bug in kbase

[dkifle16@…]

> ring r=0,(a,b),ds;
// ** redefining r **
> ideal I=a2+b;
> kbase(std(I));  //should be 1,a
_[1]=0
> kbase(std(I),1);
_[1]=a
> kbase(std(I),0);
_[1]=1

[ren]

I think your quotient is not finite dimension as a vector space over the ground field, which is a necessary requirement. A basis for your quotient would be 1,a,a2,a3,...

[Oleksandr]

Replying to ren:

I think your quotient is not finite dimension as a vector space over the ground field, which is a necessary requirement. A basis for your quotient would be 1,a,a2,a3,...

I support that. output seems correct to me

[anne@…]

Ren and Olexander are correct.

Moreover, it is the usual output of kbase to return 0, if the quotient is not a finite dimensional vector space over the ground field, even though this is the same output which is used for R/I=0.

The behaviour of kbase for 'unusual' input should be more specific (i.e. make distinction between 'R/I not zerodimensional' and 'R/I=0') and better documented. The current situation easily leads to confusion for users.