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Math Help - Finite Field

  1. #1
    Junior Member classic_phohe's Avatar
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    Finite Field

    An irreducible polynomial P(x) of degree m over GF(q) can be used to construct an extension field of GF(q). What is actually meant by constructing an extension field?

    and why is it important to ensure whether a monic polynomial of degree m over GF(2^n) is primitive???

    lastly, there are 3 bases of finite field: polynomial, normal and dual base. in what condition each of the bases is favored over the others?

    thank you very much for your help
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  2. #2
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    Quote Originally Posted by classic_phohe View Post
    An irreducible polynomial P(x) of degree m over GF(q) can be used to construct an extension field of GF(q). What is actually meant by constructing an extension field?
    If p(x) is irreducible and \mathbb{F}_q = \text{GF}(q) then \mathbb{F}_q[x]/(p(x)) is a field. Furthermore, \mathbb{F}_q can be identified as a subfield of this field by an embedding a \mapsto a + (p(x)). And so \mathbb{F}_q[x]/(p(x)) can be regarded as an extension field of \mathbb{F}_q. In fact it is an extension of degree m, do you see why?

    And I am not sure about you other question, I do not know what you mean by 'primitive'?
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  3. #3
    Junior Member classic_phohe's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    If p(x) is irreducible and \mathbb{F}_q = \text{GF}(q) then \mathbb{F}_q[x]/(p(x)) is a field. Furthermore, \mathbb{F}_q can be identified as a subfield of this field by an embedding a \mapsto a + (p(x)). And so \mathbb{F}_q[x]/(p(x)) can be regarded as an extension field of \mathbb{F}_q. In fact it is an extension of degree m, do you see why?

    And I am not sure about you other question, I do not know what you mean by 'primitive'?
    I dun quite get it yet.

    Im not quite sure what is primitive but what I read is that a monic polynomial of degree m with maximum order is said to be primitive
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  4. #4
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    Quote Originally Posted by classic_phohe View Post
    I dun quite get it yet.
    What does not make sense? Have you ever talked about talking a ring modulo an ideal in class? This is precisely what is being down. The ideal here is (p(x)) (or in notation \left< p(x)\right> is you perfer) and the ring here is \mathbb{F}_q[x]. I am asking if you form this factor ring you get a field, and this field is an extension field of \mathbb{F}_q. By this method you can construct extension fields if you are given an irreducible polynomial.
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  5. #5
    Junior Member classic_phohe's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    What does not make sense? Have you ever talked about talking a ring modulo an ideal in class? This is precisely what is being down. The ideal here is (p(x)) (or in notation \left< p(x)\right> is you perfer) and the ring here is \mathbb{F}_q[x]. I am asking if you form this factor ring you get a field, and this field is an extension field of \mathbb{F}_q. By this method you can construct extension fields if you are given an irreducible polynomial.
    erm..nope..I didnt take this subject in class, Im reading this on my own interest. So I really found a lot of doubt in it. what about composite field? what is the purpose of having composite field?
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