Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people, just like you, are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
Solved

Irreducible polynomials

Posted on 2004-08-25
3
491 Views
Last Modified: 2008-02-01
May I ask what are Irreducible polynomials?
Examples with workings will be good.

hongjun
0
Comment
Question by:hongjun
3 Comments
 
LVL 33

Assisted Solution

by:snoyes_jw
snoyes_jw earned 100 total points
ID: 11894451
Those polynomials that cannot be expressed as a product of non-trivial factors.  For example, x²-2 is irreducible over the set of rational numbers, because there are no rational numbers A and B such that x²-2 = (x+A)(x+B).

Straight from Google:

http://mathworld.wolfram.com/IrreduciblePolynomial.html
http://www.math.niu.edu/~beachy/aaol/polynomials.html
http://en.wikipedia.org/wiki/Irreducible_polynomial
0
 
LVL 84

Assisted Solution

by:ozo
ozo earned 150 total points
ID: 11894506
A polynomial f(x) is irreducible over <R> if f(x) cannot be factoored as a product of polynomials in <R>[x] of degree less than the degree of f(x)
for example, the polynomial x²+1 is
irreducible in the Reals, because x²+1 has no Real root
reducible in the Complex field because x²+1 = (x-i)(x+i)
reducible in Z2 because x²+1 = (x+1)²
reducuble in Z5 because x²+1 = (x + 3)(x+2)
0
 
LVL 4

Accepted Solution

by:
n_fortynine earned 250 total points
ID: 11918917
There is also a theorem that states that an irreducible polynimal p(x) in F[x] (where F is a field) will be reducible, i.e. have a root, in the extension field F[x]/p(x) (i.e. the field that contains all the remainders of a division of any polynomial in F[x] by p(x)). This theorem might not hold if F isn't a field.

A quick trick to recognize irreducibles of 2nd and 3rd degrees in F[x] is when they have no roots in F (F denotes a field).

For example: x^4 + x + 1 is irreducible in Z2[x], but has the root [x] in Z2[x]/(x^4 + x + 1) because [x]^4 + [x] + 1 = [x^4 + x + 1] = [0]

x^2 + x + 1 is also irreducible in Z2[x] but has the root [x^2 + x] in Z2[x]/(x^4 + x + 1) because [x^2 + x]^2 + [x^2 + x] + 1 = [x^4 + x + 1] = [0]

If you're unfamilar with rings, Z2 is the ring containing two elements [0] and [1], etc.

Hope this helps.
0

Featured Post

Announcing the Most Valuable Experts of 2016

MVEs are more concerned with the satisfaction of those they help than with the considerable points they can earn. They are the types of people you feel privileged to call colleagues. Join us in honoring this amazing group of Experts.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

Title # Comments Views Activity
Whisker Plot 1 45
Error in calculation 2 79
Math question 3 97
Energy conservation - Edward Leedskalnin 20 113
How to Win a Jar of Candy Corn: A Scientific Approach! I love mathematics. If you love mathematics also, you may enjoy this tip on how to use math to win your own jar of candy corn and to impress your friends. As I said, I love math, but I gu…
Foreword (May 2015) This web page has appeared at Google.  It's definitely worth considering! https://www.google.com/about/careers/students/guide-to-technical-development.html How to Know You are Making a Difference at EE In August, 2013, one …
This is a video describing the growing solar energy use in Utah. This is a topic that greatly interests me and so I decided to produce a video about it.
Finds all prime numbers in a range requested and places them in a public primes() array. I've demostrated a template size of 30 (2 * 3 * 5) but larger templates can be built such 210  (2 * 3 * 5 * 7) or 2310  (2 * 3 * 5 * 7 * 11). The larger templa…

856 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question