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Showing that an element is not in a field extension

  1. May 2, 2017 #1
    1. The problem statement, all variables and given/known data
    Let ##K = \mathbb{Q} (1, a_1, a_2, \dots, a_n)##, which is the smallest field containing ##\mathbb{Q}## and ##a_1, a_2, \dots, a_n##, where each ##a_i## is the square root of a rational
    number. Show that the cube root of 2 is not an element of K.

    2. Relevant equations


    3. The attempt at a solution
    I need some pointers. I am thinking about making an argument with degrees. That is, looking at the degree of ##K## over ##\mathbb{Q}## and seeing how that relates to the degree of ##\mathbb{Q}(2^{1/3})## over ##\mathbb{Q}##, such as whether the latter is a divisor of the former.
     
  2. jcsd
  3. May 2, 2017 #2

    fresh_42

    Staff: Mentor

    Sounds good. Alternatively, by foot so to say, you could prove it by contradiction. Assume ##\sqrt[3]{2} \in \mathbb{Q}(a_1, \ldots ,a_n)##. This means, there is a polynomial ##f(x_1,\ldots , x_n) \in \mathbb{Q}[x_1,\ldots , x_n]## with ##f(a_1,\ldots , a_n)^3=2##. Now use the fact, that all powers of the ##a_i## are either ##0## or ##1##, even in ##f^3##, and deduce the contradiction.
     
  4. May 2, 2017 #3
    So just to be clear, is it correct to say that ##[\mathbb{Q}(a_1, \dots ,a_n) : \mathbb{Q}] = [\mathbb{Q}(a_1, \dots ,a_n) : \mathbb{Q}(a_1, \dots ,a_{n-1})] \cdots [\mathbb{Q}(a_1) : \mathbb{Q}] = 2^n##, while ##[\mathbb{Q}(2^{1/3}) : \mathbb{Q}] = 3##, and since the latter does not divide the former, ##2^{1/3} \not\in \mathbb{Q}(a_1, \dots ,a_n)##?

    Note that ##[\mathbb{Q}(a_1, \dots ,a_n) : \mathbb{Q}]## denotes the degree of ##\mathbb{Q}(a_1, \dots ,a_n)## over ## \mathbb{Q}##
     
  5. May 2, 2017 #4

    fresh_42

    Staff: Mentor

    Yes. Only a minor thought: If some ##\sqrt{a_i}## happen to be a rational itself or already contained in other ##\mathbb{Q}(a_i)##, which you haven't excluded, then the degree is only ##2^m < 2^n##. It doesn't change the argument, but it is a possibility.

    Btw.: This is basically the reason, why one cannot construct the third of an angle only by compass and ruler.
     
  6. May 2, 2017 #5
    Also, one last thing. Why must ##[\mathbb{Q}(2^{1/3}) : \mathbb{Q}]## divide ##[\mathbb{Q}(a_1, \dots ,a_n) : \mathbb{Q}]## in order for ##2^{1/3} \in \mathbb{Q}(a_1, \dots ,a_n)##? I want to make sure I have my reasons right
     
  7. May 2, 2017 #6

    fresh_42

    Staff: Mentor

    For a tower of algebraic extensions ##\mathbb{Q} \subseteq \mathbb{Q}(\sqrt[3]{2}) \subseteq \mathbb{Q}(a_1,\ldots ,a_n)## we have
    $$
    2^m = [\,\mathbb{Q}(a_1,\ldots ,a_n)\, : \, \mathbb{Q}\,] = [\,\mathbb{Q}(a_1,\ldots ,a_n)\, : \,\mathbb{Q}(\sqrt[3]{2})\,]\,\cdot \,[\,\mathbb{Q}(\sqrt[3]{2})\, : \,\mathbb{Q}\,]\\
    = [\,\mathbb{Q}(a_1,\ldots ,a_n)\, : \,\mathbb{Q}(\sqrt[3]{2})\,]\,\cdot \,3
    $$
    (https://en.wikipedia.org/wiki/Degree_of_a_field_extension#The_multiplicativity_formula_for_degrees)
    which is impossible. If you may not or don't want to use this, then you'll have to do the work (I think):

    An induction should work and is somehow clearer to work with, than to deal with all ##a_i## in one step.
    Just write ##\sqrt[3]{2} = r_0 + r_1a_1## and show that this is impossible. All powers of ##a_1## greater than ##1## can be reduced, because ##a_1^2=r\in \mathbb{Q}##.
     
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