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How to set ring for this case of reduction to zero dimension
https://www.singular.uni-kl.de/forum/viewtopic.php?f=10&t=1854		 Page 1 of 1
Author:  Guest [ Sat Sep 04, 2010 10:54 am ]
Post subject:  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;
Author:  Guest [ Sat Sep 04, 2010 6:20 pm ]
Post subject:  Re: How to set ring for this case of reduction to zero dimension
Moreover,

if the case is ring x,y,z
how to set ring if indepset 1,0,1 and 1,0,0
Author:  steenpass [ Tue Sep 07, 2010 12:57 pm ]
Post subject:  Re: How to set ring for this case of reduction to zero dimension
Guest wrote:
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?

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

Guest wrote:
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;


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
Author:  Guest [ Tue Sep 07, 2010 2:02 pm ]
Post subject:  Re: How to set ring for this case of reduction to zero dimension
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?
Author:  steenpass [ Wed Sep 08, 2010 2:58 pm ]
Post subject:  Re: How to set ring for this case of reduction to zero dimension
Guest wrote:
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

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.

Guest wrote:
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

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

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

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
Author:  Guest [ Wed Sep 08, 2010 4:06 pm ]
Post subject:  Re: How to set ring for this case of reduction to zero dimension
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)
Author:  steenpass [ Thu Sep 09, 2010 7:18 pm ]
Post subject:  Re: How to set ring for this case of reduction to zero dimension
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
Author:  Guest [ Sat Sep 11, 2010 12:34 pm ]
Post subject:  Re: How to set ring for this case of reduction to zero dimension
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));
Author:  Guest [ Sat Sep 11, 2010 12:48 pm ]
Post subject:  Re: How to set ring for this case of reduction to zero dimension
Forget to ask why ordering is

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

How to notice this?
Author:  Guest [ Sun Sep 12, 2010 4:48 am ]
Post subject:  Re: How to set ring for this case of reduction to zero dimension
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?
Author:  Guest [ Sat Sep 18, 2010 4:06 pm ]
Post subject:  Re: How to set ring for this case of reduction to zero dimension
Hi Please tell me whether my workout is correct or not and whether my understanding is correct first.
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