Trancendental extension Galois group Let $K$ be a field and consider the extension $K(X)$ of rational functions with coefficients in $K$. It is common knowledge that $\text{Gal}(K(X)/K)$ is isomorphic to the group ${PL }_2(K)$, which is the quotient group of matrices in ${GL}_2(K)$ which identifies a matrix with its scalar multiple. I was doing some calculations in this group to verify my understanding, and I needed some help to double check my work. Is the map
$$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \mapsto \left(X \mapsto \frac{aX + b}{cX + d} \right) $$
a homomorphism or an antihomomorphism into the Galois group?
 A: An antihomomorphism "switches the order of multiplication." If we want to investigate whether that occurs with the map you describe, we can just try out the multiplication. Let $\operatorname{M}_1$, $\operatorname{M}_2\in PL_2(K)$, and $\phi$ be the map you describe:
$$ \operatorname{M}_1\operatorname{M}_2=\begin{pmatrix} a_1 & b_1 \\ c_1 & d_1 \end{pmatrix} \begin{pmatrix} a_2 & b_2 \\ c_2 & d_2 \end{pmatrix}=\begin{pmatrix} a_1a_2+b_1c_2 & a_1b_2+b_1d_2 \\ c_1a_2+d_1c_2 & c_1b_2+d_1d_2 \end{pmatrix}$$
$$\phi(\operatorname{M}_1\operatorname{M}_2)= \left(X \mapsto \frac{(a_1a_2+b_1c_2 )X + (a_1b_2+b_1d_2)}{(c_1a_2+d_1c_2 )X + (c_1b_2+d_1d_2)} \right) $$
Compared to the multiplication on the other end of the map:
$$\phi(\operatorname{M}_1)\phi(\operatorname{M}_2)=\left(X \mapsto \frac{a_1X + b_1}{c_1X + d_1} \right) \circ\left(X \mapsto \frac{a_2X + b_2}{c_2X + d_2} \right) = \left(X \mapsto \frac{a_1\frac{a_2X + b_2}{c_2X + d_2} + b_1}{c_1\frac{a_2X + b_2}{c_2X + d_2} + d_1} \right)=\left(X \mapsto \frac{a_1(a_2X + b_2) + b_1(c_2X + d_2)}{c_1(a_2X + b_2) + d_1(c_2X + d_2)}\right)=\left(X \mapsto \frac{(a_1a_2+b_1c_2 )X + (a_1b_2+b_1d_2)}{(c_1a_2+d_1c_2 )X + (c_1b_2+d_1d_2)} \right)$$
Multiplication doesn't get turned around, so the map is a homomorphism. You can be reassured of this fact here.
