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Let $P_1$,$P_2$,$P_3$,$P_4$ $\in \mathbb{P}^2$. Let $V$ be the linear system of conics passing through these points. Then show that if $P_1$,$P_2$,$P_3$,$P_4$ lie on a line, then $\dim(V)=2$ and if not, then $\dim(V)=1$.


[my attempt]

The definition of $V$ ,the linear system of conics passing through given points, is $$V:= \{ [a:b:c:d:e:f] \in \mathbb{P}^5 \mid ax^2+by^2+cz^2+dxy+eyz+fxz=0 \text{ passing through } P_1 , P_2 ,P_3, P_4\}$$

Now, let's denote $P_i=:[l_i:m_i:n_i]$ and consider the followings :

$$ \begin{pmatrix} l_1^2 & m_1^2 & n_1^2 & l_1m_1 & m_1n_1 & l_1n_1 \\ l_2^2 & m_2^2 & n_2^2 & l_2m_2 & m_2n_2 & l_2n_2 \\ l_3^2 & m_3^2 & n_3^2 & l_3m_3 & m_3n_3 & l_3n_3 \\ l_4^2 & m_4^2 & n_4^2 & l_4m_4 & m_4n_4 & l_4n_4 \\ \end{pmatrix} \begin{pmatrix} a \\ b \\ c \\ d \\ e \\ f \end{pmatrix}= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$

Suppose $P_1$,$P_2$,$P_3$,$P_4$ lie on a line (say, $\alpha x+ \beta y +\gamma z=0$).

Then $$ \begin{pmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \\ l_4 & m_4 & n_4 \\ \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \\\gamma \end{pmatrix}= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$

Therefore the rank of this matrix is less than equal $2$. (This is beacause ($\alpha , \beta , \gamma $) is a non-trivial solution of this linear system) so $P_1$,$P_2$,$P_3$,$P_4$ are linearly dependent.

However, I stuck in this point. How to solve this?

Thank you.

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This is best understood without equations IMO. The first two points are completely free choices that cut us down to a $\mathbb P^3$ of conics (i.e. the only way $P_2$ can fail to be general is if it is equal to $P_1$). If we add a third point lying on the line between the first two (call this Step 3a), then the line must be part of the conic (otherwise Bezout is violated); in this case, the remaining two points determine a second line, so we get a (generally) unique reducible conic.

If, instead, we take $P_3$ to not lie on the line $L := \overline{P_1 P_2}$ (call this Step 3b), there is no issue with Bezout so we still have a $\mathbb P^2$ of conics whose general member is irreducible. Continuing with one more general point, we get a $\mathbb P^1$ of conics whose general member is irreducible.

On the other hand, if we continue from Step 3a and impose a fourth point on the line, then we don't "lose" any conics this time; since all conics from Step 3a contain the line $L$, they all pass through $P_4$, so we still have a $\mathbb P^2$ of conics by choosing any line to be the second component.

Finally (on the third hand, I guess?), if we continue from Step 3a but let $P_4$ be a general point, then we are cut down to a $\mathbb P^1$ of reducible conics: choose a line passing through $P_4$ (there is a $\mathbb P^1$ of these) and take its union with $L$.

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  • $\begingroup$ Thank you. May I ask you something? I understand that if three points are in a line, by Bezout, the conic is reducible. However, I don't understand that " if there is no issue with Bezout , then we still have a $\mathbb{P}^2$ of conics whose general member is irreducible. " $\endgroup$ – hew Dec 4 '20 at 13:49
  • $\begingroup$ Just that if I take three points in general position, then there are exactly three reducible conics through them: there are three ways to choose a pair out of the three, hence three lines, and any two of the lines form a reducible conic. If the points are the three coordinate points, then the lines are the coordinate lines and the conics are $xy,xz,$ and $yz$. For $(a:b:c) \in \mathbb P^2$ general, $axy+bxz+cyz=0$ is irreducible. $\endgroup$ – Tabes Bridges Dec 4 '20 at 17:08

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