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I Prime Subfiellds - Lovett, Proposition 7.1.3 ...

  1. Apr 14, 2017 #1
    I am reading Abstract Algebra: Structures and Applications" by Stephen Lovett ...

    I am currently focused on Chapter 7: Field Extensions ... ...

    I need help with the proof of, or at least some remarks concerning, Proposition 7.1.3 ...


    Proposition 7.1.3 plus some introductory remarks (proof?) reads as follows:



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    In the above text from Lovett we read the following:


    "... ... However, the multiplication on these elements as defined by distributivity gives this set of elements the structure of ##\mathbb{F}_p = \mathbb{Z} / p \mathbb{Z}##. ... ... "


    ... ... BUT ... the subfield contains elements ##0, 1, 2, 3, 4, 5, \ ... \ ... \ (p -1)##


    ... and being a field, it contains divisions of these elements such as ##1/2, 3/5 \ ... \ ... \ ...##


    ... so how can this subfield be equal to ##\mathbb{Z} / p \mathbb{Z}## ... ... ?



    Hope someone can help ...

    Peter
     

    Attached Files:

  2. jcsd
  3. Apr 14, 2017 #2

    andrewkirk

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    The missing piece is that a field's characteristic must be either zero or a prime number, otherwise there will be divisors of zero (can you see why?), which would disqualify ##F## from being a field. We then use the theorem that ##\mathbb Z_p## is a field for ##p## prime, so that the fractions you listed are all modular integers in ##\mathbb Z_p##. For instance ##1/2=3## in ##\mathbb Z_5## and ##3/5=2## in ##\mathbb Z_7##.
     
  4. Apr 14, 2017 #3
    Thanks for the help Andrew ...

    Just reflecting on what you have said ...

    However, can you give me an indication of how exactly we can demonstrate that ##1/2 = 3## in ##\mathbb{Z}_5## and ##3/5 = 2## in ##\mathbb{Z}_7## ...

    Then I will get an idea of what is going on ...

    Peter
     
  5. Apr 14, 2017 #4

    andrewkirk

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    For the first one, ##3\times 2=6 = 5+1##, so that ##3_5\times 2_5=1_5## where the 5 subscript denotes 'mod 5' (technically speaking, ##k_p## denotes ##k+p\mathbb Z##).

    Similarly, ##2\times 5=10=7+3##, so that ##2_7\times 5_7=3_7##.

    A good way of thinking of it is as a clock, with the numbers 0 to p-1 marked around the outside. and a single hand. Asking 'what is 1 / 2 mod 5' is the same as asking what rotation (past how many marked numbers) must I perform 2 times, starting at 0, to end up at 1?
     
  6. Apr 14, 2017 #5
    Thanks Andrew ... really appreciate your help ..

    Peter
     
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