Automorphism of the Field of rational functions

Let $K$ be a field and let $K(x)$ be the field of rational functions in $x$ whose coefficients are in $K$. Let $\theta(x)$ $\in \operatorname{Aut}(K(x))$ such that $\theta|_K = \operatorname{id}_K$. Show that $\theta(x) =\frac{ax+b}{cx+d}$, with $ad\neq bc$.

Here is my attempt.

Let $\theta(x) = \frac{f}{g}$, $f,g \in K[x]$, with $\gcd(f, g)=1$. let $h \in K(x)$. Then $h(\frac{f}{g}) = x$ and $\frac{f(x)}{g(x)}\neq\frac{f(y)}{g(y)}$ if $x\neq y$ . Suppose $\deg f \gt 1$ or $\deg g \gt 1$, then the equation $g(x)b=f(x)$, $b\in K$, will have more than one solution for $x$. Hence $f$ and $g$ have degrees at most $1$ and thus $\theta(x)=\frac{ax+b}{cx+d}$ with inverse function $\theta(x) =\frac{dx-b}{a-cx}$.

What I'd like to know is if Ive approached the problem in the right manner. Please any input will be very much appreciated. Thanks.

Solution 1:

It is unclear to me on what grounds you claim that $g(x)b=f(x)$ will have more than one solution for $x$. Even assuming that $K$ were algebraically closed, how do you know that you don't have one solution which is a multiple solution? For a very simple example, consider the case of $K=\mathbf{F}_2$, the field with two elements, and take $g(x) = x^2+x+1$ and $f(x)=x$. Then $g(x)0 = f(x)$ has one solution ($x=0$) and $g(x)1 = f(x)$ has one solution ($x=1$), so it does not follow merely from the degrees of $f$ and $g$: there is more work to be done. I don't know for sure if you can push it through, but you are correct that the first step is showing that the degrees have to be at most one.

I know two proofs: a direct way and a clever way. The direct way is to start as you do, with $\theta(x)=\frac{f(x)}{g(x)}$, with $f$ and $g$ coprime. Then consider an $\alpha = p(x)/q(x)$, with $p$ and $q$ coprime. If you write $f(x) = a_nx^n+\cdots +a_0$ and $g(x)=b_nx^n+\cdots+b_0$, with $a_n$ and $b_n$ not both zero, and you write $p(x) = c_mx^m+\cdots+c_0$, $q(x) = d_mx^m+\cdots+d_0$, with $c_m$ and $d_m$ not both zero, then you can think about the degrees of the numerator and denominator of $\theta(\alpha)$. Try to think about what you would need for the degrees to drop after applying $\theta$ in order to conclude information about what can map to $x$ under $\theta$. From there what you want will follow. This proof basically amounts to working things through in the obvious manner until you get to where you want to go.

Details Added. Pick any $\alpha=p(x)/q(x)$ that is written as above; then $$\begin{align*} \theta(\alpha) &= \frac{c_m\theta(x)^m + \cdots + c_0}{d_m\theta(x)^m+\cdots + d_0}\\ &= \frac{c_mf^mg^0 + \cdots + c_0g^m}{d_mf^mg^0+\cdots + d_0g^m}. \end{align*}$$ I claim that this is written in lowest terms: since $p$ and $q$ are coprime, we can find polynomials $r(x)$ and $s(x)$ with coefficients in $K$ such that $r(x)p(x)+q(x)r(x) = 1$. Applying $\theta$ and clearing denominators, we obtain a linear combination of $$c_mf^mg^0+\cdots + c_0g^m\quad\text{and}\quad d_mf^mg^0+\cdots+d_0g^m$$ which is equal to some power of $g$; thus, if there is any common factor among the numerator and denominator, it must be a divisors of a power of $g(x)$. But let $k(x)$ be a divisors of $g$ that divides both numerator and denominator of $\theta(\alpha)$. Since either $c_m\neq 0$ or $d_m\neq 0$, then $k(x)$ must divide $f^m(x)$. Since $g(x)$ and $f(x)$ are relatively prime, it follows that $k(x)=1$. Thus, the rational function $\theta(\alpha)$ is already expressed in lowest terms.

The degree of the numerator of this fraction is at most $mn$, and is strictly smaller than $mn$ if and only if the coefficient of $x^{mn}$, which is $c_ma_n^m +c_{m-1}a_n^{m-1}b_m+ \cdots + c_0b_n^m$, vanishes. The denominator has degree at most $mn$, and the degree is strictly smaller if and only if the coefficient of $x^{mn}$ in the expression, which is $d_ma_n^m + \cdots + d_0b_n^m$, is equal to $0$.

Suppose first that $b_n\neq 0$. Then the numerator has degree strictly smaller than $mn$ if and only if $\frac{a_n}{b_n}$ is a root of $p(x)$; and the degree of the denominator is strictly less than $mn$ if and only if $\frac{a_n}{b_n}$ is a root of $q(x)$. Since these two polynomials are coprime, they cannot have a common root; thus, we cannot have both numerator and denominator of degree strictly less than $mn$, so at least one of them has degree exactly $mn$.

Suppose next that $b_n=0$. Then the numerator has degree strictly smaller than $mn$ if and only if $c_ma_n^m=0$, which requires $c_m=0$ since $b_n$ and $a_n$ are not both equal to zero. Similarly, the degree of the denominator is strictly smaller than $mn$ if and only if $d_ma_n^m=0$, which requires $d_m=0$. But since we cannot have both $c_m$ and $d_m$ equal to zero, at least one of the numerator and denominator has degree exactly $mn$.

Since $\theta$ is assumed to be an isomorphism, there has to be an $\alpha$ such that $\theta(\alpha)=x$. Proceeding as above, since the expression is written in lowest terms and the degree of at least one of the numerator and denominator is $mn$, it follows that we must have $mn=1$, or $m=n=1$. That is, $\theta$ must be defined as $$\theta(x) = \frac{ax+b}{cx+d},$$ with $ax+b$ and $cx+d$ coprime, $a$ and $c$ not both equal to zero.

From here it is straightforward to finish the problem. $\Box$

The second proof I know is the "clever proof". Again, take $\theta(x)=\frac{f(x)}{g(x)}$ as you do. Note that $\theta(x)\notin K$. Show that $\theta(x)$ is transcendental over $K$ by showing that $x$ is algebraic over $K(\theta(x))$, and find its irreducible polynomial. Then use the fact that $K(\theta(x))$ is isomorphic to $K(x)$ to deduce that the irreducible polynomial must be of degree $1$. The irreducible polynomial turns out to be closely related to $f$ and $g$, of course. (I also like the fact that Gauss's Lemma comes into play to show that the obvious polynomial satisfied by $\theta(x)$ is in fact irreducible in the field you are looking at).

The argument proceeds as follows: again, pick any $\beta = p(x)/q(x)$ written in lowest terms. The polynomial $\beta q(y) - p(y)\in K(\alpha)[y]$ has $x$ as a root, so $x$ is algebraic over $K(\beta)$ (since $p$ and $q$ are coprime, the polynomial is not equal to $0$). It follows that $\alpha$ is transcendental over $K$. I claim the polynomial $\beta q(y)-p(y)$ is irreducible in $K(\beta)[y]$. Indeed, this polynomial is irreducible over $K(\beta)[y]$ if and only if it is irreducible over $K[\beta][y]$ by Gauss's Lemma; but $K[\beta][y]=K[y][\beta]$, and this polynomial is trivially irreducible over $K[y][\beta]$ because it is linear in $\beta$ and the two coefficients are relatively prime in $K[y]$. Therefore, $[K(x):K(\beta)] = \max(\deg(p),\deg(q))$.

So, if $\theta$ defined by $\theta(x)=\beta$ gives an automorphism, then we must have $1=[K(x):K(\beta)] = \max(\deg(p),\deg(q))$. Hence $\theta(x) = \frac{ax+b}{cx+d}$ and the argument finishes as before. $\Box$

Solution 2:

This and related results are frequently called "Lüroth's Theorem". Searching on that term will turn up much of interest, e.g. this stimulating exercise set of George Bergman, which I append below for those who may not have easy access to a postscript viewer. See also the MO thread on elementary proofs.