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Re: primary decomposition in Q[i] |
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Dear Alberto Damiano, Quote: > I am interested in computing the Primary decomposition of ideals in C[x,y]. For example i tried the following ideal:
> I=(x^2+y^2)
> in the polinomial ring Q[i][x,y] where i^2+1=0
> the definition of the ring was as follows:
> ring r=(0,i)(x,y),dp;minpoly=i^2+1; here you've forgotten a comma, it should be rather Code: ring r=(0,i),(x,y),dp;
minpoly=i^2+1; Quote: > PrimdecGTZ returns incorrectly the ideal itself, as if it was primary, while all the other primdec command from the primdec.lib and mprimdec.lib return rerror messages. My suispect is that we cannot compute the primary decomposistion with an algebraic extension of Q as coefficient ring. Anyone has any idea?
> There should be an error in you Singular script, because the correct code Code: ring r=(0,i),(x,y),dp;
minpoly=i^2+1;
ideal I=x2+y2;
LIB "primdec.lib";
list L=primdecGTZ(I);
L;
gives always a correct answer (I've checked it extra with versions 2-0-4, 2-0-5 and even with the development version 2-1-2): Code: [1]:
[1]:
_[1]=x+(-i)*y
[2]:
_[1]=x+(-i)*y
[2]:
[1]:
_[1]=x+(i)*y
[2]:
_[1]=x+(i)*y Quote: > Thank you If the given code produces something different from the given answer, please let us know your Singular version and also the version of primdec.lib. Quote: > Alberto Damiano Best regards, Dear Alberto Damiano,
[quote]> I am interested in computing the Primary decomposition of ideals in C[x,y]. For example i tried the following ideal:
> I=(x^2+y^2)
> in the polinomial ring Q[i][x,y] where i^2+1=0
> the definition of the ring was as follows:
> ring r=(0,i)(x,y),dp;minpoly=i^2+1;[/quote]
here you've forgotten a comma, it should be rather
[code]ring r=(0,i),(x,y),dp;
minpoly=i^2+1;[/code]
[quote]> PrimdecGTZ returns incorrectly the ideal itself, as if it was primary, while all the other primdec command from the primdec.lib and mprimdec.lib return rerror messages. My suispect is that we cannot compute the primary decomposistion with an algebraic extension of Q as coefficient ring. Anyone has any idea?
>[/quote]
There should be an error in you Singular script, because the correct code
[code]
ring r=(0,i),(x,y),dp;
minpoly=i^2+1;
ideal I=x2+y2;
LIB "primdec.lib";
list L=primdecGTZ(I);
L;
[/code]
gives always a correct answer (I've checked it extra with versions 2-0-4, 2-0-5 and even with the development version 2-1-2):
[code][1]:
[1]:
_[1]=x+(-i)*y
[2]:
_[1]=x+(-i)*y
[2]:
[1]:
_[1]=x+(i)*y
[2]:
_[1]=x+(i)*y[/code]
[quote]> Thank you[/quote]
If the given code produces something different from the given answer, please let us know your Singular version and also the version of primdec.lib.
[quote]> Alberto Damiano[/quote]
Best regards,
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Posted: Thu Aug 11, 2005 8:39 pm |
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Post subject: |
primary decomposition in Q[i] |
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I am interested in computing the Primary decomposition of ideals in C[x,y]. For example i tried the following ideal: I=(x^2+y^2) in the polinomial ring Q[i][x,y] where i^2+1=0 the definition of the ring was as follows: ring r=(0,i)(x,y),dp;minpoly=i^2+1; PrimdecGTZ returns incorrectly the ideal itself, as if it was primary, while all the other primdec command from the primdec.lib and mprimdec.lib return rerror messages. My suispect is that we cannot compute the primary decomposistion with an algebraic extension of Q as coefficient ring. Anyone has any idea? Thank you Alberto Damiano email: adamiano@gmu.eduPosted in old Singular Forum on: 2004-05-05 17:17:36+02 I am interested in computing the Primary decomposition of ideals in C[x,y]. For example i tried the following ideal:
I=(x^2+y^2)
in the polinomial ring Q[i][x,y] where i^2+1=0
the definition of the ring was as follows:
ring r=(0,i)(x,y),dp;minpoly=i^2+1;
PrimdecGTZ returns incorrectly the ideal itself, as if it was primary, while all the other primdec command from the primdec.lib and mprimdec.lib return rerror messages. My suispect is that we cannot compute the primary decomposistion with an algebraic extension of Q as coefficient ring. Anyone has any idea?
Thank you
Alberto Damiano
email: adamiano@gmu.edu
Posted in old Singular Forum on: 2004-05-05 17:17:36+02
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Posted: Thu Aug 11, 2005 5:32 pm |
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