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I Field Extensions - Lovett, Theorem 7.1.10 ... ...

  1. May 1, 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 an aspect of the proof of Theorem 7.1.10 ...


    Theorem 7.1.10, and the start of its proof, reads as follows:



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    In the above text from Lovett we read the following ...

    " ... ... Let ##p(x)## be a polynomial of least degree such that ##p( \alpha ) = 0## ... ... "

    Then Lovett goes on to prove that ##p(x)## is irreducible in ##F[x]## ... ...

    ... BUT ... I am confused by this since it is my understanding that if ##p( \alpha ) = 0## then ##p(x)## has a linear factor ##x - \alpha## in ##F[x]## and so is not irreducible ... ... ?


    Can someone please help clarify this issue ... ...

    Peter
     

    Attached Files:

  2. jcsd
  3. May 1, 2017 #2

    fresh_42

    Staff: Mentor

    You can use polynomial division here. One can always write ##p(x) = q(x) \cdot (x-\alpha) + s(x)## with ##\deg s(x) < \deg p(x)##.
    Can you conclude the rest from here?
     
  4. May 1, 2017 #3

    If ##p(x) = q(x) \cdot (x-\alpha) + s(x)## and ##p( \alpha ) = 0## then we must have ##s( \alpha) = 0##

    Given this ... I do not know how to square this with D&F Proposition 9, Section 9.4, page 307 which seems to imply that if ##p ( \alpha ) = 0## for ##\alpha \in F## then ##p(x)## has a linear factor ... and is therefore not irreducible ...

    Peter
     
  5. May 1, 2017 #4

    fresh_42

    Staff: Mentor

    ##p(x)## has been chosen to be a (non-zero) polynomial of minimal degree with ##p(\alpha)=0##. But now ##s(\alpha)=0## and ##\deg s(x) < \deg p(x)## which can only hold, if ##s(x) = 0##. Going back to the equation we have ##p(x) = q(x) \cdot (x-\alpha)## or ##(x-\alpha) \mid p(x)##.
     
  6. May 1, 2017 #5

    Well ... we seem to have shown that ##p(x)## is reducible as a consequence of ##p( \alpha ) = 0## ...

    Exactly my problem with this ... :frown:

    ... hmmm ... how are we going to end up with ##p(x)## being irreducible when it has a linear factor ... ???

    Sorry to be slow about this ...

    Peter
     
  7. May 1, 2017 #6

    fresh_42

    Staff: Mentor

    The equation ##p(x)= q(x)\cdot (x-\alpha) + s(x)## doesn't exist in ##F[x]##, because ##\alpha \notin F##. It only exists in ##F[\alpha][x]##.
    Lovett shows that ##p(x)\in F[x]## is irreducible, because ##\alpha## isn't available in ##F##. However, it is reducible in ##F[\alpha][x]\;\;\;\;## so the argument with the linear factor (which we have shown is true) only holds over ##F[\alpha]##. Irreducibility depends on the ring under consideration, or likewise on the coefficients which are available in the corresponding domain of scalars.

    ##p(x) \in F[x]## irreducible (Lovett's proof)
    ##p(x) \in F[\alpha][x]## reducible, by your argument with the linear factor ##x-\alpha##

    Sorry for the confusion, which I think I might have caused, by proving your assertion with the linear factor, instead of recognizing, that there are two different rings involved here: ##F[\alpha][x]## and ##F[x]##, which makes the difference.
     
  8. May 1, 2017 #7


    Thanks so much for the help in clarifying things ... really has been most helpful ...

    Peter
     
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