Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Basic transcendental field extension question
  1. Field extensions (Replies: 5)

Loading...