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!

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

https://brainmass.com/math/basic-algebra/modern-algebra-extension-field-41042

#### 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.

\$2.49