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Basic transcendental field extension question

  1. Jan 27, 2015 #1
    My textbook says:
    If u is transcnedetal over F, it is routine to verify that:
    F(u) = {f(u)g(u)^(-1) | f,g in F[x]; g /= 0}

    However, me being the scrubbiest of all scrubs does not understand what they did here.

    First of all, I don't understand why they needed to invert the g(u) function.

    It goes on to say that F(u) is isomorphic to F(x), the field of quotients of the integral domain F[x].

    Can somebody shed some light on this stuff for me?
  2. jcsd
  3. Jan 27, 2015 #2
    By definition [itex] F(u) [/itex] is the smallest field containing [itex] F [/itex] as well as [itex] u [/itex]. Since [itex] F(u) [/itex] is a field it is closed under multiplication and addition. So, if it contains [itex] u [/itex] then it contains [itex] u^n [/itex] for any [itex] n[/itex], hence it contains [itex] au^n[/itex] for any [itex] a\in F[/itex] and so must also contain all sums of these terms ie. every polynomial of the form [itex] f(u) [/itex]. Further, a field must contain inverses of all elements, so if it contains an expression [itex] f(u)[/itex] then it must contain the inverse [itex] f(u)^{-1} [/itex] as well. Finally, since we've shown it contains [itex] f(u)[/itex] and [itex] g(u)^{-1}[/itex], for any polynomials [itex] f,g [/itex] it must contain [itex] f(u)g(u)^{-1} [/itex] since it is closed under multiplication. This shows that [itex] F(u) [/itex] must contain the collection
    [tex] K=\left\{\frac{f(u)}{g(u)}~\bigg|~ f,g\in F[x], g\neq 0 \right\} [/tex]
    The fact that the [itex] u [/itex] is transcendental is used here in that [itex] g\neq 0[/itex] implies that [itex] g(u)\neq 0 [/itex] so that it makes sense to divide by [itex] g(u) [/itex] in the larger field containing [itex] u [/itex].

    To finish it off, you just need to check that this collection of quotients of polynomials is actually a field, and then it will necessarily be the smallest field containing both [itex] F [/itex] and [itex] u [/itex].

    Intuitively, the field of quotients of [itex] F[x] [/itex] is the smallest field containing [itex] F[x] [/itex] in which every nonzero element has an inverse. So it should be the set of all quotients of [itex] f(x)/g(x) [/itex] where [itex] g\neq 0 [/itex]. But since [itex] u [/itex] is transcendental, it satisfies no polynomial relations and so we can relabel [itex] x[/itex] by [itex] u [/itex] and we should still get the same thing.

    To make this somewhat rigorous, to construct the field of quotients of [itex] F[x] [/itex], you take the collection of all pairs [itex] (f(x),g(x) )[/itex] where [itex] g(x)\neq 0 [/itex] and quotient by the equivalence relation [itex] (f(x),g(x))\sim (f'(x),g'(x)) [/itex] when [itex] f(x)g'(x)-f'(x)g(x)=0 [/itex]. Of course the pair [itex] (f(x),g(x) ) [/itex] is intuitively supposed to represent the quotient f(x)/g(x) so there is an obvious map [itex] \phi:\mathrm{Quot}(F[x])\to K [/itex] defined by
    [tex] \phi(f(x),g(x) )=\frac{f(u)}{g(u)} [/tex]
    To prove that these two things are isomorphic, check that this is well defined and an isomorphism. The fact that this is well-defined follows from the fact that the equivalence relation here is the same as cross multiplying fractions and the fact that [itex] u [/itex] is transcendental so that [itex] g(u)\neq 0[/itex] for [itex] g\neq 0[/itex]. Similarly, the fact that it is a ring homomorphism is because the addition and subtraction in the quotient field are essentially defined to copy the addition/multiplication of fractions. Finally, surjectivity is obvious and injectivity follows using the fact that [itex] u [/itex] is transcendental.
    Last edited: Jan 27, 2015
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