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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?
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?
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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".