## 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 $i$th 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 $q(x)$.