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Axiom of choice vs algebra

  1. May 23, 2006 #1

    Hurkyl

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    I was wondering: what is the proof idea behind results such as:

    (Every vector space has a basis) iff AoC
    (All bases for a vector space have the same cardinality) iff AoC
    (Every field has an algebraic closure) iff AoC

    One direction is obvious, but I have no idea how to begin the other!


    As a related question, what is the status of (all algebraic closures of a given field are isomorphic), relative to the AoC?
     
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  3. May 24, 2006 #2

    matt grime

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    I wouldn't have thought that AoC had any impact on this one. The construction of the isomorphism is explicit (and unique) in that it is defined simply by extending the isomorphism on the base field and there is no choice required to do this (as I say, it is a unique isomorphism).

    There is a book showing all the equivalences of the axiom of choice (or perhaps not all of them but a lot of them), whose name escapes me but must be something involving the axiom of choice. I'll try to locate it.

    http://www.math.purdue.edu/~jer/cgi-bin/conseq.html

    Seems to be it.
     
    Last edited: May 24, 2006
  4. May 24, 2006 #3

    Hurkyl

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    There are choices to make in extending the isomorphism. In fact, if they truly are isomorphic, then there are |Gal(F-bar / F)| different isomorphisms!

    I can't figure out how to extend the isomorphism of the base fields without at least having the two closures be well-orderable.
     
  5. May 24, 2006 #4

    matt grime

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    Ah, my mistake, it is not unique up to unique isomorphism for the trivial reason you point out.
     
  6. May 24, 2006 #5
    Could that be Equivalents of The Axiom of Choice? There is a second edition/volume? here here
     
    Last edited: May 24, 2006
  7. May 24, 2006 #6

    mathwonk

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    i have little insight about this question hence will stab at it anyhow.

    ax ch is to me a trivial statement about whether something exists or not that clearly does exist, such as a right inverse to a surjective function.

    i consider the controversy merely an argument about the formal meaning of "exist" for a set, or function.

    anyway, if one denies ax ch, then there is a surjective function with no right inverse. this is not too far from a bijective function with no inverse, or a pair of sets with the same cardinality that one cannot prove have the same cardinality.

    such nonsense may suggest the answer to your question.
     
    Last edited: May 24, 2006
  8. May 25, 2006 #7

    HallsofIvy

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    The proof I learned uses Zorn's Lemma which is equivalent to Axiom of Choice:

    Given a set X and an order relation "[itex]\le[/itex]" on X, if every chain (a subset of X such that "[itex]\le[/itex]" is a linear order) has a maximum, then X has maximal elements (That is, there exist at least one x in X such that [itex]y\le x[/tex] for every y that is "comparable" to x)

    Let V be a linear vector space and {v} be a singleton set (and so independent) and let X be the collection of all set of all independent subsets of V. Let X be "ordered by inclusion" ([itex]A\leB[/itex] if and only if A is a subset of B). If Y is a chain (a collection of subsets such that each is a subset of another) then the union of all sets in Y is also an set if independent vectors (prove this yourself) and so is a maximum for Y. By Zorn's lemma there exist at least one maximal set of independent vectors, U. This set, U, is a basis for V (It is, by definition, independent. Assume it does not span V. Then there exist a vector v in V that cannot be written as a linear combination of vectors in U. Show that adding v to U results in an independent subset of V containing U, contradicting the fact that U is maximal.).
     
  9. May 25, 2006 #8

    matt grime

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    As Hurkyl says, it is obvious how to show that AoC implies every vector space has a basis, but how does the reverse implication go? I know (at least I think I know) it was proven by Blass but I've not looked up the proof.
     
    Last edited: May 25, 2006
  10. May 25, 2006 #9

    matt grime

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    A. Blass, Existence of bases implies the axiom of choice, in Axiomatic Set Theory, Contemporary Mathematics, 31 (1984) 31 - 33. MR 86a:04001

    is the reference should anyone have that journal at hand.

    I probably have it at work: Bristol was very heavily into set theory and such things pre 1990
     
  11. May 26, 2006 #10

    Hurkyl

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    The library had a 1970 edition of it, but it doesn't seem all that helpful. :frown: It's section on algebraic equivalents to the AC are either about lattices, or universal algebras. While this vast generality is useful for, say, using the AC to prove every vector space has a basis, it isn't so helpful for the reverse direction. And I'm not even sure if any of its algebraic equivalents relate to algebraic closures of fields!
     
  12. Jun 3, 2006 #11
    that any two bases have same cardinality is not equivalent to choice; it's strictly weaker.
     
  13. Feb 3, 2008 #12
    (Every vector space has a basis) implies (Axiom of Choice)
    Ok, I typed in google:

    JSTOR axiom of choice basis

    and I got this:
    http://links.jstor.org/sici?sici=0002-9939(196606)17:3<670:BIVSAT>2.0.CO;2-Q

    This paper is dated 1966, and thus probably is the original proof. So is this the best way to find a specific proof? Go to google and type JSTOR + the key words?

    Is there a more systematic way than to rely on google? How do mathematicians find specific papers from the past? I still don't understand the system of math journals.
     
    Last edited: Feb 3, 2008
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