1. Let F be a field generated over the field K by u and v of relatively prime degrees m and n, respectively, over K. Prove that [F:K] = mn.
Solution: Since F = K(u, v) K(u) K, where [K(u):K] = m and [K(u,v):K(u)] n, we have [F:K] mn. But [K(v):K] = n is a divisor of [F:K], and since gcd(m,n) = 1, we must have [F:K] = mn.
2. Let F K be an extension field, and let u be an element of F. Show that if [K(u):K] is an odd number, then K(u^{2}) = K(u).
Solution: Since u^{2} is in K(u), we have K(u) K(u^{2}) K. Suppose that u is not in K(u^{2}). Then x^{2}-u^{2} is irreducible over K(u^{2}) since it has no roots in K(u^{2}), so u is a root of the irreducible polynomial x^{2} - u^{2} over K(u^{2}). Thus [K(u):K(u^{2})] = 2, and therefore 2 is a factor of [K(u):K]. This contracts the assumption that [K(u):K] is odd.