1

I have a projective plane: PG$(2, \mathbb{F})$, where char$\mathbb{F}\neq 2$. I have a C conics and an l line. U is a point in C'=C\l. There is an operation on C'. If A,B $\in$ C' and $H_{AB}$ $= AB \cap l$:

$A*B$ =

\begin{cases} U,& \text{if } H_{AB}U \text{ is tangent to C at U}\\ D, & \text{if} H_{AB}U \cap C = \{D,U\} \end{cases}

I have to show that (C', $*$) is isomorphic to the additive group of $\mathbb{F}$ if l is a tangent to C, or to the multiplicative group of $\mathbb{F}$ if l is a secant to C.

So far I proved that (C', $*$) is an Abelian group. To show that these groups are isomorphic, I think I have to find an isomorphism. I know that $\mathbb{F}^*=\mathbb{F}$\{$0$} is the multiplicative group and it is commutative, and that the additive group is also commutative. I am not sure how to continue. I believe we can focus on the cases where the conic is a parabola or a hyperbola, but I have yet to find the proof of these 2 statements. I would be grateful if you could help me.

  • A similar question was asked months ago, but got only partial answers: https://math.stackexchange.com/questions/3951052/groups-of-conics – Intelligenti pauca Apr 26 '21 at 17:28
  • @Intelligentipauca thank you! I think that my question is not even partially answered there, because when I proved that (c',*) is an Abelian group, I had to show that it is associative. –  Apr 27 '21 at 12:49
  • I'd like to see how you proved associativity. – Intelligenti pauca Apr 27 '21 at 13:54

1 Answers1

1

First of all, it's easy to prove that $A*U=A$ for any $A\in C'$, hence $U$ is the identity element. Then, let $K$ be the intersection between $l$ ant the tangent to $C$ at $U$: for any $A\in C'$ we can construct $A^{-1}$ as the other intersection of line $AK$ with $C$.

From $K$ a second line tangent to $C$ can be constructed, with tangency point $V$. This line is the same as $l$ if $l$ is tangent to $C$, in which case $V\not\in C'$. But if $l$ is secant to $C$, then $V\in C'$ and $V*V=U$. Hence $V$ plays the role of $-1$ in the multiplicative group.

Intelligenti pauca
  • 55,765
  • Thank you! I am not sure that I understand why it is isomorphic to the additive group in the first case. (I get the point of the second case, thank you very much for answering!) –  Apr 27 '21 at 14:21
  • 1
    Because in the first case there isn't the analogous of $-1$. – Intelligenti pauca Apr 27 '21 at 14:23
  • Can I ask you that why is the fact that there is not a '-1' element in the group means that it is isomorphic to the additive group of F (or if there is then to the multiplication group of F)? I mean is it sufficient to prove that? In that case when we just have to decide which is ismorphic to which, it is sufficient. Sorry, but I can not see why this proves that they are isomorphic in general. –  Apr 27 '21 at 17:57
  • What do you need to prove isomorphism? – Intelligenti pauca Apr 27 '21 at 20:06
  • That there exists a bijective homomorphism between the two groups. The $\phi$(identity element of (c',))=identity element (of(F,+)(or(F, ))). $\phi$(V)=$\phi$(V)$^{-1}$. –  Apr 27 '21 at 20:17
  • 1
    You can choose at will three points on $C$, the images of $0$, $1$ and $\infty$. Choosing as $l$ the line through $0$ and $\infty$ and $U=1$, you set up the multiplicative group, while choosing $l$ as the tangent at $\infty$ and $U=0$, you set up the additive group. With both groups you can construct all rational numbers, is that enough? – Intelligenti pauca Apr 27 '21 at 20:49
  • Thank you very much! Yes, I think that is enough. –  Apr 27 '21 at 21:06