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Good morning, i got stuck with these exercises.

  1. 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 $L_{PQ}\subset X$.

  2. 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.

miguemate
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  • For 2), just apply 1) directly. (An irred. curve of degree more than 1 can't contain a line.) For 1), how many complex roots (with multiplicity) does a cubic polynomial have? –  Nov 28 '13 at 15:20
  • @AsalBeagDubh thank you, so for 1) – miguemate Nov 28 '13 at 15:27
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    I don't really understand your description of what you already tried, by the way. 2) says that $F$ is already homogeneous. –  Nov 28 '13 at 15:28
  • @AsalBeagDubh you are right, i got confused, i just edited it, i am going to try again. So it is just enough to prove 1 – miguemate Nov 28 '13 at 15:34
  • Actually in the list where i took these exercises 2 was first and 1 was next, so there is another way to proof 2 without using 1? – miguemate Nov 28 '13 at 15:41
  • And for 1? I did not understand the hint given by Asal Beag Subh – Framate Nov 29 '13 at 04:01
  • @Miguemate: my hint meant the following. If you restrict a definiing equation of $X$ to the line $L$, you get a homogeneous cubic polynomial $F(x,y)$ in two variables. If $L$ is not contained in $X$, then $F$ has three zeros, counted with multiplicity. But at a singular point of $X$, you get a zero of $F$ with multiplicity at least 2. This is a contradiction. (This is a very simple version of Bezout's theorem, mentioned in another answer.) –  Nov 29 '13 at 10:39
  • @Miguemate I posted an answer for (2) that doesn't rely on one. –  Nov 30 '13 at 03:39

3 Answers3

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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.

user44322
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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.

<p>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.</p>

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.

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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 $L_{PQ} \subset X$.

Nils Matthes
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