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I Splitting Fields: Anderson and Feil, Theorem 45.5 ...

  1. Jun 22, 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 45.5 ...

    Theorem 45.5 and its proof read as follows:



    ?temp_hash=6827701059bafb016e0aaa98e096587f.png
    ?temp_hash=6827701059bafb016e0aaa98e096587f.png




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

    "... ... Now ##\alpha## and ##\beta## are roots of irreducible polynomials ##f, g \in F[x]## ... ...


    Now, we are just given that \alpha and \beta are algebraic elements of a field ##F## ... ... how, exactly, do we know that they are roots of irreducible polynomials in ##F[x] ##... .,.. ?



    "( NOTE: A&F's definition of algebraic over ##F## does not mention irreducible polynomials but says:

    "If ##E## is an extension field of a field ##F## and ##\alpha \in E## is a root of a polynomial in ##F[x]##, we say ##\alpha## is algebraic over ##F##. ...)




    Hope someone can help ...

    Peter


    Edit: Hmm ... wonder if Kronecker's Theorem has something to do with irreducible polynomials entering this discussion ...
     

    Attached Files:

  2. jcsd
  3. Jun 22, 2017 #2

    fresh_42

    Staff: Mentor

    What happens, if you consider a polynomial ##p(x) \in F[x]## that is reducible, say ##p(x)=f(x)\cdot g(x)## and ##p(\alpha)=0\,##? Remember that ##\operatorname{char} F =0##.
     
  4. Jun 22, 2017 #3

    ... ... well ... I guess ##\alpha## must be a root of either ##f## or ##g## .... and if it is a root of ##f## then ##f## is irreducible ... if it is not just continue your process until we end up with an irreducible ... so we end up with an irreducible polynomial ... so, given this, assume one at the start

    Is that correct ...?

    Peter
     
  5. Jun 22, 2017 #4

    fresh_42

    Staff: Mentor

    Yes, this is true. You just can assume a minimal (and irreducible) polynomial. I wonder how it is, if the characteristic is not zero, then we cannot conclude that either ##f(\alpha)=0## or ##g(\alpha)=0##. At least not as easy, but I don't remember this special case.

    Edit: I think I got it. If ##\operatorname{char} F = p## and ##f(\alpha)\cdot g(\alpha)=0## then ##p## divides this product, and the property of ##p## being a prime means it divides one of the factors, which in return translates to ##f(\alpha)=0## or ##g(\alpha)=0##.
     
  6. Jun 22, 2017 #5
    Thanks fresh_42 ... ...

    Hmm ... very interesting point ... cannot see reason yet ... reflecting ...

    Thanks again ...

    Peter
     
  7. Jun 22, 2017 #6

    fresh_42

    Staff: Mentor

    No, not really. Read my edit.
     
  8. Jun 22, 2017 #7

    ... oh ... OK ... yes ...

    Peter
     
  9. Jun 22, 2017 #8

    Hi fresh_42 ...

    Just confirming something ...

    You write:


    "... ... If ##\operatorname{char} F = p## and ##f(\alpha)\cdot g(\alpha)=0## then ##p## divides this product.... ... "

    I am assuming that ##p## divides ##f(\alpha)\cdot g(\alpha)## because ##p \cdot 1 = f (\alpha)\cdot g(\alpha) = 0## ... ...


    Is that correct?

    Peter
     
  10. Jun 22, 2017 #9

    fresh_42

    Staff: Mentor

    Yes, this is correct. If ##\operatorname{char}F = p##, then ##p=\operatorname{char}F = \underbrace{1+\ldots +1}_{p-times} =0## in this field. Think of ##\mathbb{Z}_2## or ##\mathbb{Z}_3## where the elements are ##\{0,1\}\, , \,\{0,1,2\},## resp., which means ##1+1=2=0## in the first case and ##1+1+1=3=0## in the second case.
     
  11. Jun 22, 2017 #10
    Thanks for all your help and support on the above posts!

    Most grateful for your help!

    Peter
     
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