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Math Help - 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

    X^4+X^2+1 has no root in \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 \mathbb{F}_2 is X^2+X+1, (exercice), and we have X^4+X^2+1=(X^2+X+1)^2.
    Therefore X^4+X^2+1 is reducible, and:

    ( \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 X^4+X+1 is irreducible in \mathbb{F}_2[X], and then, since this ring is a principal ideal domain, \mathbb{F}_2[X]/(X^4+X^2+1) is a field.


    \mathbb{F}_2[X]/(X^2+X+1) and \mathbb{F}_2[X]/(X^4+X^2+1) are fields, and [\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, \mathbb{F}_2[X]/(X^2+X+1)\cong F_4 and \mathbb{F}_2[X]/(X^4+X^2+1)\cong F_{16}

    In conclusion, ( \mathbb{F}_2[X]/(X^4+X^2+1)\cong \mathbb{F}_4\times \mathbb{F}_4)(wrong) while \mathbb{F}_2[X]/(X^4+X^2+1)\cong F_{16}
    Last edited by clic-clac; February 23rd 2009 at 05:48 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 K is a field and P,Q are two polynomials in K[X], if P and Q are relatively prime, then K[X]/(PQ)\cong K[X]/(P)\times K[X]/(Q).

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

    What remains true is that \mathbb{F}_2[X]/(X^4+X^2+1) is not an integral domain: a=X^2+X+1 is such that a^2=0 but a\neq 0. (indeed 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, a, while the second, as a product of fields, has no non-zero nilpotent element.
    Last edited by clic-clac; February 23rd 2009 at 05:51 AM.
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