On the degree of an extension of F(x)

Let F be a field, and consider the field F(x) of rational functions over F (that is, the field of fractions of the domain F[x]). Let t(x) = p(x)/q(x), with p,q \in F[x] and q \neq 0 such that the degree of q is strictly larger than the degree of p. In this exercise, we will compute the degree of F(x) over F(t(x)). (Note that if p has degree larger than or equal to the degree of q, then we can use the division algorithm to find p(x) = q(x)b(x) + r(x), and then F(t(x)) = F(r(x)/q(x)).)

  1. Show that p(y) - t(x)q(y) \in F(t(x))[y] is irreducible over F(t(x)) and has x as a root (in the extension F(x)).
  2. Show that the degree of p(y) - t(x)q(y) as a polynomial in y is the maximum of the degrees of p and q.
  3. Conclude that [F(x):F(t(x))] = \mathsf{max}(\mathsf{deg}\ p(x)), \mathsf{deg}\ q(x)).

(Fields of rational functions were introduced in Example 4 on page 264 of D&F.)

We claim that t(x) is indeterminate over F– that is, that t(x) does not satisfy any polynomial over F. Indeed, if \sum_{i=0}^n c_i t(x)^i = \sum c_i \dfrac{p(x)^i}{q(x)^i} = 0, then we have \sum c_i p(x)^iq(x)^{n-i} = 0. Let d_p and d_q denote the degrees of p and q, respectively. Since F has no zero divisors, the degree of the ith summand is d_pi + d_q(n-i). Suppose two summands have the same degree; then d_pj + d_q(n-i) = d_pj + d_q(n-j) for some i and j, which reduces to (d_p-d_q)i = (d_p-d_q)j. Since (as we assume) d_p \neq d_q, we have i = j. In this case, we can pick out the summand with the highest degree, and be guaranteed that no other summands contribute to its term of highest degree. This gives us that c_i = 0 for the highest degree summand; by induction we have c_i = 0 for all the coefficients c_i, a contradiction. (In short, no two summands have the same highest degree term. Starting from the highest of the highest degree terms and working down, we show that each c_i is 0.)

So t(x) is indeterminate over F, and in fact F[t(x)] is essentially a polynomial ring whose field of fractions is F(t(x)). By Gauss’ Lemma, the polynomial p(y) - t(x)q(y) is irreducible over F(t(x)) if and only if it is irreducible over F[t(x)]. Now F[t(x)][y] = F[y][t(x)], and our polynomial is irreducible in this ring since it is linear in t(x). So in fact p(y) - t(x)q(y) is irreducible over F(t(x)), and moreover has x as a root.

The degree of p(y) - t(x)q(y) in y is the maximum of the degrees of p and q because the coefficient of each term (in y) is a linear polynomial in t(x), which is nonzero precisely when one of the corresponding terms in p or q is nonzero. That is, we cannot have nonzero terms in p(y) and -t(x)q(y) adding to give a zero term.

To summarize, p(y) - t(x)q(y) is an irreducible polynomial over t(x) with x as a root, and so must be (essentially) the minimal polynomial of F(x) over F(t(x)). By the preceding paragraph, the degree of this extension is the larger of the degrees of p(x) and q(x).

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