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Extension field proof

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


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.