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.

Use the method found in the proof below to find a primitive element for the extension Q(i, 5^(1/4)) over Q.
*Recall that the extensionfield F of K is called a simple extension if there exists an element In this case, u is called a primitive element.
**Theorem: Let F be a finite extension of the field K. If F is separabl

Argue that every finite extenstion field of R is either R itself or is isomorphic to C.
Note: R is set of all real numbers
C is set of all complex numbers

Let F be an extension field of K. Clearly F is a vector space over K. Let u be an element of F. Show that the subspace spanned by {1, u, u^2, ...} is a field IF and ONLY IF (iff) u is algebraic over K.
Let S be the subspace of F.
Hint for the "-->" of proof. If S is a field and u is not equal to 0, then 1/u is in S. What d

Suppose that L has transcendence degree n over K and that L is algebraic over K(α1, . . . , αn). Show that α1, . . . , αn is a transcendence basis for L over K.
Might help:
Theorem - Definition: Let L be an extension of K, A a subset of L. The following are equivalent:
(1) A is a maximal algebraically

Please see the attachment to see these questions properly.
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Question 1
If [K:F] is finite and u is algebraic over K ,prove that [F(u):F] divides [K(u):F]
Hint:[F(u):F] and [K(u):F(u)] are finite by Theorems 10.4,10.7 and 10.9
Apply Theorem 10.4 to
Theorem 10.4
Le