# Scheme-theoretic proof of a correspondence between multiple tangents on a plane curve and ordinary $r$-fold points on the dual curve?

Let $X$ be a curve of degree $d$ in $\mathbb{P}^2_k$ where $k$ is an algebraically closed field of characteristic $0.$ We say that a line of $\mathbb{P}^2_k$ is a multiple tangent of $X$ if it is tangent to $X$ at more than one point. If $L$ is a multiple tangent of $X,$ tangent to $X$ at the points $P_1, \ldots, P_r$ and if none of the $P_i$ is an inflection point, show that the corresponding point of the dual curve $X^*$ is an ordinary $r$-fold point, meaning that it is a point of multiplicity $r$ with distinct tangent directions.

This is obvious from a geometric point-of-view, but I would like to give a rigorous proof of this fact, using say some scheme-theory. I would be grateful for explanations on how to prove this, or references.

Update 22 nov: So, I see how one can do this if one assumes the (non-obvious) fact that the map $X \rightarrow X^*$ is the normalization of $X^*.$ To prove this one needs to show that the Gauss map $X \rightarrow X^*$ is birational, and I can prove that this is the case if I know that there are finitely many multiple tangents. So I'm back at step one. I think the above should have an elementary proof since it is in Hartshorne, but I can't seem to find the right approach.

Update 23 nov: So I received an answer, but this is not the sort of answer I was after (I don't think it is rigorous either). Let me sketch the proof I had in mind if I knew that the Gauss map was birational. If so, we could identify $X \rightarrow X^*$ with the normalization, and if so, we can identify $X$ with blowing-up at singular points of $X^*.$ Then saying that the blow-up of $X^*$ along $L$ gives $r$ points such that the induced map on tangent spaces is injective is precisely the fact that $L$ is an ordinary $r$-fold point. This is a purely algebraic proof in some sense, and doesn't use the analytic category.

Update 24 nov:
The previous answer I received was deleted and then replaced with basically the same answer, but changed some small details. This is still not the sort of proof I have in mind, so let me be extra clear. I would want a purely algebraic proof of the fact that multiple tangents of $X$ corresponds to ordinary $r$-fold points. Preferably, I would want a proof that would not pass to the analytic category and not use the full strength of the biduality theorem. The reason I think the biduality theorem should not be used is that Hartshorne never mentions it, so I would suspect he had something other in mind. If it is possible, I would also want an answer that works in arbitrary characteristic. So, the answer I have received is not what I was after.

• This is (part of) Hartshorne's Algebraic Geometry, Exercise IV 2.3. Seems worth mentioning... – Ben Nov 17 '16 at 17:52

The birationality of the Gauss map is, in fact, not hard to prove in the case at hand. It is a consequence of the biduality theorem, which has a simple proof for plane curves in characteristic $0$. Please see p. 308 in the following article of Kleiman here.

Kleiman, Steven L. The enumerative theory of singularities. Real and complex singularities (Proc. Ninth Nordic Summer School/NAVF Sympos. Math., Oslo, 1976), pp. 297–396. Sijthoff and Noordhoff, Alphen aan den Rijn, 1977.

Let me sketch the proof I had in mind if I knew that the Gauss map was birational. If so, we could identify $X \rightarrow X^*$ with the normalization, and if so, we can identify $X$ with blowing-up at singular points of $X^*.$ Then saying that the blow-up of $X^*$ along $L$ gives $r$ points such that the induced map on tangent spaces is injective is precisely the fact that $L$ is an ordinary $r$-fold point. This is a purely algebraic proof in some sense, and doesn't use the analytic category.

Yes, you can salvage your proof, as you say, by using the birationality of the Gauss map. The biduality theorem is just what you are looking for, and its proof is elementary.

The biduality theorem asserts not only that the Gauss map $X \to X^*$ is birational, but that its inverse is the Gauss map $X^* \to (X^*)^*$. The computation is formal; you can use formal power series in place of convergent power series to compute the formal derivatives involved in each of the two Gauss maps.

As I said, this proof seems to go to the analytic category (or at least, by passing to completions).

Yes, but as you know, what is involved is just the implicit function theorem of undergraduate vector calculus; moreover, there is no need to show the convergence of the power series giving the implicit function.

Ideally, I want something that works in characteristic $p>0$ as well.

The Gauss map is not always birational in characteristic $p > 0$. For example, consider the Fermat curve$$X : x^{p + 1} + y^{p + 1} + z^{p + 1} = 0.$$Its Gauss map is the Frobenius $p$th power map$$(x, y, z) \mapsto (x^p, y^p, z^p),$$which is of degree $p$. Moreover, every tangent to $X$ is inflectional.

• Thank you for your answer. As I said, this proof seems to go to the analytic category (or at least, by passing to completions). Since the statement that multiple tangents correspond to inflection points should be true in characteristic $p>0,$ I don't think this is the proof I am after. Ideally, I would want a proof that is characteristic-free and doesn't use the biduality theorem. I don't think the exercise of Hartshorne really warrants that proof, but I might be wrong. – Dedalus Nov 24 '16 at 10:43
• I probably have not been clear enough in the problem description what I was after. However, this is not what I wanted, but mea culpa for not being clear enough. To clarify, maybe more what I am after (and I invite you to post an answer if you know how to prove this!): 1) The proof should not use the strength of the biduality theorem. This is an exercise in Hartshorne, so probably this is not what was in mind. 2) The proof should not pass to the analytic category, preferably not to completions either. Ideally, I want something that works in characteristic $p>0.$ as well. – Dedalus Nov 24 '16 at 10:46
• Thank you for your thoughtful answer. I think we misunderstood each other when I said "I want something that works in characteristic $p>0$ as well." I meant that I thought that the statement that "A multiple tangent of $X,$ tangent to the points $P_1, \ldots, P_r$ with no $P_i$ an inflection point, corresponds to an ordinary $r$-fold point on $X^*.$ " – Dedalus Nov 26 '16 at 12:22
• This is what I would be very grateful for a method to show, given that it is true. In your example, every tangent is an inflection point so it is not a counterexample. Maybe one could use something similar in characteristic $p>0$ as the previous method, but I'm not entirely convinced. – Dedalus Nov 26 '16 at 12:24