How to set ring for this case of reduction to zero dimension

a1 and a2 are random guess
Would like to reduce to zero dimensional case.
indepSet(G);
// 1, 0
is 1 corresponding to x? 0 corresponding to y?

why ring S=(x,0),(y), lp; get error?
How to set ring for this case?

option (redSB);
ring R=0, (x,y), lp;
ideal a1 = 20x3+30x2+10x+1;
ideal a2 = y2-2y+1;

ideal I = intersect(a1, a2);
ideal G = std(I);
indepSet(G);
// 1, 0
ring S=(x,0),(y), lp;

Moreover,

if the case is ring x,y,z
how to set ring if indepset 1,0,1 and 1,0,0

The Singular manual answers your question:
indepSet ( ideal_expression ) computes a maximal set U of independent variables (in the sense defined in the note below) of the ideal given by a standard basis. If v is the result then v[i] is 1 if and only if the i-th variable of the ring, x(i), is an independent variable. Hence, the set U consisting of all variables x(i) with v[i]=1 is a maximal independent set.
See http://www.singular.uni-kl.de/Manual/la ... ng_218.htm



Which ring do you want to work in? "ring S=(x,0),(y), lp;" is definitely wrong, see http://www.singular.uni-kl.de/Manual/la ... .htm#SEC39 for examples of valid ring declarations. If you want to "turn off" the variable x, try

ring S = 0,y,lp;

Then use the commands "fetch" or "imap" to pass your objects to the new ring.

Regards,
Andreas

I follow the example of reducing high dimension ideals to zero dimensional case in Book "A Singular Introduction to Commutative Algebra"
page 275

indepset(G) -> 0, 1 then the example
set ring S = (0,y), (x), lp

It seems that 0 means turn off x, then set ring S = (0,y), (x), lp

then i try other ideals,
if the case is indepset -> 1, 0, i think it tell me to turn off y
it should be S= (x,0), (y), lp

However, it is wrong,

I have tried ring S = x, 0, lp, it get error again.
So, how to set ring this case 1,0

if meet another case such as 1,0,1 and 1,0,0
how to set ring to reduce to zero dimensional case?

No, "0, 1" means that y is an independet variable while x is not. In order to understand the ring declaration, please have a look at http://www.singular.uni-kl.de/Manual/la ... htm#SEC192. "ring S = (0,y), (x), lp;" declares the ring K(y)[x] as stated in Greuel/Pfister.


The right form is "ring S = (0,x),(y),lp;".


Assuming that you you set "ring R = 0,(x,y,z),lp;" previously, this is then "ring S = (0,x,z),(y),lp;" and "ring S = (0,x),(y,z),lp;", respectively.

Regards,
Andreas

Thanks. How about the next step of the following.

ring S = (0,x),(y),lp;
ideal G = imap(R,G);
G;

G[1]=(20x3+30x2+10x+1)*y2+(-40x3-60x2-20x-2)*y+(20x3+30x2+10x+1)

setring R;
poly h1=20x3+30x2+10x+1; // LCM of leading coefficient

ideal I2 = std(I + ideal(h1));
indepSet(I2);

which is not in zero dimensional case, How to do?
actually i do not understand why std(I + ideal(h1)

Not all of the steps of the example in Greuel/Pfister can be taken as they are to compute your example. The first step is to compute a primary decomposition of quotient(I,h), but in Greuel/Pfister, there's nothing to compute as quotient(I,h) = x. So they continue straightforward with the computation of DECOMP(<I,h>), cf. the fourth point in Algorithm 4.3.4. This is where the line

ideal I2 = std(I+ideal(h));

comes from. But in your case, quotient(I,h) = y2-2y+1 and the decomposition is not just the quotient itself, so to follow the algorithm, you have to do

setring S;

and to compute ZeroDecomp(<G>).

Another remark: Be careful about the ordering, cf. Algorithm 4.3.2. In your case, you have to change to the ring RR = 0,(y,x),lp;

Regards,
Andreas

It seems that successfully, let me share my successful code

option (redSB);
ring R=0, (x,y), lp;
ideal a1 = 20x3+30x2+10x+1;
ideal a2 = y2-2y+1;

ideal I = intersect(a1, a2);
ideal G = std(I);
indepSet(G);

ring S = (0,x),(y),lp;
ideal G = imap(R,G);
G;

G[1]=(20x3+30x2+10x+1)*y2+(-40x3-60x2-20x-2)*y+(20x3+30x2+10x+1)

setring R;
poly h1=20x3+30x2+10x+1;

ideal I1 = quotient(I,h1);

ring RR = 0,(y,x),lp;

map phi=RR, x, x+y;
map psi=RR, x, -x+y;
ideal I1 = y2-2y+1;
I1 = std(phi(I1));

factorize(I1[1]);

[1]:
_[1] = 1
_[2] = x-1
[2]:
1,2

I1 = std(psi(I1));
I1;
y2-2yx-2y+x2+2x+1

factorize(I1[1]);

associated prime ideal is -y+x+1

ideal h1 = -y+x+1;
ideal I2 = std(I + ideal(h1));
indepSet(I2);

0,0

list fac = factorize(I2[1]);
fac;

[1]:
_[1]=1
_[2]=y-1
_[3]=20y3-30y2+10y+1
[2]:
1,2,1

ideal J1 = std(I2, (y-1)^2);

J1[1]=y2-2y+1
J1[2]=x-y+1

setring R;
map phi=R, x, x+y;
map psi=R, x, -x+y;

ideal K1 = std(phi(J1));
factorize(K1[1]);

[1]:
_[1]=1
_[2]=y-1
[2]:
1,1

K1=std(psi(K1));
K1;

K1[1]=y2-2y+1
K1[2]=x-y+1

associated prime ideals are y-1 and x-y+1

is it correct?

1. How to know whether it is in general position? I observe that general position is see factorize result of std(phi(I)) and i.e. std(I, (y+1)^3), but do not know how to know whether it is in general position from factorize result

for example std(I, (y+1)^3) which y+1 get from factorize result of std(phi())
it seems that we have assumed that std(I, (y+1)^3) must be in general position

2. map phi=R, x, x+y; and map psi=R, x, -x+y; why always use these map?
is it that when not in general position, we use phi, for example, ideal K1=std(phi(J1))?

when in general position, we use psi to get the final result? for example, Q1=std(psi(Q1));

Forget to ask why ordering is

ring RR = 0,(y,x),lp;

How to notice this?

Moreover, from the examples, there are ideal a1, a2, a3

what condition make group a1, a2, a3? a1, a2, a3 are different input, or different data type?

Any application examples?

Hi Please tell me whether my workout is correct or not and whether my understanding is correct first.