} | a3+2abd+bde+2acg+bfg+cdh+cgi a2d+bd2+ade+de2+cdg+afg+efg+dfh+fgi a2g+bdg+cg2+adh+deh+fgh+agi+dhi+gi2 |
{0} | a2b+b2d+abe+be2+bcg+ach+ceh+bfh+chi abd+2bde+e3+bfg+cdh+2efh+fhi abg+beg+bdh+e2h+cgh+fh2+bgi+ehi+hi2 |
{0} | a2c+bcd+abf+bef+c2g+cfh+aci+bfi+ci2 acd+cde+bdf+e2f+cfg+f2h+cdi+efi+fi2 acg+bfg+cdh+efh+2cgi+2fhi+i3 |
3 3
o31 : Matrix R <--- R |
i32 : I = flatten N
o32 = {0} | a3+2abd+bde+2acg+bfg+cdh+cgi a2b+b2d+abe+be2+bcg+ach+ceh+bfh+chi a2c+bcd+abf+bef+c2g+cfh+aci+bfi+ci2 a2d+bd2+ade+de2+cdg+afg+efg+dfh+fgi abd+2bde+e3+bfg+cdh+2efh+fhi acd+cde+bdf+e2f+cfg+f2h+cdi+efi+fi2 a2g+bdg+cg2+adh+deh+fgh+agi+dhi+gi2 abg+beg+bdh+e2h+cgh+fh2+bgi+ehi+hi2 acg+bfg+cdh+efh+2cgi+2fhi+i3 |
1 9
o32 : Matrix R <--- R |
(actually this produces a 1 x 9 matrix of
of forms, not the ideal: {\tt J = ideal I};
the matrix will be more useful to us).
But the trace is not in $I$! This is obvious
from the fact that the trace has degree $1$,
but the polynomials in $I$ are of degree $3$.
We could also check by division with
remainder:
i33 : Tr = trace M
o33 = a + e + i
o33 : R |
i34 : Tr //I -- the quotient, which is 0
o34 = 0
9 1
o34 : Matrix R <--- R |
i35 : Tr % I -- the remainder, which is Tr again
o35 = a + e + i
o35 : R |
(Here {\tt Tr} is an element of $R$, not a matrix.
We could do the same thing with a $1 \times 1$ matrix
with {\tt Tr} as its element.)
This is of course because the entries of $I$ do
NOT
generate the ideal of all forms
vanishing on $X$ -- this we may find with
{\tt J = radical ideal I},
(but this takes a while: see the documentation for
{\tt radical} on a faster way to find this)
which shows that the radical is generated by
the trace, the determinant, and the sum of
the principal $2 \times 2$ minors, that is, by the
coefficients of the characteristic polynomial.
In particular, we can try the powers of the
radical:
i36 : Tr^2 % I
2 2 2
o36 = a + 2a*e + e + 2a*i + 2e*i + i
o36 : R |
i37 : Tr^3 % I
2 2 3 2 2 2 2 3
o37 = 3a e + 3b*d*e + 3a*e + 3e + 3b*f*g + 3c*d*h + 6e*f*h + 3a i + 6a*e*i + 3e i + 3c*g*i + 6f*h*i + 3a*i + 3e*i + 3i
o37 : R |
i38 : Tr^4 % I
2 2 2 3 4 2 2 2 2 2 3 2 2 2 2 2 2 2 3 3 4
o38 = 6a e + 6b*d*e + 6a*e + 6e + 6b*e*f*g + 6c*d*e*h - 6b*d*f*h + 6a*e*f*h + 12e f*h - 6c*f*g*h - 6f h + 12a e*i + 12b*d*e*i + 12a*e i + 12e i + 12c*e*g*i + 6b*f*g*i + 6c*d*h*i + 6a*f*h*i + 36e*f*h*i + 6a i + 12a*e*i + 6e i + 6c*g*i + 12f*h*i + 6a*i + 12e*i + 6i
o38 : R |
i39 : Tr^5 % I
2 2 2 3 4 2 2 2 2 2 2 2 2 2 2 3 2 2 2 2 2 3 3 3 2 3 3 3 4 4 5
o39 = 30a e i + 30b*d*e i + 30a*e i + 30e i + 30c*e g*i + 30a f*h*i + 30b*d*f*h*i + 60a*e*f*h*i + 90e f*h*i + 30c*f*g*h*i + 30f h i + 30a e*i + 30b*d*e*i + 30a*e i + 30e i + 30c*e*g*i + 60a*f*h*i + 120e*f*h*i + 30a i + 30b*d*i + 30a*e*i + 30e i + 30c*g*i + 90f*h*i + 30a*i + 30e*i + 30i
o39 : R |
i40 : Tr^6 % I
2 2 2 2 2 3 2 4 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 2 3 3 3 3 3 3 2 4 4 4 2 4 4 4 5 5 6
o40 = 90a e i + 90b*d*e i + 90a*e i + 90e i + 90c*e g*i + 90a f*h*i + 90b*d*f*h*i + 180a*e*f*h*i + 270e f*h*i + 90c*f*g*h*i + 90f h i + 90a e*i + 90b*d*e*i + 90a*e i + 90e i + 90c*e*g*i + 180a*f*h*i + 360e*f*h*i + 90a i + 90b*d*i + 90a*e*i + 90e i + 90c*g*i + 270f*h*i + 90a*i + 90e*i + 90i
o40 : R |
i41 : Tr^7 % I
o41 = 0
o41 : R |
The seventh power is the first one in the
ideal! (Bernard Mourrain has worked out a
formula for which power in general.)
In this case
i42 : Tr^6 // I
o42 = {3} | a3+6a2e+3bde+15ae2+22e3+3bfg+3cdh-12efh+6a2i+30aei+60e2i+3cgi-30fhi+15ai2+60ei2+4i3 |
{3} | 18de2+18efg+18dei+18fgi |
{3} | 18deh+18egi+18dhi+18gi2 |
{3} | -27abe-63be2-36ach+9ceh+18bfh-72abi-162bei+90chi |
{3} | -2a3+15a2e+21bde+6ae2+e3-6bfg-6cdh-36afh+15efh+60a2i+72bdi+30aei+6e2i+66cgi+141fhi-30ai2-75ei2-26i3 |
{3} | 63beg+36a2h+18bdh-18aeh+18e2h+9cgh+18fh2+117bgi-162ahi-27ehi+45hi2 |
{3} | -18a2c+18ace+18abf-18bef+36cfh+45aci+18cei-81bfi+153ci2 |
{3} | -18cde-36a2f-36bdf+72aef+9e2f-45cfg-18f2h-108cdi+162afi+9efi+54fi2 |
{3} | 16a3+18abd-30a2e-42bde-30ae2-44e3+18acg-18ceg+3bfg+3cdh+18afh-57efh-75a2i-90bdi-60aei-75e2i-87cgi-165fhi-84ai2-84ei2-89i3 |
9 1
o42 : Matrix R <--- R |
is not 0. It is a matrix that makes the
following true:
i43 : Tr^6 == I * (Tr^6 // I) + (Tr^6 % I)
o43 = true |
\beginsection{ D. Elimination Theory}\par
Consider the problem of projecting the
``twisted cubic'', a curve in $\P^3$ defined
by the three $2 \times 2$ minors of a certain
$2 \times 3$ matrix into the plane.
Such problems can be solved in a very
simple and direct way using Groebner bases.
The technique lends itself to many extensions,
and in its developed form can be used to find
the closure of the image of any map of
affine varieties.
In this section we shall first give a
simple direct treatment of the problem above,
and then show how to use Macaulay2's
general tool to solve the problem.
We first
clear the earlier meaning of {\tt x} to make it
into a subscripted variable
i44 : erase global x
o44 = x
o44 : Symbol |
and then set
i45 : R = KK[x_0..x_3]
o45 = R
o45 : PolynomialRing |
the homogeneous coordinate ring of $\P^3$
and
i46 : M = map(R^2, 3, (i,j)->x_(i+j))
o46 = {0} | x_0 x_1 x_2 |
{0} | x_1 x_2 x_3 |
2 3
o46 : Matrix R <--- R |
i47 : I = gens minors(2,M)
o47 = {0} | -x_1^2+x_0x_2 -x_1x_2+x_0x_3 -x_2^2+x_1x_3 |
1 3
o47 : Matrix R <--- R |
a matrix whose image is
the ideal of the twisted cubic.
As projection center we
take the point defined by
i48 : pideal = ideal(x_0+x_3, x_1, x_2)
o48 = ideal (x + x , x , x )
0 3 1 2
o48 : Ideal of R |
To find the image we must intersect the ideal
$I$ with the subring generated by the
generators of {\tt pideal}. We make a change of
variable so that these generators become
the last three variables in the ring; that
is, we write the ring as $KK[y_0..y_3]$
where
$$y_0 = x_0, y_1 = x_1, y_2 = x_2, y_3 = x_0+x_3$$
and thus
$x_3 = y_3-y_0$, etc.
We want the new ring to have an ``elimination
order'' for the first variable.
i49 : erase global y
o49 = y
o49 : Symbol |
i50 : S = KK[y_0..y_3,MonomialOrder=> Eliminate 1]
o50 = S
o50 : PolynomialRing |
Here is one way to make the substitution
i51 : I1 = substitute(I, matrix{{y_0,y_1,y_2,y_3-y_0}})
o51 = {0} | y_0y_2-y_1^2 -y_0^2+y_0y_3-y_1y_2 -y_0y_1-y_2^2+y_1y_3 |
1 3
o51 : Matrix S <--- S |
The elimination of 1 variable from the
matrix of Groebner basis elements proceeds
as follows:
i52 : J = selectInSubring(1,gens gb I1)
o52 = {0} | y_1^3+y_2^3-y_1y_2y_3 |
1 1
o52 : Matrix S <--- S |
and gives (a matrix with element)
the cubic equation of a rational
curve with one double point in the plane.
However, we are still in a ring with 4
variables, so if we really want a plane
curve (and not the cone over one) we must
move to yet another ring:
i53 : S1 = KK[y_1..y_3]
o53 = S1
o53 : PolynomialRing |
i54 : J1 = substitute(J, S1)
o54 = {0} | y_1^3+y_2^3-y_1y_2y_3 |
1 1
o54 : Matrix S1 <--- S1 |
This time we didn't have to give so much
detail to the {\tt substitute} command because of
the coincidence of the names of the variables.
Having shown the primitive method, we
now show a much more flexible and transparent
one: we set up a ring map from the polynomial
ring in $3$ variables (representing the plane)
to $R/I$, taking the variables $y$ to the three
linear forms that define the projection
center. Then we just take the kernel of
this map! (``Under the hood'',
Macaulay2 is doing a more refined version
of the same computation as before.)
Here is the ring map
i55 : Rbar = R/(ideal I)
o55 = Rbar
o55 : QuotientRing |
i56 : f = map(Rbar, S1,
matrix(Rbar,{{x_0+x_3, x_1,x_2}})
)
o56 = map(Rbar,S1,{x + x , x , x })
0 3 1 2
o56 : RingMap Rbar <--- S1 |
and the desired ideal
i57 : J1 = ker f
3 3
o57 = ideal(y - y y y + y )
2 1 2 3 3
o57 : Ideal of S1 |
\beginsection{ E. Quotients and saturation}\par
Another typical application of
Groebner bases and syzygies is to the
computation of ideal quotients and
saturations. Again we give an easy example
that we can treat directly, and then
introduce the tool used in Macaulay2 to
treat the general case.
If $I$ and $J$ are ideals in a ring $R$, we define
$(I:J)$, the ideal quotient, by
$$(I:J) = \{f \in R \mid fJ \subset I\}.$$
In our first examples we consider
the case where $J$ is
generated by a single element $g$.
This arises in practice, for example, in the
problem of homogenizing an ideal. Suppose
we consider the affine space curve
parametrized by
$t \mapsto (t,t^2,t^3)$. The ideal of polynomials
vanishing on the curve is easily seen to
be $(b-a^2, c-a^3)$ (where we have taken
$a,b,c$ as the coordinates of affine space).
To find the projective closure of the curve
in $\P^3$, we must homogenize these equations
with respect to a new variable d, getting
$dc-a^2, d^2c-a^3$. But these forms do NOT
define the projective closure! In general,
homogenizing the generators of the ideal $I$ of
an affine variety one gets an ideal $I_1$ that
defines the projective closure UP TO
a component supported on the hyperplane
at infinity (the hyperplane $d=0$). To see
the ideal of the closure we must remove
any such components, for example by
replacing $I_1$ by the union $I_2$ of all the
ideals $(I_1:d^n)$, where $n$ ranges over positive
integers. This is not so hard as it seems:
First of all, we can successively compute
the increasing sequence of ideals
$(I_1:d), (I_1:d^2), \ldots $ until we get two
that are the same; all succeeding ones
will be equal, so we have found the union.
A second method involves a special property
of the reverse lex order, and is much more
efficient in this case. We shall illustrate
both. First we set up the example above:
i58 : R = KK[a,b,c,d]
o58 = R
o58 : PolynomialRing |
i59 : I1 = ideal(d*c-a^2, d^2*c-a^3)
2 3 2
o59 = ideal (- a + c*d, - a + c*d )
o59 : Ideal of R |
How to compute the ideal quotient:
If $I$ is generated by $f_1,\ldots,f_n$, we see that
$s \in (I:g)$ iff there are ring elements
$r_i$ such that
$$\sum_{i=1}^{n} r_i f_i + s g = 0. $$
Thus it suffices to compute the kernel
(syzygies) of the $1 \times (n+1)$ matrix
$$(f_1, ... ,f_n, g)$$
and collect the coefficients of $g$, that is,
the entries of the last row of a matrix
representing the kernel.
Thus in our case we may compute $(I_1:d)$
by concatenating the matrix for $I_1$
with the single variable $d$
i60 : I1aug = (gens I1) | matrix{{d}}
o60 = {0} | -a2+cd -a3+cd2 d |
1 3
o60 : Matrix R <--- R |
i61 : augrelations = gens ker I1aug
o61 = {2} | -a d |
{3} | 1 0 |
{1} | ac-cd a2-cd |
3 2
o61 : Matrix R <--- R |
There are 3 rows (numbered 0,1,2 !) and
2 columns, so to extract the last row we
may do
i62 : I21 = submatrix(augrelations, {2}, {0,1})
o62 = {1} | ac-cd a2-cd |
1 2
o62 : Matrix R <--- R |
But this is not an ``ideal'' properly speaking:
first of all, it is a matrix, not an ideal,
and second of all its target is not $R^1$
but $R(-1)$, the free module of rank 1 with
generator in degree 1. We can fix both
of these problems by
i63 : I21 = ideal I21
2
o63 = ideal (a*c - c*d, a - c*d)
o63 : Ideal of R |
This is larger than the original ideal, having
two quadratic generators instead of a
quadric and a cubic, so
we continue. Instead of doing the same
computation again, we introduce the built-in
command
i64 : I22 = I21 : d
2 2
o64 = ideal (c - c*d, a*c - c*d, a - c*d)
o64 : Ideal of R |
which is again larger than {\tt I21}, having
three quadratic generators. Repeating,
i65 : I23 = I22 : d
2 2
o65 = ideal (c - c*d, a*c - c*d, a - c*d)
o65 : Ideal of R |
we get another ideal with three quadratic
generators. It must be the same as {\tt I21},
but the generators are written differently
because of the route taken to get it, so
(being suspicious) we might check with
i66 : (gens I23) % (gens I22)
o66 = 0
1 3
o66 : Matrix R <--- R |
which returns 0, showing that {\tt I23} is
contained in (gives remainder 0 when divided
by) {\tt I22}. Thus the homogeneous ideal {\tt I2} of
the projective closure is equal to {\tt I23}
(this is the homogeneous ideal of
the twisted cubic, already encountered above).
A more perspicuous way of approaching the
computation of the union of the $(I:d^n)$,
which is called the saturation of $I$ with
respect to $d$, and written $(I:d^\infty)$,
is first to compute a reverse lex Groebner
basis.
i67 : gens gb I1
o67 = {0} | a2-cd acd-cd2 c2d2-cd3 |
1 3
o67 : Matrix R <--- R |
This yields {\tt (a2-cd, acd-cd2, c2d2-cd3)},
meaning
$$(a^2-cd, acd-cd^2, c^2d^2-cd^3).$$
We see that the second generator is divisible
by $d$, and the third is divisible by $d^2$.
General theory says that we get the right
answer simply by making these divisions,
that is, the saturation is
$$(a^2-cd, ac-cd, c^2-cd),$$
as previously computed. The same thing
can be accomplished in one line by
i68 : I2 = divideByVariable(gens gb I1,d)
o68 = {0} | a2-cd ac-cd c2-cd |
1 3
o68 : Matrix R <--- R |
This saturation may be found directly in Macaulay2:
i69 : saturate(I1, d)
2 2
o69 = ideal (c - c*d, a*c - c*d, a - c*d)
o69 : Ideal of R |



