Show that field extensions $K(\alpha)$ and $K(\frac{a\alpha + b}{c\alpha + d})$ are equal

Question

In my Algebra/Galois theory lecture I encountered the following statement about field extensions:

Let $L|K$ be a field extension, $\alpha\in L\setminus K$ and $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in Gl_n(K)$. Then $K(\alpha) = K\left(\frac{a\alpha + b}{c\alpha + d}\right)$.

The Inclusion $K\left(\frac{a\alpha + b}{c\alpha + d}\right) \subset K(\alpha)$ is obvious to me. But I struggle with showing the opposite direction. I really can't find a way to write $\alpha$ as a term involving the more complicated fraction.

Can anyone help me here?

Answer 1

To finish Jyrki's comment...

Fix a matrix $A \in Gl_2(K)$. Notice that we have an action of $A$ on $L$ given by $A \cdot \alpha := \tfrac{a\alpha + b}{c\alpha + d}$. Given $\alpha \in L - K$, we can consider the element $t := A \cdot \alpha \in L$. Our goal is to compare $K(t)$ with $K(\alpha)$.

• Let $F \subseteq L$ be a field containing $K \cup \{\alpha\}$. It will also contain $A \cdot \alpha = t$, since $a\alpha + b \in F$, $c \alpha + d \in F - \{0\}$, and $\tfrac{a\alpha+ b}{c\alpha + d} \in F$. Thus $K \cup \{t\} \subseteq F$.
• Let $F \subseteq L$ be a field containing $K \cup \{t\}$. Notice that $tc-a \in F$, $b-td \in F$, and if $tc - a \neq 0$ then $\tfrac{b-td}{tc-a} = \alpha \in F$. If $tc = a$, then we would have $$ tc = \frac{ac \alpha + bc}{c\alpha + d} = a \implies ac\alpha + bc = ac\alpha + ad \implies bc = ad \implies \det(A) = 0, $$ which cannot happen by assumption. Thus $K \cup \{\alpha\} \subseteq F$.

Since every field containing $K \cup \{\alpha\}$ must contain $K \cup \{t\}$ and vice versa, we have that $K(t) = K(\alpha)$.

After this, one probably would want to explore the following.

Exercise: Let $G$ be a group acting on $L$ which fixes $K$, in the sense that $G \cdot K \subseteq K$. Is it true that for any $g \in G$ and any $\alpha \in L-K$ we have $K(g \cdot \alpha) = K(\alpha)$?

Answer 2

It may help to know that in the projective line $\mathbb PL^1$, for any invertible $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\operatorname{GL}_2(L)$, the map $\varphi_A:x\mapsto\frac{ax+b}{cx+d}$ is called a Möbius transformation, and it is invertible with $(\varphi_A)^{-1}=\varphi_{A^{-1}}$. You can check this yourself.

Anyway, you can check that if $\alpha\in L$ and $A\in\operatorname{GL}_2(K)$ and $\beta\in K(\alpha)$, then $\varphi_A(\beta)\in K(\alpha)$. Use these in conjunction with $\varphi_A^{-1}=\varphi_{A^{-1}}$ to show that $\alpha\in K(\varphi_A(\alpha))$.