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I Splitting Fields: Anderson and Feil, Theorem 45.4 ...

  1. Jun 21, 2017 #1
    I am reading Anderson and Feil - A First Course in Abstract Algebra.

    I am currently focused on Ch. 45: The Splitting Field ... ...

    I need some help with some aspects of the proof of Theorem 45.4 ...

    Theorem 45.4 and its proof read as follows:



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    My questions on the above proof are as follows:


    Question 1


    In the above text from Anderson and Feil we read the following:


    "... ... This means that ##f = ( x - \alpha)^k g##, where ##k## is an integer greater than ##1## and ##g## is a polynomial over ##K## ... ...


    Since ##f## is in ##F[x]## ... that is ##f## is over ##F## ... shouldn't ##g## be over ##F## not ##K##?

    (I am assuming that ##f## being "over ##F##" means the coefficients of ##f## are in ##F## ... )




    Question 2


    In the above text from Anderson and Feil we read the following:


    "... ... We then have that ##x - \alpha## is a factor of both ##f## and ##f'##. But if we use term-by-term differentiation instead, it is clear that ##f'\in F[x]##. ... ... "


    What do Anderson and Feil mean by term-by-term differentiation in this context ... ... and if they do use term-by-term differentiation (what ever they mean) how does this show that ##f'\in F[x]## ... ... ?



    Hope someone can help ...

    Help will be much appreciated ... ...

    Peter
     

    Attached Files:

  2. jcsd
  3. Jun 21, 2017 #2

    andrewkirk

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    If ##\alpha\in K-F## then ##g## will not necessarily be in ##F## as its coefficients will be functions of ##\alpha##.

    Consider ##F=\mathbb R,\ K=\mathbb C## and ##f(x)=(x^2+1)^2=(x-i)^2(x+i)^2##, which has roots ##i## and ##-i##, both of multiplicity 2. Taking ##\alpha=i## we get ##g(x)=(x+i)^2##, which is not in ##\mathbb R[x]##.
     
  4. Jun 21, 2017 #3

    andrewkirk

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    I was going off to have dinner, thinking I didn't have time to work out the second one. Then I realised it's actually the easier of the two.

    By 'term-by-term' they mean writing the polynomial in expanded, unfactorised form as ##f(x)=\sum_{k=0}^n c_kx^k## and then differentiating to get ##f'(x)=\sum_{k=1}^n kc_kx^{k-1}##. The coefficients are of the form ##kc_k## which, since ##k## is in ##\mathbb N## rather than ##F##, means ##\sum_{j=1}^k c_k##, which must be in ##F## since ##c_k## is. Hence ##f'\in F[x]##.
     
  5. Jun 21, 2017 #4

    Thanks Andrew ... good example ... makes it pretty clear ...

    Peter
     
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