Back to Forum | View unanswered posts | View active topics





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 - unspecified minpoly
Author Message
pankajsejwal
  Post subject:  Re: unspecified minpoly  Reply with quote
ring r1 = (2^3,a),x,dp;
is generated by a3+a+1 working as maximal ideal and it is a known fact and also stored in Singular CAS,
https://github.com/Singular/Sources/tre ... y/gftables
Post Posted: Mon May 30, 2016 8:28 am
shengyushen
  Post subject:  unspecified minpoly  Reply with quote
Dear all:

I am learning using singular , and I run the following script in singular
> ring A1=(2^3,a),x,dp;
> minpoly ;
1*a^3+1*a^1+1*a^0
> ring A2=(2,a),x,dp;
// ** redefining A2 **
> minpoly ;
0


I am confused by it,
I have not specified a monopoly for both A1 and A2 rings, but the A1 have a monopoly already, but not the A2.

Does this means that the A1 have only 1 monopoly so we don't need to specify it?

Shengyu Shen
Post Posted: Fri Feb 06, 2015 11:11 am


It is currently Fri May 13, 2022 10:56 am
cron
Powered by phpBB © 2000, 2002, 2005, 2007 phpBB Group