Post a reply
Username:
Note:If not registered, provide any username. For more comfort, register here.
Subject:
Message body:
Enter your message here, it may contain no more than 60000 characters. 

Smilies
:D :) :( :o :shock: :? 8) :lol: :x :P :oops: :cry: :evil: :twisted: :roll: :wink: :!: :?: :idea: :arrow: :| :mrgreen:
Font size:
Font colour
Options:
BBCode is ON
[img] is ON
[flash] is OFF
[url] is ON
Smilies are ON
Disable BBCode
Disable smilies
Do not automatically parse URLs
Confirmation of post
To prevent automated posts the board requires you to enter a confirmation code. The code is displayed in the image you should see below. If you are visually impaired or cannot otherwise read this code please contact the %sBoard Administrator%s.
Confirmation code:
Enter the code exactly as it appears. All letters are case insensitive, there is no zero.
   

Topic review - Groebner Basis and extending partial solutions
Author Message
  Post subject:  Groebner Basis and extending partial solutions  Reply with quote
Suppose that a finite set of polynomials in C[x,y,z] has a finite number of solutions (i.e. the generated ideal is 0-dimensional).
Suppose also that the Groebner basis with respect to lex order x>y>z is

[f(z), g(y,z), h(y,z), k(x,y,z)]

As well known, the system can be now easily solved: choose a root z0 of f, plug it into g and h and look for a common root (y0) etc.

The question is the following: Is it true that for EVERY root z0 of f there exist y0, z0 such that (x0,y0,z0) satisfy the system?

In all the examples I have seen this is true, but I don't know whether this is true in general or there is a counterexample.

Note that an extension from (y0,z0) to (x0,y0,z0) is not always possible (there is an "Extension theorem" which must be used).
The problem here is to extend from (z0) to (x0,y0,z0) which seems to be always possible. Is it?
Post Posted: Sat Mar 04, 2017 3:21 pm


It is currently Fri May 13, 2022 10:54 am
cron
Powered by phpBB © 2000, 2002, 2005, 2007 phpBB Group