Community

Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
Login Register Memberlist Search AoPS Blogs Contests Galleries Forum Index
The time now is Mon Nov 23, 2009 12:56 am
All times are UTC - 8
 		View posts since last visit
View unanswered posts
Home » Olympiad Section » Algebra » Algebra Theorems and Formulas
 
Field extension
Moderators: High School Olympiad Moderators, Arne, darij grinberg, harazi, mathmanman, Megus, N.T.TUAN, orl, pbornsztein
Post new topic   Reply to topic View previous topicView next topic
2 Posts • Page 1 of 1
Author Message
Los
Poincare Conjecture
Poincare Conjecture


Offline
Joined: 29 Dec 2007
Posts: 228
Finland
To rate posts you must be logged in
#1
Field extension
Hello,

if p(x)=x^3+2x+1 is a polynomial in \mathbb{Z}_3[x], is \mathbb{Z}_3[x]/p(x) a field?


Thanks in advance,


Los

PostPosted: Thu Aug 13, 2009 8:22 am  Back to top 
  ProfilePM
MeKnowsNothing
Riemann Hypothesis
Riemann Hypothesis

Online
Joined: 27 Feb 2007
Posts: 357
To rate posts you must be logged in
#2
Do you mean \mathbb{Z}_3[x]/\langle p(x)\rangle? This is a field as \langle x^3 + 2x + 1\rangle is a maximal ideal of \mathbb{Z}_3[x]. This follows from the fact that p(x) is irreducible in \mathbb{Z}_3[x].

PostPosted: Fri Aug 14, 2009 4:31 am  Back to top 
  ProfilePM
Display posts from previous:   Sort by:   
2 Posts • Page 1 of 1
Post new topic   Reply to topic View previous topicView next topic
Home » Olympiad Section » Algebra » Algebra Theorems and Formulas
 
Jump to:  


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum
You cannot attach files in this forum
You can download files in this forum
You cannot post calendar events in this forum


© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us