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Fundamental theorem of algebra

  1. May 6, 2010 #1

    lavinia

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    I would like a reference for a purely algebraic proof of the fundamental theorem of algebra - or if you would like to supply a proof that would be even better.
     
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  3. May 6, 2010 #2

    Landau

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    I don't think a purely algebraic proof is possible. The completeness of R or C has to come up somewhere, i.e. some analytic aspect has to be invoked, like the intermediate value theorem.

    See here for several proofs.
     
  4. May 7, 2010 #3

    TMM

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    Here's one:

    Let f(x) be a polynomial with real coefficients of degree n with splitting field K/R. K(i) is a Galois extension of R, so let G=Aut(K(i)/R). If P is any 2-sylow subgroup of G, then the fixed field of P is an odd extension of R, which must be real, and since complex roots come in pairs, it follows this extension is trivial. It follows that Gal(K(i)/C) is a 2-group. Since 2-groups have subgroups of all orders, if the extension K(i)/C was non-trivial, then we would have a quadratic extension of C. Since we can easily show with the quadratic formula that quadratics over C have roots in C, this is a contradiction, hence K is a subfield of C. Q.E.D.

    This proof is due to Artin and is purely algebraic.
     
  5. May 7, 2010 #4

    Landau

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    Are you sure?
    Here you are using the fact that every polynomial with real coefficients of odd degree has a real root. Either this follows from FTA* (because of FTA there are an odd number of roots, complex roots come in conjugate pairs, hence at least one root is real), or from the intermediate value theorem. Obviously the first argument is not allowed in a proof of the FTA, so you still have to use analysis.

    *FTA = the fundamental theorem of algebra
     
  6. May 7, 2010 #5

    TMM

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    It's a really trivial application of the IVT. Monic, odd polynomials are large and positive for large x and large and negative for small x, hence they cross the axis somewhere. I hardly consider this an analytic proof.
     
  7. May 7, 2010 #6

    Landau

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    Of course this is an analytic proof! The intermediate value ultimately rests on the completeness of R, which is almost by definition an analytic property. This is also what I already said:
    Then you reply saying you have a "purely algebraic proof", while you are doing exactly what I predicted. The point is, a purely algebraic proof does not exist, whether you call the result that you use from analysis trivial or not.
     
    Last edited: May 7, 2010
  8. May 7, 2010 #7

    TMM

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    I suppose I also used numbers so it's a number theory proof. And I had to found it in ZF so it's a set theory proof. It might as well also be called a combinatorics proof since that's how Sylow's theorem is proven.

    When asked for an analytic proof of the theorem, I would've supplied the complex analytic one in your link. I don't think the OP is a picky as you are.
     
  9. May 7, 2010 #8

    Landau

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    Question: does there exist a purely algebraic proof of FTA?

    Landau 1: no, a purely algebraic proof does not exist, every proof must use some analytic aspect like the IVT.
    TMM 1: yes, here is a purely algebraic proof.
    Landau 2: but hey, you are still using IVT, like I said.
    TMM 2: yes I did.
    I understand that some proofs require more analysis. But the question was about a purely algebraic proof, not about an algebraic proof. Anyway, let us wait and see which proof in the wikipedia link (which includes your Artin proof) TS likes best.
     
  10. May 7, 2010 #9
  11. May 7, 2010 #10

    Landau

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    Yeah, this one (linked to at MathOverflow) is cool, using martingales.
     
  12. May 7, 2010 #11

    Hurkyl

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    Nope. The IVT is a theorem of, for example, real closed fields.
     
  13. May 7, 2010 #12

    Landau

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    I don't see the connection between your first and second sentence. The (usual) IVT cannot be proven without completeness of R; in fact, they are equivalent! So the notion of real closed fields is not really relevant here.
     
  14. May 7, 2010 #13

    Hurkyl

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    There exist real closed fields that are not complete, yet satisfy the polynomial IVT.

    In any case, "real closed field" is a buzzword for you to go searching; there are a variety of ways to characterize the notion. Being an ordered field, and satisfying the polynomial IVT is one of them. If you are asking something that makes sense, then your answer probably lies there.


    P.S. one could argue that < is analytic, not algebraic. If one wants to be so pedantic, then you problem is ill-posed, since you cannot talk about "real number" without having access to some notion equivalent to <.

    P.P.S. I would not so argue -- I think it's ridiculous to try and draw a hard line between the two.
     
  15. May 7, 2010 #14

    Landau

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    That is irrelevant. We were talking about the IVT for the field R. And the IVT for the field R is equivalent to completeness of R. That's all I claimed.
    I am aware of this.

    Hyrkyl, I appreciate your input, but I think it is inappropriate to have a discussion about model theory in this thread. I am not trying to draw a hard, rigoruous line between 'analytic' and 'algebraic': as explained, 'analytic' in this context just means that the completeness of R is involved, that's all. So there is no problem, let alone an ill-posed problem.
     
  16. May 7, 2010 #15

    Hurkyl

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    I mixed your name up with the OP, and thought you wrote both posts, so that confused things a little.


    (p.s. the theory of real closed fields isn't solely confined to model theory, and I really have seen texts that develop the theory algebraically, without even referencing formal logic)
     
  17. May 7, 2010 #16

    Landau

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    Ok, I can see how that would be confusing :)
     
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