Opened 9 years ago

Closed 9 years ago

## #558 closed bug (not a bug)

# bug in kbase

Reported by: | Owned by: | somebody | |
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Priority: | minor | Milestone: | 3-2-0 and higher |

Component: | dontKnow | Version: | 3-1-6 |

Keywords: | Cc: |

### Description

> 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

### Change History (4)

### comment:1 follow-up: 2 Changed 9 years ago by

### comment:2 Changed 9 years ago by

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

### comment:3 Changed 9 years ago by

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.

### comment:4 Changed 9 years ago by

Resolution: | → not a bug |
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Status: | new → closed |

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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,...