Field extensions - rational functions

Question

Let K(x) be the field of rational functions in an indeterminate x. Show that every element of K[x] which is not in K is trancedental over K.

Answer 1

Herein we show that every $r(x) \in K(x) \setminus K$ is transcendental over $K$; this of course covers the case $r(x) \in K[x]$.

First of all, $x$ itself is transcendental over $K$ by definition; that is, there is no $h(x) \in K[x]$ with

$h(x) = \displaystyle \sum_0^n h_i x^i = 0. \tag 1$

Now suppose

$r(x) = \dfrac{p(x)}{q(x)} \in K(x) \setminus K \tag 2$

were not transcendental over $K$; here

$p(x), q(x) \in K[x], \tag 3$

and we may assume

$\gcd(p(x), q(x)) = 1; \tag 4$

furthermore, at least one of the two assertions

$p(x) \notin K, \; q(x) \notin K \tag 5$

must bind, lest $r(x) \in K$. Now $r(x)$, not being transcendental over $K$, must satisfy some polynomial

$R(x) = \displaystyle \sum_0^{\deg(R(x))} R_i x^i \in K[x], \tag 6$

with

$\deg(R(x)) \ge 1; \tag 7$

that is,

$R(r(x)) = 0; \tag 8$

(7) implies that the leading coefficient of $R(x)$,

$R_{\deg(R(x))} \ne 0, \tag 9$

and, taking $R(x)$ to be of minimal degree amongst all such polynomials saisfying (7)-(9), we may assume

$R_0 \ne 0 \tag{10}$

as well; otherwise, we could write

$R(x) = R_1(x)x^k \tag{11}$

for some $k$, $1 \le k \le \deg(R(x))$, and $R_1(x) \in K[x]$ with $\deg(R_1(x)) < \deg(R(x))$, and then

$R_1(r(x)) (r(x))^k = R(r(x)) = 0, \tag{12}$

which, since $K(x)$ is a field, yields

$R_1(r(x)) = 0, \tag{13}$

contradicting the minimality of $\deg(R(x))$. Now expanding out (8) using (6) yields

$\displaystyle \sum_0^{\deg(R(x))} R_i (r(x))^i = 0; \tag{14}$

we insert (2) into (9) to find

$\displaystyle \sum_0^{\deg(R(x))} R_i \left( \dfrac{p(x)}{q(x)} \right )^i = 0, \tag{15}$

or

$\displaystyle \sum_0^{\deg(R(x))} R_i \dfrac{(p(x))^i}{(q(x))^i} = 0; \tag{16}$

recalling

$\deg(R(x)) \ge 1, \tag{17}$

we multiply (11) through by $(q(x))^{\deg(R(x))}$, obtaining

$\displaystyle \sum_0^{\deg(R(x))} R_i (q(x))^{\deg(R(x))}\dfrac{(p(x))^i}{(q(x))^i} = 0, \tag{18}$

or

$\displaystyle \sum_0^{\deg(R(x))} R_i (q(x))^{\deg(R(x) - i} (p(x))^i = 0; \tag{19}$

if $q(x) \in K$, then by (5), $p(x) \notin K$, $\deg(p(x)) \ge 1$, and, by virtue of the fact that $R_{\deg(R(x))} \ne 0$, (19) is a polynomial in $x$ of degree $(\deg(p(x)))(\deg(R(x))) \ge 1$, which contradicts the nonexistence of $h(x)$ as in (1); similarly, if $p(x) \in K$,then $q(x) \notin K$, and, now via $R_0 \ne 0$, (19) is of degree $(\deg(q(x)))(\deg(R(x))) \ge 1$, again a contradiction to the fact that $x$ is transcendental over $K$; so then in the case $p(x), q(x) \notin K$ we write (19) in the form

$R_{\deg(R(x))} (p(x))^{\deg(R(x))} = - \displaystyle \sum_0^{\deg(R(x)) - 1} R_i (q(x))^{\deg(R(x) - i} (p(x))^i, \tag{20}$

whence, since $R_{\deg(R(x))} \ne 0$, is equivalent to

$ (p(x))^{\deg(R(x))} = - R_{\deg(R(x))}^{-1} \displaystyle \sum_0^{\deg(R(x)) - 1} R_i (q(x))^{\deg(R(x) - i} (p(x))^i; \tag{21}$

now let $q_1(x) \in [x]$ be an irreducible, hence prime (since $K[x]$ is a principal ideal domain), factor of $q(x)$; then $\deg q_1(x) \ge 1$ and, since $q(x)$ is a factor in every term on the right of (21), we conclude that

$q(x) \mid (p(x))^{\deg(R(x))}, \tag{22}$

and hence that

$q_1(x) \mid (p(x))^{\deg(R(x))}; \tag{23}$

since $q_1(x)$ is prime this yields

$q_1(x) \mid p(x); \tag{24}$

we now have a $q_1(x)$ with $\deg(q_1(x)) \ge 1$ and

$q_1(x) \mid q(x), q_1(x) \mid p(x), \tag{24}$

in contradiction to (4). Thus there is no rational function $r(x) \in K(x)$ satisfying (6)-(8); non-constant rational functions must be transcendental over $K$.