Exercise 3.1.11 (Homomorphism $\psi : \begin{pmatrix} a & b\\0&c \end{pmatrix} \mapsto (a,c)$ from $G$ onto $F^\times \times F^\times$)

Proof. Let

We know that $G$ is a subgroup of ${\mathrm{GL}}_2(F)$.

(a)

Consider the map $$\phi \{\begin{array}{ccc}\hfill G\hfill & \hfill \to \hfill & F^{\text{×}}\hfill \\ \hfill \left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill c\hfill \end{array}\right)\hfill & \hfill \mapsto \hfill & a\hfill \end{array}.$$

(Since $\mathit{ac}\ne 0$ for every $\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill c\hfill \end{array}\right)\in G$, then $a\ne 0$ so $a\in F^{\text{×}}$.)

Let $A=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill c\hfill \end{array}\right)$ and $B=\left(\begin{array}{cc}\hfill a^{\prime }\hfill & \hfill b^{\prime }\hfill \\ \hfill 0\hfill & \hfill c^{\prime }\hfill \end{array}\right)$ be elements of $G$. Then

$$\begin{array}{lll}\hfill \mathit{AB}=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill c\hfill \end{array}\right)\left(\begin{array}{cc}\hfill a^{\prime }\hfill & \hfill b^{\prime }\hfill \\ \hfill 0\hfill & \hfill c^{\prime }\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill a{a}^{\prime }\hfill & \hfill a{b}^{\prime }+b{c}^{\prime }\hfill \\ \hfill 0\hfill & \hfill c{c}^{\prime }\hfill \end{array}\right).& & \hfill \text{(1)}\end{array}$$

Therefore

$$\phi (\mathit{AB})=a{a}^{\prime }=\phi (A)\phi (B),$$

so $\phi$ is a homomorphism.

If $a\in F\ast$, then $a=\phi (M)$, where $M=\left(\begin{array}{cc}\hfill a\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\in G$, so $\phi$ is a surjective homomorphism.

Let $A=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill c\hfill \end{array}\right)\in G$. Then

$$A\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )⟺a=1.$$

So

$$\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=\left\{\left(\begin{array}{cc}\hfill 1\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill c\hfill \end{array}\right)\mid b,c\in F,c\ne 0\right\}.$$

More generally, if $\alpha \in F^{\text{×}}$, then $A\in {\phi }^{-1}(a)⟺a=\alpha$, so the fiber above $\alpha$ is

$${\phi }^{-1}(\{\alpha \})=\left\{\left(\begin{array}{cc}\hfill \alpha \hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill c\hfill \end{array}\right)\mid b,c\in F,c\ne 0\right\}.$$

(This is also the translate $\left(\begin{array}{cc}\hfill \alpha \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\cdot \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$.)

(b)

Consider now the map $$\psi \{\begin{array}{ccc}\hfill G\hfill & \hfill \to \hfill & F^{\text{×}}\times F^{\text{×}}\hfill \\ \hfill \left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill c\hfill \end{array}\right)\hfill & \hfill \mapsto \hfill & (a,c)\hfill \end{array}.$$

(Since $\mathit{ac}\ne 0$, $(a,c)\in F^{\text{×}}\times F^{\text{×}}$.)

Let $A=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill c\hfill \end{array}\right)$ and $B=\left(\begin{array}{cc}\hfill a^{\prime }\hfill & \hfill b^{\prime }\hfill \\ \hfill 0\hfill & \hfill c^{\prime }\hfill \end{array}\right)$ be elements of $G$. Then by(1)

$$\psi (\mathit{AB})=(a{a}^{\prime })(c{c}^{\prime })=(\mathit{ac})(a^{\prime }{c}^{\prime })=\psi (A)\psi (B),$$

so $\psi$ is a homomorphism.

Let $(a,c)\in F^{\text{×}}\times F^{\text{×}}$. Then $(a,c)=\psi (M)$, where $M=\left(\begin{array}{cc}\hfill a\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill c\hfill \end{array}\right)\in G$, so $\psi$ is a surjective homomorphism.

Let $A=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill c\hfill \end{array}\right)\in G$. Then

$$A\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )⟺a=c=1,$$

so

$$\ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )=\left\{\left(\begin{array}{cc}\hfill 1\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\mid b\in F\right\}.$$

More generally the fiber above $(\alpha ,\beta )\in F^{\text{×}}\times F^{\text{×}}$ is

$${\phi }^{-1}(\alpha ,\beta )=\left\{\left(\begin{array}{cc}\hfill \alpha \hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill \beta \hfill \end{array}\right)\mid b\in F\right\}.$$

(c)

Let $H=\left\{\left(\begin{array}{cc}\hfill 1\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\mid b\in F\right\}$ (so $H=\ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )$).

Consider the maps

$$\chi \{\begin{array}{ccc}\hfill F\hfill & \hfill \to \hfill & H\hfill \\ \hfill b\hfill & \hfill \mapsto \hfill & \left(\begin{array}{cc}\hfill 1\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\hfill \end{array},\xi \{\begin{array}{ccc}\hfill H\hfill & \hfill \to \hfill & F\hfill \\ \hfill \left(\begin{array}{cc}\hfill 1\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\hfill & \hfill \mapsto \hfill & b\hfill \end{array}.$$

Then $\chi \circ \xi ={\mathrm{id}}_H$ and $\xi \circ \chi ={\mathrm{id}}_F$. Therefore $\chi$ is bijective (and $\xi ={\chi }^{-1}$). Moreover, for all $b,b^{\prime }\in F$,

$$\chi (b)\chi (b^{\prime })=\left(\begin{array}{cc}\hfill 1\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\left(\begin{array}{cc}\hfill 1\hfill & \hfill b^{\prime }\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill 1\hfill & \hfill b+b^{\prime }\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)=\chi (b+b^{\prime }),$$

so $\chi$ is an isomorphism, and

$$H=\ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )\simeq F.$$