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 extension field 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 separable over K, then it is a simple extension of K.
Proof: Let F be a finite separable extension of K. If K is a finite field, then F is a finite field. We must also have F = K(u) for any generator u of the cyclic group , the non-zero elements of the field. If we can prove this result in the case , then it is clear how to extend it by induction to the case . Thus we may assume that K is infinite and Let f(x) and g(x) be the minimal polynomials of u and v, and assume their degrees to be m and n respectively. Let E be an extension of F over which both f(x) and g(x) split. Since F is separable, the roots and of f(x) and g(x) are distinct. If , then the equation has a unique solution . Therefore, since K is infinite, there must exist an element We will show that for It is clear that If we can show that , then it will easily follow that , giving us the desired equality. Our strategy is to show that the minimal polynomial p(x) of v over K(t) has degree 1, which will force v to belong to K(t). Let This polynomial has coefficients in K(t) and has v as a root since Since we also have g(v) = 0, it follows that p(x) is a common divisor of h(x) and g(x). Now we consider all three polynomials p(x), h(x), and g(x) over the extension field E. Since it follows that Thus vj is not a root of h(x), for j = 2,3,...,n we can conclude that over E we have gcd(h(x),g(x)) = x - v. But p(x) is a common divisor of h(x) and g(x) over E as well as over K(t), and so p(x) must be linear.
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Use the method found in the proof below to find a ...
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