How to Determine Irreducible Polynomials in (Z/2Z)[x]?

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Homework Help Overview

The discussion revolves around identifying monic polynomials of degrees 2 and 3 in the polynomial ring (Z/2Z)[x] and determining their irreducibility.

Discussion Character

  • Exploratory, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the nature of monic polynomials and their roots in Z/2Z. There is an attempt to clarify how to express polynomials and their factors. Some participants list potential polynomials of degrees 2 and 3, while others reflect on the combinations of terms involved.

Discussion Status

Participants are actively sharing polynomial examples and exploring the structure of these polynomials. There is a sense of progress as one participant expresses confidence in continuing the work, although no consensus on irreducibility has been reached yet.

Contextual Notes

One participant expresses uncertainty about the requirements of the task and how to approach writing polynomials via modulus. There is also a mention of a preference for hints rather than complete answers.

silvermane
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Homework Statement


Find all monic polynomials of degrees 2 and 3 in (Z/2Z)[x]. Determine which ones are irreducible, and write the others as products of irreducible factors.



The Attempt at a Solution


I know that factors of degree 1 correspond to roots in Z/2Z and that monic polynomials are polynomials where the top term coefficient is equal to 1. I think I'm not understanding what I'm supposed to do, or how to write them via modulus.

Any hints or tips are greatly appreciated! :)
Please no answers!
 
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Here are the polynomials of dregree 2:


x²+1
x²+x
x²+x+1

Can you now find all the polynomials of degree 3? There are 8 of them.
 
So It's pretty much just combinations of functions with an X^2 for degree 2, and other terms such as x and 1 for the lesser degrees.

Furthermore, for x^3, we would have this:

x^3
X^3 +x^2
X^3 +x^2+x
X^3 +x^2+x+1
X^3 +x+1
X^3 +x^2+1
X^3 +1
X^3 +x

Thanks for your help!
I think I can handle it from here :)
 
Yes!

Good luck!
 

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