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Extensions fields

  1. Feb 16, 2008 #1
    [SOLVED] extensions fields

    1. The problem statement, all variables and given/known data
    Can someone help me with these true or false problems:
    1) Every algebraic extension is a finite extension.
    2)[tex]\mathbb{C}[/tex] is algebraically closed in [tex]\mathbb{C}(x)[/tex], where x is an indeterminate
    3)[tex]\mathbb{C}(x)[/tex] is algebraically closed, where x is an indeterminate
    4)An algebraically closed field must be of characteristic 0

    Recall that an extension field E of a field F is a finite extension if the vector space E over F has finite dimension.


    2. Relevant equations



    3. The attempt at a solution
    1) the converse is true (it was a theorem in my book). this direction is probably not true (or else it would have also been a theorem in my book). But I need a counterexample
    2) I am confused about the notation. My book has always denoted the ring of polynomials of a field f as F[x], never F(x). Furthermore, when a is an element of an extension field of F, then F(a) means F adjoined to a. But I have absolutely no idea what this means when x is an indeterminate?
    3) same as 2
    4) A field of characteristic 0 must contain a copy of the rationals. Does that help?
     
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  3. Feb 16, 2008 #2

    Hurkyl

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    For any field F, the notation F(x) denotes the field of all rational functions in x with coefficients in F. (Just like F(a), for a in an extension field, is the field of rational functions in a with coefficients in F)


    What sort of major theorems do you know about algebraic extensions and algebraic closures?
     
    Last edited: Feb 16, 2008
  4. Feb 16, 2008 #3
    So, you are saying F(x) is the field of quotients of F[x]?

    Also, F(a) means the smallest field that contains a and F. It is only the isomorphic to the field of quotients of F[x] if a is transcendental over a. If a is algebraic over F, the F(a) is is just F[x] evaluated at a. So, I guess by indeterminate, they really mean an element that is transcendental over F. It would be nice if they had said that explicitly or defined it somewhere.
    EDIT: actually my book did define that; I just skipped that part
    Please confirm that.

    1) Every field has an algebraic closure.
    2) A field F is algebraically closed iff every nonconstant polynomial in F[x] factors in F[x] into linear factors.
    3)Let E be an algebraic extension of a field F. Then there exist a finite number of elements, [tex]\alpha_1,\alpha_2,...\alpha_n[/tex] such that [tex]E = F(\alpha_1,...,\alpha_n)[/tex] iff E is a finite finite extension of F. So, I guess that the algebraic closure of Q would be a counterexample for 1 because you need to add [tex]\sqrt{p_i}[/tex] where p_i is the sequence of prime numbers and it is not hard to prove that there an infinite number of prime numbers. Please confirm that.
     
    Last edited: Feb 16, 2008
  5. Feb 17, 2008 #4

    Hurkyl

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    Yes, that is equivalent to what I said.


    That sounds like a possibility. Can you show that each [itex]\sqrt{p}[/itex] is not contained in Q adjoined with the square roots of other primes?
     
  6. Feb 17, 2008 #5
    Almost. Say [tex]\sqrt{p} = \sum_{i=0}^n q_i \sqrt{n_i}[/tex] where the q_i are rational and n_i can be 1 or have factors of primes that are not p. Squaring both sides gives you a rational equal to a finite sum of numbers, some of which must be irrational. Why must some of them be irrational? I am not really sure, but I think it has to do with the fact that you must have cross-terms when you square the RHS, but please help with that. Also please help me prove that a finite sum of irrational numbers cannot be equal to a rational number.
     
  7. Feb 17, 2008 #6

    Hurkyl

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    I can't; it's false. It's easy to find a counterexample if you work backwards.



    As for proving the [itex]\sqrt{p_i}[/itex] are algebraically independent -- at the moment I only have ideas that use more advanced mathematics. (e.g. p-adic analysis) While I suspect it might be instructive to work on this some more... for now I suggest seeking an easier way to show that [itex]\bar{\mathbb{Q}} / \mathbb{Q}[/itex] is not a finite extension.
     
  8. Feb 18, 2008 #7
    Hmmm. I wouldn't think these problems would require advanced mathematics or else it would be rather cruel to make them only TF. I can't really think of any way of proving [itex]\bar{\mathbb{Q}} / \mathbb{Q}[/itex] is not a finite extension without proving that an infinite sequence of rationals are algebraically independent. It seemed like the primes would be the easiest.
     
  9. Feb 18, 2008 #8

    morphism

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    [itex]\bar{\mathbb{Q}} / \mathbb{Q}[/itex] contains [itex]\mathbb{Q}(\sqrt[n]2)[/itex] as an intermediate field for each n>1.
     
  10. Feb 18, 2008 #9
    It sure seems like it would be easy to show that [tex]\sqrt[n]{2}[/tex] is not contained [tex]\mathbb{Q}(\sqrt[n-1]{2},...,\sqrt[2]{2})[/tex], but again I am stuck. I am not even sure how to begin. I would know how to find a basis for [tex]\mathbb{Q}(\sqrt[m]{2})[/tex] and there is that theorem that says that if K is a finite extension of E and E is a finite extension of F, then you just multiply the basis elements of K over E by those for E over F...but that doesn't seem like it will help here.
     
  11. Feb 18, 2008 #10

    morphism

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    That doesn't really matter. Just the fact that [itex]\mathbb{Q}(\sqrt[n]2)[/itex] sits between [itex]\bar{\mathbb{Q}}[/itex] and [itex]\mathbb{Q}[/itex] is enough to let you conclude that [itex][\bar{\mathbb{Q}}:\mathbb{Q}] \geq [\mathbb{Q}(\sqrt[n]2) : \mathbb{Q}][/itex], which in turn tells you...
     
  12. Feb 18, 2008 #11
    Why does [itex][\bar{\mathbb{Q}}:\mathbb{Q}] \geq [\mathbb{Q}(\sqrt[n]2) : \mathbb{Q}][/itex] let you conclude anything? It is obviously true since [tex]\sqrt[n]2 \in \bar{\mathbb{Q}}[/tex]. Do you mean to say [itex][\bar{\mathbb{Q}}:\mathbb{Q}] > [\mathbb{Q}(\sqrt[n]2) : \mathbb{Q}][/itex]?

    EDIT: never mind, I see what you are arguing; you're argument is the following:
    EDIT: x^n-2 is irreducible by Eisenstein's criterion which implies that deg[tex](\sqrt[n]2,\mathbb{Q})[/tex] is n
    EDIT: then take n to infinity. Very nice!
    EDIT: So 1 is F.
     
    Last edited: Feb 18, 2008
  13. Feb 18, 2008 #12
    Do you have any insight on 2-4?
     
  14. Feb 18, 2008 #13

    morphism

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    What exactly does it mean for a field to be algebraically closed in another one?

    For 4, what is the algebraic closure of, say, the field with 2 elements?

    Edit: And for 3, does 1/x have a square root in C(x)? [I.e. does the polynomial y^2 - 1/x in (C(x))[y] have a root in C(x)?]
     
    Last edited: Feb 18, 2008
  15. Feb 18, 2008 #14
    Sorry. Let E be an extension field of F. Then the algebraic closure of F in E is {[itex]\alpha \in E | \alpha[/tex] is algebraic over F}.

    F is algebraically closed in E if it is its own algebraic closure in E.
     
  16. Feb 29, 2008 #15
    Is it something like the subfield of C generated by {0,1,i,-i}? And what would the characteristic of that field be?

    Very nice example. Please confirm this proof:

    Since, [itex]\mathbb{C}[/tex] In element of [itex]\mathbb{C}(x)[/itex] must be expressible as
    [tex] \frac{\prod_{i=0}^n (x-c_i)^{q_i}}{\prod_{j=0}^m (x-c_j)^{q_j}} [/tex]
    where the c_j and the c_i are different.

    When you square that, the quotient will still be in "lowest terms". Thus we must have n=0. And the square of the denominator must be equal to x, but that is impossible because squaring multiplies the degree of the polynomial by two and 1 is not equal to two times anything.

    As for 2, I think it is obviously true because C is its own algebraic closure period which means that all the elements that are algebraic over C are contained in C, right?
     
    Last edited: Mar 1, 2008
  17. Mar 1, 2008 #16

    morphism

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    It's not going to be a subfield of C. Try to think about it this way: What is the characteristic of a field F, really? If it's 0, then F contains a copy of the rationals. Otherwise, if it's a prime p, then F contains a copy of Z/pZ. (Q and Z/pZ are, in the respective cases, the prime subfields of F.) Now, what can you say about the algebraic closure of Z/pZ?

    I think you have the right idea, but the way you're expressing it is kind of sketchy. Try this: Suppose p(x) and q(x) are polynomials in C[x] such that (p(x)/q(x))^2 = 1/x. Then x(p(x)^2) = q(x)^2, which is absurd by the degree argument you mentioned.

    Yup. In fact a field is algebraically closed iff it has no proper algebraic extension, or in your terminology, iff it's its own algebraic closure in any extension field.
     
  18. Mar 1, 2008 #17
    I have no idea what the algebraic closure of Z/pZ looks like! It is obviously a subfield of Q but which one? My book really should have done that example.

    EDIT: Actually the next section is called "Finite Fields" maybe I will learn that in there.
     
    Last edited: Mar 1, 2008
  19. Mar 1, 2008 #18

    morphism

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    It can't be a subfield of Q! The algebraic closure of Z/pZ will contain Z/pZ as a subfield, so its characteristic will be p.
     
  20. Mar 1, 2008 #19
    Sorry--you're right! I'll post back if I cannot figure it out after reading the next section although there should be a way to do this without using the material in the next section...
     
  21. Mar 1, 2008 #20
    I don't see how the statement "its characteristic will be p" follows from the fact that "The algebraic closure of Z/pZ will contain Z/pZ as a subfield".
     
    Last edited: Mar 1, 2008
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