Krull Dimension

58 views
Skip to first unread message

Loreno Heer

unread,
Feb 5, 2013, 6:25:21 PM2/5/13
to sage-a...@googlegroups.com
I tried the following:

sage: R.<x,y> = CC[]
sage: R
Multivariate Polynomial Ring in x, y over Complex Field with 53 bits of precision
sage: I = (x^2+y,x^3*y^2)*R
sage: S= R.quo(I)
sage: S
Quotient of Multivariate Polynomial Ring in x, y over Complex Field with 53 bits of precision by the ideal (x^2 + y, x^3*y^2)
sage: S.krull_dimension()
---------------------------------------------------------------------------
NotImplementedError Traceback (most recent call last)

/root/sage-5.6-linux-32bit-ubuntu_12.04.1_lts-i686-Linux/<ipython console> in <module>()

/root/sage-5.6-linux-32bit-ubuntu_12.04.1_lts-i686-Linux/local/lib/python2.7/site-packages/sage/rings/ring.so in sage.rings.ring.CommutativeRing.krull_dimension (sage/rings/ring.c:11127)()

NotImplementedError:


Any chance this will be implemented? Is this functionality available in Singular?

David Kohel

unread,
Feb 6, 2013, 8:37:41 AM2/6/13
to sage-a...@googlegroups.com
Hi,

It seems that I.dimension() returns the Krull dimension, but nobody
has tied this to the quotient R/I, even though someone defined a
default function.

Personally I find dimension ambiguous, since it can refer to the
Krull dimension or dimension (or rank) of R/I as a k-algebra,
which is finite for Krull dimension 0.

Cheers,

David

Volker Braun

unread,
Feb 6, 2013, 12:07:27 PM2/6/13
to sage-a...@googlegroups.com, David...@univ-amu.fr
On Wednesday, February 6, 2013 1:37:41 PM UTC, David Kohel wrote:
Personally  I find dimension ambiguous, since it can refer to the
Krull dimension or dimension (or rank) of R/I as a k-algebra,
which is finite for Krull dimension 0. 

The latter is .vector_space_dimension()
 
Reply all
Reply to author
Forward
0 new messages