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Thread: Quotient ring structure

  1. #1
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    Quotient ring structure

    Problem: Describe the ring structure of F_2[x]/(x^4 + x^2 + 1).

    Quotient rings are very difficult for me. What are the elements of this quotient ring? Is this polynomial irreducible? How do I determine this? Do I try to divide out by the degree 1 polynomials in this quotient ring? If the poly is irreducible, then this is a field, right? How do the elements in this ring differ from F_2[x]/(x^4 + x + 1)?

    Sorry for being so stupid! :<
    Sam
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  2. #2
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    Hi

    $\displaystyle X^4+X^2+1$ has no root in $\displaystyle \mathbb{F}_2,$ so it can't be written as a product of irreducible polynomials of degree 1 and 3. The other and last possibility for it to be reducible is to be the product of two irreducible polynomials of degree 2. The only irreducible polynomial of degree 2 in $\displaystyle \mathbb{F}_2$ is $\displaystyle X^2+X+1,$ (exercice), and we have $\displaystyle X^4+X^2+1=(X^2+X+1)^2.$
    Therefore $\displaystyle X^4+X^2+1$ is reducible, and:

    ($\displaystyle \mathbb{F}_2[X]/(X^4+X^2+1)\cong \mathbb{F}_2[X]/(X^2+X+1)\times \mathbb{F}_2[X]/(X^2+X+1)$)(wrong) which is not a field, since it's not an integral domain.

    With the same idea of proof, you can show that $\displaystyle X^4+X+1$ is irreducible in $\displaystyle \mathbb{F}_2[X],$ and then, since this ring is a principal ideal domain, $\displaystyle \mathbb{F}_2[X]/(X^4+X^2+1)$ is a field.


    $\displaystyle \mathbb{F}_2[X]/(X^2+X+1)$ and $\displaystyle \mathbb{F}_2[X]/(X^4+X^2+1)$ are fields, and $\displaystyle [\mathbb{F}_2[X]/(X^2+X+1):\mathbb{F}_2]=2,\ [\mathbb{F}_2[X]/(X^4+X^2+1):\mathbb{F}_2]=4$

    By unicity up to isomorphism for finite fields, $\displaystyle \mathbb{F}_2[X]/(X^2+X+1)\cong F_4$ and $\displaystyle \mathbb{F}_2[X]/(X^4+X^2+1)\cong F_{16}$

    In conclusion, ($\displaystyle \mathbb{F}_2[X]/(X^4+X^2+1)\cong \mathbb{F}_4\times \mathbb{F}_4$)(wrong) while $\displaystyle \mathbb{F}_2[X]/(X^4+X^2+1)\cong F_{16}$
    Last edited by clic-clac; Feb 23rd 2009 at 04:48 AM.
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  3. #3
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    Thank you so much again clic-clac! This is by far the most helpful forum I've found; I'm donating to it right away.
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  4. #4
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    I'm sorry I can't write that like that. When $\displaystyle K$ is a field and $\displaystyle P,Q$ are two polynomials in $\displaystyle K[X],$ if $\displaystyle P$ and $\displaystyle Q$ are relatively prime, then $\displaystyle K[X]/(PQ)\cong K[X]/(P)\times K[X]/(Q).$

    But of course, $\displaystyle X^2+X+1$ and itself are not relatively prime polynomials.

    What remains true is that $\displaystyle \mathbb{F}_2[X]/(X^4+X^2+1)$ is not an integral domain: $\displaystyle a=X^2+X+1$ is such that $\displaystyle a^2=0$ but $\displaystyle a\neq 0.$ (indeed $\displaystyle X^2+X+1\notin (X^4+X^2+1)$ ).


    One more thing: to see that the "isomorphism" quoted is wrong, one may note that the first ring has a non-zero nilpotent element, $\displaystyle a$, while the second, as a product of fields, has no non-zero nilpotent element.
    Last edited by clic-clac; Feb 23rd 2009 at 04:51 AM.
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