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 $\mathbb CP^2$

  • 2
    $\begingroup$ 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. $\endgroup$ Feb 27, 2013 at 18:55

1 Answer 1


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 $\mathbb{P}^2$.

  • $\begingroup$ That is very cute. Just to be sure I get it, the image of this elliptic curve has exactly two cusps? $\endgroup$
    – aglearner
    Feb 27, 2013 at 22:42
  • 1
    $\begingroup$ 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... $\endgroup$
    – aglearner
    Feb 27, 2013 at 23:35
  • 2
    $\begingroup$ @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,...). $\endgroup$ Feb 28, 2013 at 8:21
  • $\begingroup$ 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 :) ) $\endgroup$
    – aglearner
    Feb 28, 2013 at 9:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.