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I'm trying to follow a proof of Segre's Theorem, and I'm unsure of how some of the results are obtained.

In $PG(2,q)$, $q$ odd, we have an oval ($q+1$ points, no three collinear). We pick $3$ of them and set our basis such that they have coordinates $P=(1,0,0)$, $Q=(0,1,0)$ and $R=(0,0,1)$.
The line $QR$ is therefore $X=0$, $PR$ is $Y=0$ and $PQ$ is $Z=0$.

(1) The tangent line at $P$ is $Y=aZ$.

Also more generally,

(2) The tangent lines to an oval of even order meet at a common point, and the tangent lines of an oval of odd order do not.

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    $\begingroup$ Can you state what theorem of Segre's you are interested in and/or give more context? $\endgroup$ – Mariano Suárez-Álvarez May 30 '11 at 1:50
  • $\begingroup$ It's Beniamino Segre's 1955 theorem here under Odd q $\endgroup$ – Zeophlite Jun 1 '11 at 9:32
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    $\begingroup$ @Zeophite: I was suggesting you add it to the question body, really. $\endgroup$ – Mariano Suárez-Álvarez Jun 1 '11 at 17:01
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This is to a great extent educated guessing, as the Wikipedia page was kinda terse on the definition department :-)

Anyway I think I have an explanation to your first question. As the abstract oval is not (originally) defined by a polynomial equation, it is kind of difficult to come up with a definition of a tangent. The following seems to fit the bill (correct me if I'm wrong). Fix a point $X$ on the oval. There are $q+1$ lines going thru it. There are $q$ other points on the oval. Each and every one of them belongs to exactly one of the $q+1$ lines. Furthermore, no two points of the oval (other than $X$) belong to the same line, as that would violate the assumption that no three points are collinear. Therefore the $q$ non-$X$ points of the oval are on $q$ distinct lines thru $X$. So there is exactly one line thru $X$ that does not intersect the oval elsewhere. I assume that this line is referred to as the tangent of the oval at $X$.

On with the first question. With the coordinate system now set up (we are free to do it in this way, because the group of projective linear transformations is 3-transitive), we see that the $q+1$ lines thru the point $P$ are the lines $Z=0$ (1 choice) and $Y=aZ$ ($q$ choices, one for each value of the constant $a$). Because the line $Z=0$ goes thru $Q$ also, it cannot be the tangent of the oval at $P$. Therefore the tangent is of the other form, for some $a\in GF(q)$. [Edit] Because the point $R$ cannot lie on the tangent either, we can say that $a\neq0$. [/Edit] I dare guess that's all there is to your question 1), but it is hard to tell without access to the textbook.

[Edit] Question 2) is harder. Kinda makes sense, if the Wikipedia article mentions the list as a published paper. It is obviously true in the Fano plane ($q=2$), where our answer to question 1) tells us that the tangent lines must be $Y=Z$, $X=Y$ and $X=Z$ and they all meet at the point with projective coordinates $(1:1:1)$. A curious result anyway. [/Edit]

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