MATH 601: Abstract Algebra II 3rd Homework Solutions Hungerford, Exercise V.3.6

Jonathan Rosenberg

assignment due Monday, February 19, 2001

(a) We are given f(x), g(x) relatively prime in K[x], such that \frac{f(x)}{g(x)} ∉ K. In particular, f(x) and g(x)

each have degree at least 1. Now x is a root of the polynomial φ(y) = (\frac{f(x)}{g(x)}) g(y) − f(y) ∈ K (\frac{f(x)}{g(x)}) [y],

since φ(x) = (\frac{f(x)}{g(x)}) g(x) − f(x) = 0, so x is algebraic over \frac{f(x)}{g(x)}. Furthermore, as a polynomial in y, φ has as its degree max(deg f, deg g).

Now (\frac{f(x)}{g(x)}) is transcendental over K, since if it were a root of a polynomial h(z) = z^{n} + a_{n−1}z^{n−1} + ⋯ + a_{0} ∈ K[z], we could clear denominators in

(\frac{f(x)}{g(x)})^{n} + a_{n−1} (\frac{f(x)}{g(x)})^{n−1} + ⋯ + a_{0} = 0

by multiplying by g(x)^{n} to get a polynomial equation with coefficients in K,

f(x)^{n} + a_{n−1}f(x)^{n−1}g(x) + ⋯ + a_{0} g(x)^{n} = 0,

satisfied by x, contradicting the fact that x is transcendental over K. So let z = \frac{f(x)}{g(x)}; we have K (\frac{f(x)}{g(x)}) = K(z), the field of rational functions in the transcendental element z. We want to show φ(y) = zg(y) − f(y) is irreducible in K(z)[y]. Note that in fact φ(y) ∈ K[z][y], and it’s primitive as a polynomial over K[z], since its coefficients are the coefficients of g, multiplied by z, and the coefficients of f, so the only common factors of all the coefficients are non-zero elements of the ground field K, which are units in K[z]. So by Gauss’s Lemma, φ can have a nontrivial factorization only if it factors in K[z][y] = K[y, z]. Since φ has degree 1 in z, if it had such a factorization φ(y) = φ_{1}(y)φ_{2}(y), then one could assume φ_{1} had degree 1 in z and φ_{2} had degree 0 in z, i.e., were an element of K[y]. But then we’d have φ_{2}(y) dividing zg(y) − f(y), hence dividing both g(y) and f(y), which is impossible unless φ_{2} is a constant, since f and g were assumed relatively prime. Thus φ is irreducible and x has degree precisely max(deg f, deg g) over K(z).

(b) Assume K ⊂≠ E ⊆ K(x) with E a field. Then E contains some element \frac{f(x)}{g(x)} of K(x) not in K.

Hence we can apply part (a) to conclude that [K(x) : K (\frac{f(x)}{g(x)})] < ∞, and so [K(x) : E] < ∞, since

K (\frac{f(x)}{g(x)}) ⊆ E.

(c) We saw above that any element z = \frac{f(x)}{g(x)} of K(x) which is not in K is transcendental over K. So K(z) ≅ K(x) and there is a monomorphism σ: K(x) → K(z) ⊆ K(x) which is the identity on K and sends x ↦ z. For any rational function h, this monomorphism sends h(x) ↦ h(y). Note that σ is a Kautomorphism of K(x) if and only if it is surjective. In this case, we have K(z) = K(x). Since, by part (b), [K(x) : K(z)] = max(deg f, deg g), we see σ is an automorphism of K(x) if and only of max(deg f, deg g) = 1.

(d) By part (c), Aut_{K} K(x) can be identified precisely with the maps x ↦ \frac{f(x)}{g(x)}, where max(deg f, deg g) = 1. Thus these are the maps

x ↦ \frac{ax + b}{cx + d}

where a, b, c, d ∈ K, a and c are not both 0, b ≠ 0 if a = 0 and d ≠ 0 is c = 0, and ax + b and cx + d are not multiples of each other. The criteria on a, b, c, d translate into saying that the vectors (a, b) and (c, d) in K^{2} are linearly independent, or that

det (\begin{matrix}a & b \\ c & d \end{matrix}) = ad − bc ≠ 0.