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 $a^2 - b^2d = x$.
