Doubts on the proof about the classification of irreducible projective conics

Question

There are some details in the proof of corollary 3.12 on Frances Kirwan's book complex algebraic curves that I can't understand.

Corollary 3.12 Any irrecducible projective conic $C$ in $P_2$ is equivalent under a projective transformation to the conic $x^2=yz$ and in particular is nonsingular.

Proof: Since $C$ has at most finitely many singular points, by applying a suitable transformtion we may assume that $[0,1,0]$ is a nonsingular point of $C$ and that the tangent line to $C$ at $[0,1,0]$ is the line $z=0$. Then $C$ must be defined by a polynomial of the form $ayz+bx^2+cxz+dz^2$ for some complex nunbers $a,b,c,d$. Since $C$ is irreducible, $a$ and $b$ are both nonzero. The projective transformation $[x,y,z]$$\to$$[\sqrt bx, ay+cx+dz, -z]$ takes $C$ to the conic $x^2=yz$. Since this conic is nonsigular it follows that $C$ is nonsingular as well.

1) For any two points in the projective plane there exists a projective transformation carrying one to the other, so we can just assume that $[0,1,0]$ is a nonsingular point of $C$, but how can we also guarantee something like "the tangent line to $C$ at $[0,1,0]$ is the line $z=0$" ?

2) The author says that "Since $C$ is irreducible", but here the $C$ is the image of the original irreducible conic by a projective transformation, which means that we have to show the irreducible polynomials are preserved under such a transformation. Denote the projective transformation by $[x,y,z]\to[f_1, f_2, f_3]$, here $f_i$ is linear in $x,y,z$ and can be written in the form $f_i=a_ix+b_iy+c_iz$. I think this can be done by the ring isomorphism between $K[x,y,z]$ and $K[f_1, f_2, f_3]$ where $K$ is the base field. Am I correct?

3) Since $C$ is irreducible, $b$ must be nonzero, or the polynomial of $C$ will be divisible by $z$. However, if $a$ is zero, the $C$ is in the form $bx^2+cxz+dz^2$, anything wrong occurs in this case? Will $C$ necessarily become reducible?

4) At the end of the proof the author says that $C$ is nonsingular follows from the fact that $x^2=yz$ is nonsingular. So I think that the singular points will also be preserved under the projective transformation, that is, $A$ is a singular point if and only if $F(A)$ is a singulat point, where $F$ is the projective transformation. Let $P(x,y,z)$ be the original curve and $A=(a,b,c)$ be its singular point, so $P_x, P_y, P_z$ are zero at $[a,b,c]$. Under the projective transformation $F$, $P(x,y,z)$ becomes $P(f_1, f_2, f_3)=P'(x,y,z)$ where $f_i=a_ix+b_iy+c_iz$, so I wish to prove that $P'_x,P'_y,P'_z$ are zero at the point $[f_1(A), f_2(A), f_3(A)]$ by chain rule but I don't know how to do that excatly. Since $P$ is homogeneous, will the Euler relation help?

Sorry for so many questions, I'm a self learner and just new to these materials so I think ther're not totally obvious to me. Can someone give me some hints? Many thanks.

Answer 1

1. You can always perform something like a rotation around that point. In the most common embeddings $[0,1,0]$ would be seen as a point at infinity, but you might as well pick an embedding where that is the origin. Or use the unit sphere with antipodal points identified as a model of the projective plane, in which point you can rotate around any point.

   As discussed in this answer of mine, a projective transformation is uniquely defined by the images of four points, no three of which are collinear. So you can easily pick one point of the conic and map it to $[0,1,0]$, then pick any other point on the tangent in that first point and map it to any other point on $z=0$, and you still have two points to map as you choose.

2. Yes, ring isomorphism sounds like a fair approach.

   Personally I'd view “irreducible quadric” as “full rank matrix” and “projective transformation” as “matrix conjugation”, and if I know that conjugation resp. change of basis does not change the rank, then I know that irreducibility is preserved by projective transformations. But the question of which considerations feel natural depend a lot on what your background is, so go with what came to your mind and I'm just stating my view to offer some diversity.

3. I'll switch to my preferred matrix notation.

   $$2\left(ayz+bx^2+cxz+dz^2\right)=[x,y,z]\cdot\begin{bmatrix}2b&0&c\\0&0&a\\c&a&2d\end{bmatrix}\cdot\begin{bmatrix}x\\y\\z\end{bmatrix}=0$$

   The quadric is reducible iff the determinant of that matrix is zero. So you care about

   $$0=\begin{vmatrix}2b&0&c\\0&0&a\\c&a&2d\end{vmatrix}=-2a^2b$$

   So this confirms the claim algebraically.

   To build some geometric intuition, consider the following: $a$ corresponds to the only term that contains $y$. So if you have $a=0$, then for any point on the conic, any point which differs from that first point only in its $y$ coordinate will be on the conic as well. So you end up with a pair of lines that have the $y$ direction in common, or projectively speaking a pair of lines through $[0,1,0]$.

   If you want to actually split the matrix into a pair of lines, this question has some background. Note that it involves taking square roots, so if your underlying field is not algebraically closed this operation might not even be feasible.

4. I'm inclined to view a projective transformation as essentially the same as a change in vector space basis. Both of these can be expressed by a matrix multiplication. With this view in mind, you can build on various results from vector algebra about how key properties don't depend on the choice of basis. I confess that with this high level view in mind, I forget about the underlying mechanics that make this work in each incident. I hope some other answer will complement mine in this regard.