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I Splitting Fields: Anderson and Feil, Theorem 45.6 ...

  1. Jun 23, 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 an aspect of the proof of Theorem 45.6 ...

    Theorem 45.6 and its proof read as follows:



    ?temp_hash=e4d3c679eb07bb16c3e25a7846ea7af0.png




    At the start of the proof of Theorem 45.6 we read the following:

    "Suppose that ##K## is a normal extension of ##F##, a field with characteristic zero. Then by Theorem 45.5, ##K = F( \alpha )##, where ##\alpha## is algebraic over ##F##. ... .. "



    Can someone please explain exactly how ##K = F( \alpha )## follows in the above statement ... ?


    The quote mentions Anderson and Feil's Theorem 45.5 and also mentions that K is a normal extension so I am providing the statement of Theorem 45.5 and Anderson and Feil's definition of a normal extension as follows ... ...



    ?temp_hash=e4d3c679eb07bb16c3e25a7846ea7af0.png





    ?temp_hash=e4d3c679eb07bb16c3e25a7846ea7af0.png



    Hope someone can help ...

    Peter
     

    Attached Files:

  2. jcsd
  3. Jun 23, 2017 #2

    fresh_42

    Staff: Mentor

    Gather what you have: ##F \subseteq K## a finite extension. This means ##\dim_FK =: n < \infty##.
    Now what does that mean for ##\{1,a,a^2,a^3,\ldots ,a^n\}\,##, if we take an arbitrary element ##a \in K\,##?
    Can you deduce from that, that the conditions for Theorem 45.5 are met?
    And what does ##F \subseteq K## simple algebraic extension mean?
     
  4. Jun 24, 2017 #3
    Thanks for the advice to "Gather what you have: ... "...

    I was myopically focused on the first sentence of the proof where we are given

    (i) ##K## is a normal extension of ##F##
    (ii) ##F## is a field with characteristic zero

    and was wondering about how to get from this to the initial conditions of Theorem 45.5 that

    (i) ##F## is a field with characteristic zero
    (ii) ##K## was a finite algebraic extension of ##F##


    BUT ... in my tunnel vision I neglected that ##K## was given as a finite extension ... sorry for such a simple oversight ...


    ... HOWEVER ... still concerned that we need to establish that ##K## is algebraic over ##F## ... that is all elements of ##K## are algebraic over ##F## ... before applying Theorem 45..5 ...


    Peter
     
  5. Jun 24, 2017 #4

    fresh_42

    Staff: Mentor

    That's why I wrote - with ##n## being the dimesion of ##K## over ##F##
    How many elements has this set? Can they be linear independent, and if not, what does that mean?
     
  6. Jun 24, 2017 #5

    Just to answer your questions ...

    You write:

    "Now what does that mean for ##\{1,a,a^2,a^3,\ldots ,a^n \},##, if we take an arbitrary element ##a \in K\,##?"

    It means that ##\{1,a,a^2,a^3,\ldots ,a^{n - 1} \} ,## is a basis for the vector space ##K## over ##F## ... and that ##\{1,a,a^2,a^3,\ldots ,a^n \},## is a linearly dependent set ... ... is that correct?




    You write:

    "Can you deduce from that, that the conditions for Theorem 45.5 are met?"

    ... regarding conditions for Theorem 45.5 to be met ... see my previous post ...



    You write:

    "And what does ##F \subseteq K## simple algebraic extension mean?""


    Means that ##K## is of the form ##K = F( \alpha )## ...

    Peter
     
  7. Jun 24, 2017 #6

    As mentioned in the previous post the set is linearly dependent ... so this means that any element of the set can be expressed as a linear combination of the others ...

    Peter
     
  8. Jun 24, 2017 #7

    fresh_42

    Staff: Mentor

    I don't know. I don't care either. This would exclude e.g. all ##a \in F \subseteq K## without reason.

    It means the set ##\{1,a,a^2,a^3,\ldots ,a^{n}\}## is linear dependent - in any case. Linear dependency means, we have a non trivial expression of zero: ##0 = c_0\cdot 1 + c_1\cdot a + c_2\cdot a^2 + \ldots + c_n\cdot a^n## with not all ##F \ni c_i=0##. If we define ##p(x)=c_0\cdot 1 + c_1\cdot x + c_2\cdot x^2 + \ldots + c_n\cdot x^n## we get a polynomial in ##F[x]## with ##p(a)=0## and ##a## is algebraic.

    ##K=F(\alpha)## for a certain ##\alpha \in K## is the result of theorem 45.5.
     
  9. Jun 24, 2017 #8
    Thanks for all your help fresh_42 ...

    Gradually building a basic understanding of Galois theory, thanks to your support ...

    Peter
     
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