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Finite Field of Characterisitic p ... ...

  1. May 17, 2017 #1
    1. The problem statement, all variables and given/known data

    I need help with Exercise 1 of Dummit and Foote, Section 13.2 : Algebraic Extensions ..

    I have been unable to make a meaningful start on the problem ... ...

    Exercise 1 of Dummit and Foote, Section 13.2 reads as follows:


    ?temp_hash=be203a92aafcc0e065b898a92755429b.png


    2. Relevant equations

    A relevant definition is the definition of the characteristic of a field ...

    This definition along with some remarks by D&F reads as follows:



    ?temp_hash=be203a92aafcc0e065b898a92755429b.png


    A proposition which may be relevant is D&F Proposition 1 of Section 13.1 ... ...

    This proposition reads as follows:


    ?temp_hash=be203a92aafcc0e065b898a92755429b.png



    3. The attempt at a solution

    I have been unable to make a meaningful start on this problem ...

    BUT ... further I hope to understand why D&F put this exercise at the end of a section on algebraic extensions ... the exercise seems a bit remote from the subject matter ... ...

    Hope someone can help ...

    Peter
     

    Attached Files:

  2. jcsd
  3. May 17, 2017 #2

    andrewkirk

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    The problem is solved by considering the field as a vector space over the field ##\langle 1_F\rangle \cong \mathbb Z_p##. If the vector space has dimension ##n## then it has a basis of size ##n## and we can show that the order of the vector space is ##p^n## by considering the number of different ##n##-tuples whose components are in ##\mathbb Z_p##.

    I think the reason it comes at the end of the foregoing chapter is that that chapter (IIRC) has shown how a field can be regarded as a vector space.

    The vector space will be one-dimensional if the field is isomorphic to ##\mathbb Z_p##, and so will have order ##p## and ##\{1_F\}## is a basis. If it is not isomorphic, the field generated by ##1_F## will be a subfield (call it ##S##) of order ##p##. The other elements of the basis can be found by considering what elements are needed to extend ##S## to ##F##.
     
  4. May 17, 2017 #3


    Thanks Andrew ... I vaguely follow ...

    Are you able to formalise the proof a bit ...

    Peter
     
  5. May 17, 2017 #4

    andrewkirk

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    Use the distributive law to prove that the vector space over ##\mathbb Z_p## generated by ##1_F## is a subfield of ##F## of order ##p##, which we can label as ##S##.
    Hence ##F## is a field extension of ##S##.
    In the previous work (somewhere in D&F, else see the wiki article on Field Extensions) we've seen that a field extension is a vector space over the subfield (ie we mean the subfield ##S## is the scalar field from which scalar multiplication in the vector space ##F## is defined).
    Since ##F## is finite, it must be a finite-dimensional vector space and hence, from linear algebra, we know it must be isomorphic to ##S^n## for some finite ##n\in\mathbb N##.
    The canonical element of ##S^n## is an ordered ##n##-tuple of elements of ##S##. Since there are ##p## distinct elements of ##S##, that means there are ##p^n## distinct elements of ##S^n##.
    Hence, via the vector space isomorphism, we know that ##|F|=|S^n|=p^n##.

    You might find it helpful to play around with a couple of examples.
    (1) What are the elements of the field ##\mathbb Z_3(\sqrt 2)##, and what is the characteristic and order of this field?
    (2) What are the elements of the field ##\mathbb Z_3(\sqrt{-1})##, and what is the characteristic and order of this field?
     
    Last edited: May 17, 2017
  6. May 18, 2017 #5

    Thanks so much Andrew ... much clearer now ...

    BUT ... I'm struggling a bit when I come to the following step ... ... :

    " ... ... Since ##F## is finite, it must be a finite-dimensional vector space and hence, from linear algebra, we know it must be isomorphic to ##S^n## for some finite ##n\in\mathbb N##. ... ... "


    Can you give an indication of why this statement is true ... ... ?

    Can you indicate where it might be proved in full ...?

    Peter
     
  7. May 18, 2017 #6

    andrewkirk

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    There is an argument in this wiki section.
    The general theorem is that an ##n## dimensional vector space ##V## over field ##S## is isomorphic to the vector space that is the direct sum of ##n## copies of ##F##, aka ##F^n##.
    We can prove it by, given a basis ##b_1,...,b_n## for ##V##, defining a map from ##V## to ##S^n## that is the linear extension of the map that takes ##b_k## to the element of ##S^n## whose components are all ##0_S## except for a ##1_S## in the ##k##th position.

    Normally one would use ##F## for the field, but I've used ##S## to be consistent with the notation used above, since ##F## is already being used for something else (the vector space ##V##).

    It is straightforward, but instructive, to show that the map is an isomorphism.
     
  8. May 19, 2017 #7
    Thanks for all your help Andrew ...

    ... I think I now understand this exercise, thanks to you ...

    I have a solution that is a very slight variant of your solution ...

    ============================================================================

    ##F## finite field of characteristic ##p##


    ##\Longrightarrow## prime subfield of ##F## is isomorphic to ##\mathbb{F}_p## and ##p## must be prime ... (Lidl and Niederreiter, Introduction to Finite Fields ... ... , Theorem 1.78 ... ... )


    also


    ##F## finite means that ##F## has finite dimension over ##\mathbb{F}_p##, say dimension of ##F## over ##\mathbb{F}_p = n##


    ##\Longrightarrow## Basis for ##F## over ##\mathbb{F}_p## has ##n## elements


    ##\Longrightarrow## all elements of ##F## are uniquely expressible as

    ##c_1 v_1 + c_2 v_2 + \ ... \ ... \ \ c_n v_n## ... ... ... (1)


    where ##c_1, c_2, \ ... \ ... \ \ , c_n \in \mathbb{F}_p ##

    and

    ##v_1, v_2, \ ... \ ... \ \ , v_n \in F##



    ##\Longrightarrow## number of elements in ##F, |F| = p^n##

    since each ##c_i## in expression (1) has ##p## possibilities ... ...


    =========================================================================


    Is the above correct?


    Could someone please critique solution/proof ... is it rigorous ...?


    Peter
     
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