# Kernel is a group of order $q+1$

Hello I hope you can help me with this doubt I have.

Let $$F_{q}$$ a field and consider its quadratic extension $$F_{q^{2}}=F_{q}(\sqrt{d})$$ for $$d$$ square free in $$F_{q}$$ now we consider the next morphism of groups.

$$\varphi:F^{*}_{q^{2}}\rightarrow GL_{2}(F_{q})$$ given by $$\varphi(\alpha)=M_{\alpha}$$.

Where $$M_{\alpha}:F_{q^{2}}\rightarrow F_{q^{2}}$$ it is the nultiplication by $$\alpha$$.

So I do the composition with $$det:GL_{2}(F_{q})\rightarrow F^{*}_{q}$$ and I have to check $$Ker(det\circ\varphi)$$ is a group of order $$q+1$$.

I have done it giving explicit the composition, for $$\alpha=a+b\sqrt{d}$$ we have that the matrix that represent $$M_{\alpha}$$ is $$\left( \begin{array}{cc} a & bd\\ b & a \\ \end{array} \right)$$, it follows that $$(det\circ\varphi)(\alpha)=a^{2}-b^{2}d$$ therefore an element in the kernel of $$(det\circ\varphi)$$ satisfies $$a^{2}-b^{2}d=1$$.

I do not know hot to proceed with this how to prove $$Ker(det\circ\varphi)$$ is a group of order $$q+1$$?, Is there a explicit way for the elements of the kernel?

Thank you for your time and I am sorry for the inconvenience!

Show that the map $$\det \circ \varphi$$ is surjective. Then the first isomorphism theorem tells you that $$\frac{F_{q^2}^{\times}}{\ker(\det \circ \varphi)} \cong F_q^{\times}.$$ The two groups above must therefore have the same order, thus $$\frac{q^2 - 1}{\vert \ker(\det \circ \varphi) \vert} = q - 1,$$ thus $$\vert \ker(\det \circ \varphi) \vert = q + 1$$, which is what you want.
One can show the map is surjective using a counting argument. Let $$S = \{a^2 : a \in F_q\}$$ be the set of squares in $$F_q$$ and $$T = \{x + b^2d : b \in F_q\}$$. Since you are assuming $$d$$ is square free the characteristic of $$F_q$$ must be odd, so we have $$\vert S \vert = \vert T \vert = \frac{q - 1}{2} + 1 > \frac{q}{2},$$ thus $$S$$ and $$T$$ must intersect, which guarantees you a solution to the equation $$a^2 - b^2d = x$$.