An irreducible curve of degree 3 has one singular point Good morning, i got stuck with these exercises.


*

*Let $X$ be an hypersurface of degree 3 and suppose that $X$ has two singular points $P$ and $Q$. Let $L_{PQ}$ the line containing $P$ and $Q$. Show that $L_{PQ}\subset X$.

*Let $F(x,y,z)$ be an homogeneous polynomial, $k$ algebraic closed and let $C=Z(F)\subset\mathbb P_k^2$ be an irreducible curve. Prove that is $deg(F) =3$ then $X$ has at most one singular point.
 A: For 2.
Assume that there are 2 distinct singular points $p$ and $q$. Then the multiplicity of $p$ and $q$ are each greater than $2$. Let $L$ be a line  joining $p$ and $q$. Then by Bezout's Theorem, $3=(deg(F))(deg(L))\geq (\text{multiplicty of } L\cap C\text{ at }p)+(\text{multiplicty of } L\cap C\text{ at }q)\geq2+2=4$. This is a contradiction.
A: For (1): If $L_{PQ} \subsetneq X$ We have that $L_{PQ} \cap X$ is a finite set of points, say $P,Q,R_1,\ldots,R_k$. Theorem 7.7 of Hartshorne now gives that
$$i(L_{PQ},X,P) \deg P + i(L_{PQ},X,Q)\deg Q + \sum_{i=1}^k i(L_{PQ},X,R_i)\deg R_i = (\deg L_{PQ})(\deg X).  $$
Now the degree of a line is one, while the degree of $X$ is three. So the right hand side is $3$ while on the left, the sum $$i(L_{PQ},X,P) \deg P + i(L_{PQ},X,Q)\deg Q \geq 4$$ which is a contradiction.


Second Proof (we assume $k$ is algebraically closed): We can reduce to the case that $X \subseteq \Bbb{P}^2$ as follows. Assume $P = [1:0:\ldots : 0]$ and $Q = [0:1:0: \ldots 0]$. Cut $X$ with the hyperplane $x_n = 0$. We will then have a hypersurface in $\Bbb{P}^{n-1}$, whose defining equation is the same as $X$ but we set the variable $x_n = 0$. Continue cutting with hyperplanes and we will have a hypersurface $X' \subseteq \Bbb{P}^2$ whose defining equation is still some cubic curve in the variables $x_0,x_1,x_2$. It is now enough to show that $X'$ contains the line $l_{pq}$ joining $p= [1:0:0]$ and $q = [0:1:0]$. This is because each time we cut with the hyperplane $x_i = 0$ for $i \geq 2$, the points $P,Q$ are always in these hyperplanes.
If $X'$ did not contain $l_{pq}$ Bezout's theorem says $$(\deg l_{pq})(\deg X') \geq \sum (\text{intersection multiplicities}).$$ The left hand side is $1\cdot 3$ while the right hand side is at least $4$ since $p,q$ singular means their multipicities are at least $2$ each. This is a contradiction.


Proof of (2) without using (1): Say the singular points are $[1:0:0]$ and $[0:1:0]$. Then the equation for your cubic necessarily has no $x^3$ and $y^3$ terms. Then using the condition that all the partials simultaneously vanish at both these points we get that your cubic is an equation in the variables $z^3,xz^2, yz^2$, contradicting irreducibility.
A: Let's prove $1.$ ($2.$ can be proved along the same lines).
Applying a projective change of coordinates, we may assume without loss of generality that
$$
P=[1:0:...:0],Q=[0:1:0:...:0].
$$
Let
$$
F(X_1,X_2,...,X_n)=a_1X_1^3+a_2X_1^2X_2+a_3X_1X_2^2+a_4X_2^3+... \in k[X_1,...,X_n]
$$
be the polynomial defining the hypersurface $X \subset \Bbb P^n$.
Since $P$ and $Q$ are singular points on $X$, we have $F(P)=F(Q)=0$ and all partial derivatives of $F$ vanish at both $P$ and $Q$ as well. A direct calculation shows that this implies
$$
a_1=a_2=a_3=a_4=0,
$$
which yields $L_{PQ} \subset X$.
