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Let $k$ be a characteristic zero field and let $L/k$ be a quadratic extension. Write $L = k(\sqrt{p})$.

Let $q$ be a non-square in $k^\star$ and let $r \in k^\star$ be any constant.

Consider the fraction field $F$ of field $k[X,Y]/(X^2 - qY^2 - r)$ (function field of a conic).

Under which conditions on $p$, $q$ and $r$ are the fields $L$ and $F$ linearly disjoint?

(Does there exist a "nice" answer to this question?)

If I'm right (I hope I am?), this question is equivalent to:

when do there exist coprime $F,G \in k[X,Y]$ such that $X^2 - qY^2 - r$ divides $F^2 - pG^2$?

This is a concrete question, but I don't know whether it has a concrete answer.

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I have very good and very concrete news for you, Evariste: the fields $L$ and $F$ are always linearly disjoint! Here is why.

First, linear disjointness here means that $\sqrt p \notin F$ and that's what I'll prove.
Also, observe that $F=k(y,\sqrt {qy^2+r})$ is a degree $2$ extension of $k(y)$.
Now if we had $\sqrt p \in k(y,\sqrt {qy^2+r})$, we would have the tower $k(y)\subset k(y,\sqrt p)\subset k(y,\sqrt p,\sqrt {qy^2+r})=F$ and degree considerations would force $k(y,\sqrt p)=k(y,\sqrt p,\sqrt {qy^2+r})$.
In other words, we would have $\sqrt {qy^2+r}\in k(y,\sqrt p)$ .

However the ring $k[y,\sqrt p]$ is integrally closed and since $\sqrt {qy^2+r}$ is both integral over that ring and in its fraction field $k(y,\sqrt p)$, we would have already $\sqrt {qy^2+r}\in k[y,\sqrt p]$, which leads to an equality $qy^2+r=(\alpha y+\beta)^2$ with $\alpha, \beta \in k[\sqrt p]$. This is clearly impossible .

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    Cher @Evariste, votre enthousiasme et votre gentillesse font chaud au coeur!2012-01-25