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Algebraic closure of finite field

  1. Feb 5, 2004 #1

    matt grime

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    EDIT: forgot to say, this is in response to another thread which had wandered way off its original question. Arguably this could be in at least 3 other forums
    end EDIT

    EDIT contains some gross errors, will correct them some timeEDIT

    OK, just wrote a long thread on this only to hit ctrl w in mozilla - try it and see what happens.

    So I will assume, cos I'm lazy, that the reader knows what I mean when I say F_p is the field with p - elements - it is the set 0,1,2,...,p-1 with addition and multiplication defined remainder p, of course p must be a prime - if it is composite then there are non-zero elements x and y with x*y=0, which can't happen in a field.

    We can define polynomials F_p[x] one variable with coefficients in F_p. Not every poly will have roots in F_p. THE canonical example is x^2+x+1 in F_2. put x=0 or 1 in there and the outcome is always 1. Notice that means that as FUNCTIONS from F_2 to F_2 this polynomial and the constant polynomial H(x)=1 are indistinguishable. They are different polynomials though. This doesn't happen over the complex numbers: suppose f(x)=g(x) for all x in C, then f-g is always zero. Depending on how much you like to use sledgehammers to crack peanuts, it is then obvious, as it is an analytic function, that it is the zero function, and that f=g.


    Anyway, just as we can define C to be R(i), that is the set of all things a+ib with a and b real, and multiplication defined according to the rule i.i=-1 and so on, we can define F_2(j), where j satisfies j^2+j+1=0.

    Then F_2(j) is all things a+bj, for a and b in F_2, with j^2=j+1.

    Check that this new thing is a field and has 4 elements.

    Claim, if F is a field with finitely many elements, it has p^r elements for some p a prime, moreoever there is some polynomial with coefficients in F that has no root in F. Proof can be found in most basic algebra books, particularly anything to do with galois theory. If there is a call for it, I will provide more details here, but they aren't important.



    This says that no finite field is algebraically closed, that is possesses all the roots to all polynomials.

    We will assume that all finite fields can be constructed by adjoining elements to some F_p like we do in the case of R(i) to get C. The difference there is that once we adjoin i, all polynomials have all their roots in C. We had F_2(j), but we could equally add things to F_p. We can add on more roots of more polys, if we let F_q be the field with q elements, then there is a sequence

    F_p--->F_(p^2)--->F)(p^3)--> ...


    there is something we can think of as the limit of this sequence, this is an algebraic closure. To construct all those F_q, we take F_p, find a polynomial of smallest degree with no root, add in this artificially, take some other poly with smallest degree that has no root in the new field, and add it on, keep on doing this using a poly of smallest degree at each point

    Example.

    F_2, all degree 1 polys have roots, X^2+X has roots, so does X^2, as does X^2+1, but X^2+X+1 does not, add in the j as above. Now every degree two or fewer poly has its roots. What about cubics? Can you find a cubic with no root in 0,1,j,j+1? Add in any cubic poly roots, check all the 4th degree polys etc... or just take it as read that you can always find some other poly without roots, so there is one of smallest degree.

    So we can construct this 'limit' and any polynomial, because it has finite degree must have its roots in one of the F_q, and by technical results we're omitting, has a root in the limit.

    Note, this limit IS NOT UNIQUE. There were lots of choices made, if we'd made them in a different order - for instance if there were two polys of a given degree with no roots, we'd have to pick one first then the other to add the roots in, and the results might not be equal (but would be equivalent). It so happens that the choices we made mean that although there are two different limits, they are isomorphic as fields, but that this isomorphism is not unique, so we really must speak of AN algebraic closure. But, any algebriac closure of F_p contains all the F_(p^r) just embedded differently.

    Ok, that's confusing isn't it? But we know the case R(i)=C. The thing is that i and -i are indistinguishable. The standard thought experiment for this is that, if aliens came to earth and they'd developed a similar number system to ours, but instead of picking i to be sqrt(-1), they'd picked j and their j was our -i, we couldn't tell the difference. Why not? Because the only thing we have to define i is that it's the root of x^2+1, well, send x to -x in that and the equation is unchanged!

    Want another way to think about this? Well, there is a map from C to C sending i to -i, or if you like, taking the conjugate. This is an automorphism of C that fixes the reals. From here it is a short step to Galois Theory, where Ian Stewarts book gives a much better intro than this.

    Comments? Bits that need more explaining? Less explaining? Tangents to go down?
     
    Last edited: Feb 6, 2004
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  3. Feb 5, 2004 #2

    Hurkyl

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    I'm trying to remember; it's not generally true that the algebraic closure of a field is itself algebraically closed, correct? But it is possible to iterate the process and eventually arrive at a fixed point?
     
  4. Feb 5, 2004 #3

    matt grime

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    This is not something I want to leap in and make a definite claim about because it is a long time since I did this. In fact I was deliberately glossing over this point.

    I will try and figure it our tonight, but as a warning to others, the algebraic closure of a finite field is countable, yet there exist uncountable algebraically closed fields that have non-zero characteristic. The closure as the limit above is clearly countable. This is why I'm being careful - there are large fields you see, but I don't remember how to construct them.
     
  5. Feb 5, 2004 #4

    matt grime

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    Major correction needed


    GF(p^r) < GF(p^s) iff r|s,

    need to add that in to get the correct sequence.

    This is where the axiom of choice comes in to define the closure - we can take the limit for even exponent, the limit for mutiples of three, that agrees with the mutliples of 2 when they overlap, then for mutliples of 5 which agree the with mutliples of 2 and 3 etc..
     
    Last edited: Feb 6, 2004
  6. Feb 5, 2004 #5

    Hurkyl

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    Ok, things are coming back to me now. :smile: I was mixing up "K is a (minimal) field that contains all of the roots of F" with "K is the algebraic closure of F".

    At the moment, I see clearly how to do the former by well ordering all polys over F.

    Wikipedia says that the algebraic closure of GF(p) is simply the union of all of the GF(p^n)


    I feel like I'm missing something really trivial to prove that the union of the GF(p^n) is algebraically closed, or that the two phrases I mentioned in my first paragraph are equivalent.


    edit: fixed omission of "the union of the" in the last paragraph
     
    Last edited: Feb 5, 2004
  7. Feb 5, 2004 #6

    matt grime

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    Damn, you edited that post, here's the answer still for the E<H<K thing.
    This is sort of like integral is integral from algebriac integers. As stated I don't think it is true - it would be if each were algebraic over the smaller field.

    And the wikipedia thing misses the point that it is the nested union that you take.


    EDIT removed the incorrect assrtion about splitting.
     
    Last edited: Feb 6, 2004
  8. Feb 5, 2004 #7

    Hurkyl

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    Sorry about that! I realized that question as asked wasn't particularly helpful. :smile: What I did want to ask was the case where they were algebraic extensions, and it turns out I was overlooking something simple. :frown:

    In any case, now it's clear why the union of the GF(p^n) is algebraically closed.


    (I'm implicitly assuming the GF(p^n) are represented so that m | n iff GF(p^m) | GF(p^n))


    So, in general, if F[x] is countable, then we can construct the algebraic closure of F by inductively adjoining to F the roots of every polynomial in F[x].


    When F[x] is uncountable, we can still create a field K that contains all of the roots of F via transfinite induction, but the proof that K is algebraically closed now fails... and you appear to be confirming my recollection that there are examples where this K is not algebraically closed.

    The fun question, now, is to come up with such an example. :smile:

    EDIT: corrected a mistake.
     
    Last edited: Feb 6, 2004
  9. Feb 6, 2004 #8

    matt grime

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    it should read m|n, I got it wrong.

    As for nasty algebraic closures, well, I wouldn' want to imagine what the closure of the p-adics is.
     
  10. Feb 6, 2004 #9
    the neat thing about wikipedia is that anyone who wants to can edit any of the articles. so if you think they are missing the point of something, you can just add it!

    you can even just say a few words about something you think that should be added, and in most likelihood, someone else will come along and fill in the details
     
  11. Feb 6, 2004 #10

    matt grime

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    Didn't know that about wikipedia. It's an implicit thing that one is contained in the other, though. I mean A_n is inside S_n, but what is A_nUS_n? And it's not quite nested, it's.. a little strange I think.
     
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