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§ 6.2 Finite and algebraic extensions: solved problems

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(u2) = K(u).

Solution: Since u2 is in K(u), we have K(u) K(u2) K. Suppose that u is not in K(u2). Then x2-u2 is irreducible over K(u2) since it has no roots in K(u2), so u is a root of the irreducible polynomial x2 - u2 over K(u2). Thus [K(u):K(u2)] = 2, and therefore 2 is a factor of [K(u):K]. This contracts the assumption that [K(u):K] is odd.