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I Existence and Uniqeness of Finite Fields ...

  1. Jun 5, 2017 #1
    I am reading David S. Dummit and Richard M. Foote : Abstract Algebra ...

    I am trying to understand the example on Finite Fields in Section 13.5 Separable and Inseparable Extensions ...


    The example reads as follows:



    ?temp_hash=7f262d4c0e613b648031240adf83746b.png
    ?temp_hash=7f262d4c0e613b648031240adf83746b.png




    My questions are as follows:



    Question 1


    In the above text from D&F we read the following:

    " ... ... If ##\mathbb{F}## is of dimension ##n## over its prime subfield ##\mathbb{F}_p##, then ##\mathbb{F}## has precisely ##p^n## elements. ... ... "


    Can someone please explain why, exactly, this follows?




    Question 2




    In the above text from D&F we read the following:

    " ... ... Since the multiplicative group ##\mathbb{F}^{ \times }## is (in fact cyclic) of order ##p^n - 1##, we have ##\alpha^{ p^n - 1 } = 1## for every ##\alpha \ne 0 in \mathbb{F}## ... ... "


    Can someone give me the exact reasoning concerning why ##\mathbb{F}^{ \times }## being of order ##p^n - 1## implies that ##\alpha^{ p^n - 1} = 1## for every ##\alpha \ne 0## in ##\mathbb{F}## ... ... ?


    (I am guessing that for some reason I cannot explain, that ##\mathbb{F}^{ \times }## being of order ##p^n - 1 ## implies that the characteristic is ##p^n - 1## ... ... but why does it mean this is the case ...? )




    Hope someone can help ...

    Peter
     
  2. jcsd
  3. Jun 5, 2017 #2

    fresh_42

    Staff: Mentor

    What does it mean "of dimension ##n##"? If you write down all possible linear combinations of basis vectors, how many possibilities do you get?
    If you have any finite group ##G##, say ##|G|=m## and take an element ##g \in G##, then consider the subgroup ##U:=\langle g \rangle \subseteq G##. What can be said about the order of ##U## and therewith the order of ##g##? Now with that, what do you get for ##g^m##?
    No. The characteristic of a field always has to be either ##0## or prime, and ##p^n-1## is only in one single case prime whereas the statement is generally valid. The characteristic is the smallest number ##n## such that ## \underbrace{1+ \ldots + 1}_{n-times} =0## or zero, if ##n## isn't finite. The order of ##\mathbb{F}^\times## is ##m=|\mathbb{F}|-|\{0\}|=p^n-1## not the characteristic.
     
    Last edited: Jun 5, 2017
  4. Jun 6, 2017 #3

    Hi fresh_42 ... ... thanks for the help ...

    Regarding Question 1, you write the following"

    "What does it mean "of dimension n" role="presentation">n"? If you write down all possible linear combinations of basis vectors, how many possibilities do you get?"

    Well, dimension ##n## means there are ##n## basis vectors, and each linear combination of the basis vectors has ##n## coefficients with ##p## possibilities for each ... ... so we get a total of ##p^n## unique elements in the field ##\mathbb{F}## ... ... Is that the correct analysis/reasoning?

    Still thinking about Question 2 ...

    Peter
     
  5. Jun 6, 2017 #4

    fresh_42

    Staff: Mentor

    Yes. And they are all different, because ##\alpha_1 b_1 + \ldots + \alpha_n b_n = \beta_1 b_1 + \ldots + \beta_n b_n## means, that each ##\alpha_i=\beta_i## if ##\{ b_1,\ldots ,b_n\}## is a basis. So there cannot be two different combinations resulting in the same vector, i.e. all ##p^n## combinations are different elements.
     
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