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Math Help - Polynomial rings -- F2[x]

  1. #1
    Super Member Bernhard's Avatar
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    Polynomial rings -- F2[x]

    In Dummit and Foote section 7.4 Exercise 15 reads as follows:

    "Let  x^2 + x + 1 be an element of the polynomial ring  E = {\mathbb{F}_2 [x] ... ..."

    I am unsure of the meaning of the notation  {\mathbb{F}_2 - can someone please help?

    Peter
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Polynomial rings -- F2[x]

    Hi Bernhard.

    F_2 = \{0, 1\} is the field \mathbb Z/2 \mathbb Z.
    Calculations are modulo 2.
    Last edited by ILikeSerena; December 27th 2012 at 04:23 PM.
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  3. #3
    Super Member Bernhard's Avatar
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    Re: Polynomial rings -- F2[x]

    Thanks!

    So  {\mathbb{F}}_p would mean Z/<p> where p is prime and the notation would not be used when p is composite since then Z/<p> would not be a field

    Is that correct?

    Peter
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    Re: Polynomial rings -- F2[x]

    yes, it turns out that for finite fields, the possibilities are more severely restricted than for other structures: there are NO fields of order 6, for example.

    here is why: every field has a minimal subfield, called its prime field, which is the subfield generated by 1. if a field is finite, then the (additive) order of 1 is likewise finite.

    this number (called the characteristic of F....if 1 is of infinite order, we use 0 as the characteristic) must be a prime number for a finite field, because if mn*1 = 0, then (m*1)(n*1) = 0, so we have zero divisors.

    furthermore, any finite field K which contains a prime field isomorphic to Z/(p), is a vector space (of necessarily finite dimension n) over Z/(p), and hence has exactly pn elements, meaning finite fields must be of prime power order.

    moreover, every two such fields are isomorphic as fields, meaning a finite field is essentially (up to isomorphism) determined by its size. it is customary to refer to such a field as Fk, or sometimes as GF(k) (GF for "galois field", in honor of evariste galois).
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