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For separable extensions, why may we argue as if they're finite?

  1. May 17, 2012 #1
    I'm reading the following article by Maxwell Rosenlicht:

    http://www.jstor.org/stable/2318066

    (The question should be clear without the article, but I present it here for reference.)

    In the beginning of the article he discusses differential fields (i.e. a field [itex]F[/itex] with a map [itex]F\to F[/itex], [itex]a\mapsto a'[/itex] such that [itex](a+b)'=a'+b'[/itex] and [itex](ab)'=a'b+ab'[/itex]). He presents the result that given a differential field [itex]F[/itex] and a separable extension [itex]K/F[/itex], there exists a unique differential structure on [itex]K[/itex] that extends that on [itex]F[/itex]. After showing uniqueness quite easily, he proceeds to show existence. His first sentence toward showing existence is, "Using the usual field-theoretic arguments, we may assume that [itex]K[/itex] is a finite extension of [itex]F[/itex], so that we can write [itex]K=F(x)[/itex], for a certain [itex]x\in K[/itex]."

    Of course if the extension is finite, then it is simple, so I understand the second part of the sentence. But what are the "usual field-theoretic arguments" he's referring to? It's not clear to me why we can assume that this extension is finite.

    If this is too complicated to explain quickly, perhaps you could give me a reference in Dummit and Foote or an internet link.
     
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  3. May 17, 2012 #2

    Hurkyl

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    Presumably, they're referring to the method of writing an infinite extension as a union of finite extensions.

    IMO it's clearest with transfinite induction; start with F and keep adjoining elements one at a time until you have all of K.

    BTW, I assume K/F is supposed to be algebraic?
     
  4. May 17, 2012 #3

    micromass

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  5. May 17, 2012 #4

    mathwonk

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    i agree with THoSE ANSWERS. by zorn there is a maximal extension. if that is not defined on all of K, add in one more element of K and use the primitive element theorem to get a further extension. contradiction to maximality.
     
  6. May 17, 2012 #5
    OK, I agree that an algebraic extension is a union of finite extensions. I'm not quite seeing how that allows us to argue as if [itex]K/F[/itex] is finite. (Perhaps I'm misunderstanding.)
    Definitely--that is explicitly stated in the article. I was under the impression that separable implied algebraic.
    Right, the primitive element theorem applies to finite extensions as I understand it. My question is why may [itex]K/F[/itex] be assumed finite.

    Sorry, maximal with respect to what property? Finiteness? I don't think that's what you mean since it seems the hypotheses of Zorn's Lemma are not satisfied (there are totally ordered subsets with no upper bound which is a finite extension).
     
  7. May 17, 2012 #6

    micromass

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    Let's say you've proven it for finite extensions, then the results follows from Zorn's lemma.

    Look at

    [itex]\mathcal{F}=\{L~\vert~F\subseteq L \subseteq K~\text{and there is a differentiable structure on}~L\}[/itex]

    This is a partially ordered set with respect to inclusion (and probably you need the differentiable structure to agree as well). By Zorn's lemma it has a maximal element. If this maximal element is different from K, then a finite extension of the maximal element contradicts the maximality.
     
  8. May 17, 2012 #7
    Thank you! Perfect!
     
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