Explore BrainMass

Explore BrainMass

    Extension field proof

    Not what you're looking for? Search our solutions OR ask your own Custom question.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    Please see attachment.

    P/S: To show subfield please show
    a) closed under addition, and multiplication
    b) additive identity and additive inverse
    c) closed under reciprocal

    © BrainMass Inc. brainmass.com March 4, 2021, 6:20 pm ad1c9bdddf


    Solution Preview

    Due to the limits of the typing system, I'll be using Roman letters instead of Greek:
    t = tau, a = alpha, B = beta
    Also no subscripts, a sub zero is a0; and superscripts, x to the exponenet of 3 is x^3. Multiplication is *
    root2 = square root of 2

    1. Let E be an extension field of F. As discussed in class, let G = Aut[F]E = {automorphisms of E that fix F}. If t is an element of G, show that E^t = {x element of E| t(x) = x} is a subfield of E that contains F. Do not assume that E is algebraic or finite over F.

    (As a result, if H is a subgroup of G, then E^H = {x element of E|t(x) = x for all t element of H} = intersection over all t of E^t is a subfield of E that contains F. You needn't show this.)

    Since by definition t fixes F, then for every x element of F, t(x) = x, so certainly F is a subset of E^t, or equivalently E^t contains F. All we need to show then is that E^t is a field.

    Automorphism are a special case of isomorphisms, and isomorphisms are a special case of homomorphisms, so we know that all t elements of G follow all the rules for homomorphisms of fields; specifically we know t(u + v) = t(u) + t(v) and t(u*v) = t(u) * t(v)

    If x1, x2 are elements of E^t, then t(x1 + x2) = t(x1) + t(x2) by homomorphism = x1 + x2 ...

    Solution Summary

    This is a proof regarding extension fields. Cartesian and extension fields are analyzed.