Fixed field of $\mathbb Q$-Automorphism of $\mathbb Q(X)$

Question

I have a few questions regarding this example in my textbook:

Let $L=\mathbb Q(X)$ and $K=\mathbb Q$. Let's look at the group $G$ generated by the $\mathbb Q$-automorphism $$\tau_a:L\to L,\ \frac{p}{q}\mapsto\frac{p(xa)}{q(a)}$$ for a fixed $a\in\mathbb Q\backslash\{0,\pm 1\}$. Then, where $\mathcal F(\cdot)$ is denoting the fixed field, for $p/q\in\mathcal F (G)$ with $\gcd(p,q)=1$ we have $p(ax)q(x)=p(x)q(ax)$ for all $x\in\mathbb Q$. Now, a comparison of the coefficients yields $\deg(p)=\deg(q)$ since $a\neq 0,\pm 1$.Furthermore, from $\gcd(p,q)=1$ we obtain $\gcd(p(ax),q(ax))=1$ and thus there must be a constant $b\in\mathbb Q$ such that $p(ax)=b\cdot p(x)$. So we have $p(x)=c\cdot x^{\deg(p)}$ and $q(x)=dx^{\deg(p)}$. In total this means that $\deg(p)=\deg(q)=0$, so $p/q\in\mathbb Q$.

1. Why does the comparison yield $\deg(p)=\deg(q)$?
2. Why does there exist such a $b$?
3. Why is $p$ of the claimed form then?

Answer 1

First I guess we have $$\tau_a:L\to L,\ \frac{p}{q}\mapsto\frac{p(ax)}{q(ax)}$$

as only then having $p(x)q(ax) = p(ax)q(x)$ makes sense. Nevertheles at the beginning let $p(x) = b_nx^n + \cdots + b_1x+b_0$ and $q(x) = c_mx^m + \cdots + c_1x+c_0$.

Regarding your first question compare the leading coefficients to get the answer. You will get that the leading coefficient of the left side is $b_nc_ma^n$, while the leading coefficient of the right side is $b_nc_ma^m$. Direct comparison yields $a^m = a^n$, as $b_n, c_m \not = 0$. As $a\not=0$ we have that $a^{m-n} = 1$, which means that $m=n$, as $|a| \not = 1$.

For the second question note that $\gcd(p(ax),q(ax)) = 1$. Also from the relation we have that $p(ax) \mid p(x)q(ax)$. From above we must have that $p(ax) \mid p(x)$. So $\exists f(x) \in L$ s.t. $p(x) = f(x)p(ax)$. However $p(ax)$ and $p(x)$ have the same degree so we have that $\deg f =0$ and so $p(x) = b\cdot p(ax)$ for some $b \in \mathbb{Q}$. Multiply both sides by $b^{-1}$ to get the wanted result.

For the final part again it comes down to comparing coefficients. We have that

$$p(ax) = b \cdot p(x)$$ $$b_na^nx^n + b_{n-1}a^{n-1}x^{n-1} + \cdots+ b_1ax + b_0 = bb_nx^n + bb_{n-1}x^{n-1} + \cdots + bb_1x + bb_0$$

Now as $b_n \not = 0$ we must have $b=a^n$. However for the next ones we have that $b_ia^i = bb_i$ and as $a^i \not = b$ we must have that $b_i=0$ for $i < n$. Hence $p(x) = b_nx^n$. Similar reasoning yields that $q(x) = c_mx^m = c_nx^n$ and hence:

$$\frac{p(x)}{q(x)} = \frac{b_nx^m}{c_nx^n} = \frac{b_n}{c_n} \in \mathbb{Q}$$

Answer 2

Well, your automorphism of $\Bbb Q(X)$ seems to be translation by $a$, namely the automorphism of your field induced by $X\mapsto X+a$. But the only rational functions that are translation-invariant are the constants. I see that in the body of your analysis of the situation, you seem to speak instead of the automorphism that sends $X$ to $aX$. But no matter, everything I say below applies to that situation just as well, so long as $a\ne\pm1$. The only rational functions $f$ with $f(X)=f(aX)$ are equally the constants.

Indeed, if the invariant subfield were anything other than $\Bbb Q$ itself, such field would be a transcendental extension of $\Bbb Q$ in $L$, so necessarily $\Bbb Q(f)$ for a single rational function $\,f$, and the field extension degree $\bigl[\Bbb Q(x):\Bbb Q(f)\bigr]$ will be the maximum of the degrees of the numerator and the denominator of $f$, in particular a finite number. But the field extension in question can have only finitely many automorphisms, while your group is infinite.

Thus the answers to your specific questions are:
(1) The degrees of your $p$ and $q$ are equal because they must both be zero;
(2) There’s such a $b$ because $p$ and $q$ are already constants; and
(3) The unusual form arises from the fact that $\deg(p)=0$.