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

  • $\begingroup$ @JyrkiLahtonen Oh of course, thanks. I have edited the answer. $\endgroup$ – Ethan Alwaise Mar 13 at 12:12
  • $\begingroup$ Thank you for the answer and by curiosity is there a explicit way for the elements of the kernel or the matrix they represent? $\endgroup$ – Liddo Mar 13 at 13:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.