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Let E be an algebraic over F, F is perfect. Show that E is perfect

  1. Aug 12, 2007 #1
    let E be an algebraic over F where F is perfect. Show that E is perfect. :uhh:
  2. jcsd
  3. Aug 12, 2007 #2


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    Well, what have you tried? Our purpose here is to help you work through a problem, not to be an answer book!
  4. Aug 12, 2007 #3


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    what doies perfect mean? all algebvraic extensions are separable? sounds trivial from the definitions doesnt it? have you thought about it?

    if so, and it still eludes you, think abiout the relation between the irreducible polynomial for a given element over a base field as opposed to over a larger field.
    Last edited: Aug 12, 2007
  5. Aug 12, 2007 #4
    @mathwonk: Yes "perfect" means all extensions are seperable.

    @barbie: Do the following
    1)Let [tex]K/E<\infty[/tex].
    2)Let [tex]\alpha \in E[/tex]
    3)Show [tex]\mbox{irr} \left< \alpha, E \right>[/tex] has zeros of multiplicity 1 (Hint: Show that [tex]\mbox{irr}\left< \alpha, E \right> | \mbox{irr} \left< \alpha, F \right>[/tex]*

    *)If K algebraic over E algebraic over F then K algebraic over F.
  6. Aug 13, 2007 #5
    Just out of interest: What does the notation irr<a,E> mean?

    I though maybe it was a notation for the minimal polynomial of a over E, but as a was in E that doesn't make much sense your 3 wouldn't make much sense.
  7. Aug 13, 2007 #6
    It is the monic irreducible polynomial such that contains "a" as a zero. This notation is non-standard, but saves space which I have seen in one book.
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