# Injective morphism from an elliptic curve to $\mathbb CP^2$.

Let $E$ be the elliptic curve $x^3+y^3+z^3=0$.

Question. Are there injective morphisms $E\to \mathbb CP^2$ of arbitrary high degree?

Comments. 1) There are injective morphisms $E\to \mathbb CP^2$ of degree $3$ (the obvious one) and of degree $6$ - whose image is the curve dual to the cubic. 2) Clearly for $\mathbb CP^1$ there are injective morphisms to $\mathbb CP^2$ of arbitrary high degree. 3) I don't know any (smooth projective) curve for which on can prove that it does not admit an injective morphism to $\mathbb CP^2$ of arbitrary high degree.

This question is related to Injective morphism from curves to $\mathbb CP^2$

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Just to point it out to everyone, your example of the dual curve to the cubic implies that you want injective on the level of point sets, not a closed immersion. The dual of the cubic has $9$ cusps. – David Speyer Feb 27 '13 at 18:55

The answer is yes. And more generally, one has the following result:

Lemma: for any elliptic curve $X$, the degree of injective morphisms $X\to \mathbb{P}^2_{\mathbb{C}}$ (or $\mathbb{C}P^2$ if you want) is not bounded.

Proof: The curve $X$ is isomorphic to a plane cubic of equation $X^3+Y^3+Z^3=\lambda XYZ$ for some $\lambda$ (Hessian form). Then, the curve has nine inflexion points. These are points $[0:1:\omega]$, $[1:0:\omega]$, $[1:\omega:0]$ with $\omega^3=1$. Choose three of them, $p_1,p_2,p_3$ such that the three corresponding tangent lines $L_1,L_2,L_3$ intersect at three points $L_1\cap L_2$, $L_1\cap L_3$, $L_2\cap L_3$, different from $p_1,p_2,p_3$.

We change the coordinates and assume that $L_1,L_2,L_3$ are the lines $x=0$, $y=0$ and $z=0$. The points $p_1,p_2,p_3$ become points of the form $[0:1:a_1]$, $[a_2:0:1]$, $[1:a_3:0]$ with $a_1a_2a_3\not=0$. Take now the birational map of $\mathbb{P}^2$ given locally by $(x,y)-->(x,x^ny)$, and globally by $[x:y:z]-->[xz^n:x^ny:z^{n+1}]$, for $n\ge 2$. Its base-points are $[0:1:0]$ and $[1:0:0]$, which do not belong to the curve. The curves contracted are $z=0$ and $x=0$, and $y=0$ is fixed. Outside of the triangle $xyz=0$, the map is an automorphism. In consequence, the image of the curve has degree $3n$ and only cuspidal singularities. This yields an injective map from $X$ to $\mathbb{P}^2$.

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That is very cute. Just to be sure I get it, the image of this elliptic curve has exactly two cusps? – aglearner Feb 27 '13 at 22:42
Do I understand correctly that the same reasoning applies to the curve $x^{2n}+y^{2n}+z^{2n}=0$ for any $n$. Namely, lines $x=z$, $x=-z$ and $y=z$ have tangency of order $2n$ with such a curve and we can apply birational transformation of the type your proposed with respect to these three lines. I am asking this, since I had a secret hope that curves of (fixed) higher genus can not admit injections in $\mathbb CP^2$ of arbitrary high degree. But it looks now they can... So this tells me that it will be harder to answer in negative the original question 122645... – aglearner Feb 27 '13 at 23:35
@aglearner, 1) yes the curve has exactly two cusps, the images of the lines $z=0$ and $x=0$. 2) yes it works for special curves admitting three lines (in fact two suffice) such that each line intersects the curve at only one point. For the curve $x^{n}+y^{n}+z^{n}$ is such a curve, but not with the lines you suggest. You can take $x=ay$, $y=az$ with $a^n=-1$. Your "secret hope" could maybe work for general curves of high genus, but there are special ones which will admit these injections (hyperrelliptic, or with inflexion pts,...). – Jérémy Blanc Feb 28 '13 at 8:21
Thank you Jeremy (I was in fact thinking of $x^{2n}+y^{2n}-z^{2n}=0$, but, of course I agree with your choice of tangent lines :) ) – aglearner Feb 28 '13 at 9:06