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I Fields and Field Extensions - Dummit and Foote, Ch. 13 ... .

  1. May 3, 2017 #1
    I am reading Dummit and Foote, Chapter 13 - Field Theory.

    I am currently studying Theorem 3 [pages 512 - 513]

    I need some help with an aspect the proof of Theorem 3 ... ...

    Theorem 3 on pages 512-513 reads as follows:


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    ?temp_hash=fe36f5a47fc27d01f829e59ddf52c9e4.png



    In the above text from Dummit and Foote, we read the following:

    " ... ... We identify ##F## with its isomorphic image in ##K## and view ##F## as a subfield of ##K##. If ##\overline{x} = \pi (x)## denotes the image of ##x## in the quotient ##K##, then

    ##p( \overline{x} ) = \overline{ p(x) }## ... ... (since ##\pi## is a homomorphism)

    ... ... "


    My question is as follows: ... where in the proof of ##p( \overline{x} ) = \overline{ p(x) }## does it depend on ##\pi## being a homomorphism ... ...

    ... indeed, how does one formally and rigorously demonstrate that ##p( \overline{x} ) = \overline{ p(x) }## ... ... and how does this proof depend on ##\pi## being a homomorphism ...



    To make my question clearer consider the case of ##p(x) = x^2 - 5## ... ...

    Then ...

    ##p( \overline{x} ) = \overline{x}^2 - 5_K##

    ##= ( x + ( p(x) ) ( x + ( p(x) ) - ( 5 + ( p(x) )##

    ##= ( x^2 + ( p(x) ) - ( 5 + ( p(x) )##

    ##= (x^2 - 5) + ( p(x) ) = 0##

    ##= \overline{ p(x) }##


    ... ... in the above case, my question is ... where does the above calculation depend on ##\pi## being a homomorphism ... ?



    Hope someone can help ...

    Peter
     

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    Last edited: May 3, 2017
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  3. May 3, 2017 #2

    andrewkirk

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    The proof does not demonstrate that ##p(\overline x)=\overline{p(x)}## *. It is left 'as an exercise for the reader'.

    Here are the missing steps. We write ##p(x)=\sum_{k=0}^n a_kx^k##:

    \begin{align*}
    p(\overline x)&=
    \sum_{k=0}^n a_k\overline x^k\\
    &=\sum_{k=0}^n a_k\pi(x)^k\\
    &=\pi\left(\sum_{k=0}^n a_kx^k\right)\\
    &\quad\quad\quad\textrm{[HERE we have used the homomorphism property to move the additions and multiplications inside the }\pi\textrm{ operator}]\\
    &=\left(\sum_{k=0}^n a_kx^k\right) + (p(x))\\
    &=p(x)+(px(x))\\
    &=(p(x))\\
    &=0_K
    \end{align*}

    * Indeed, depending on how one interprets the author's statement that ##\overline x=\pi(x)##, the symbol string ##\overline{p(x)}## may not even be defined.
     
  4. May 4, 2017 #3


    Thanks for the help, Andrew ...

    Just now reflecting on what you have written... but at first sight, it seems very clear ...

    Thanks again,

    Peter
     
  5. May 4, 2017 #4
    Hi Andrew,

    Thanks for you help ... BUT ... just a further issue on this topic ...

    Why do we need to consider a mapping/homomorphism at all in proving that ##p( \overline{x} ) = \overline{ p(x) }## ...

    Surely we can just prove that ##p( \overline{x} ) = \overline{ p(x) }## in the field ##F[x] / ( p(x) )## by simply considering the nature of the cosets of the quotient ... and in particular the rules for adding and multiplying cosets ...

    I wonder whether we need ##\pi## at all in the proof of Theorem 3 ...

    Can you comment ... ?

    Peter
     
  6. May 4, 2017 #5

    andrewkirk

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    The difficulty is that ##\overline x## is defined as ##\pi(x)##, so we cannot even state the claim ##p( \overline{x} ) = \overline{ p(x) }##, let alone prove it, unless we have defined ##\pi##.

    Also, recall that the key claim of Theorem 3 is that ##K## contains an isomorphic copy of ##F##. The easiest way to demonstrate that is to create an injective homomorphism into ##K## from a field that is isomorphic to ##F##. That injective homomorphism is ##\pi|_{F'}## where ##F'## is the subring of ##F[x]## consisting of polynomials of degree 0, which is isomorphic to ##F##.

    So we have ##F\cong F'\cong Im\ \pi|_{F'}##. In the book they identify ##F'## with ##F## but I find it clearer in this context to highlight the difference.
     
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