# Multiple tangents on a plane curve corresponds to ordinary $r$-fold points on the dual curve

Let $X$ be a (smooth, irreducible) 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 a question from Hartshorne, and while I believe I know how to prove it (with different methods), I don't think is the kind of solution the exercise warrants. Let me briefly outline the methods I know of how to prove this:

1. Assume that we know that the Gauss map $X \rightarrow X^*$ is birational. Then it follows that $X$ is the normalization of $X^*$ and we can identify the Gauss map with successive blow-ups at singularities of $X^*.$ Then one checks that essentially by definition that the point $L$ is an ordinary $r$-fold point. To know that the Gauss map is birational, one can argue by the biduality theorem for plane curves in characteristic $0.$
2. Use the Lefschetz principle to reduce to when $X$ is a curve over $\mathbb{C}$ and argue using the analytic topology / by geometry.
3. Pass to completions and try to define the Gauss map locally.

All of these methods seems to me to be somewhat against the spirit of the material that Hartshorne has introduced so far in the book. The first of the above items uses the biduality theorem, which, while not extremely hard to prove, still is quite a lot for this The second is not algebraic at all the last seems somewhat of a stretch.

I would be very grateful for a purely algebraic solution of this exercise. Let me explain what I mean by a purely algebraic solution. In some sense, it should only use elementary machinery from scheme theory and not go to the analytic category. If completions are neccessary, they can be used but I would prefer something not using that. Further, I would (if possible) want it to be a proof that would hold true in characteristic $p>0$ as well.

This question has previously been posted to math.stackexchange, but the answers I received was not what I was looking for so I thought I would post it here. I apologize if this is not the right forum for this question, but I am really curious.

You should really try to do this exercise yourself. Here is a hint. Because the characteristic is $0$, by generic smoothness the map from the curve $X$ to the dual projective space has nonzero derivative except at finitely many points. Thus, for a general point $p$ of $X$, the tangent line $L$ to $X$ at $p$ is not an inflection line, and the same holds for any other point $q$ of $X$ such that $L$ is the tangent line to $X$ at $q$.

Now choose homogeneous coordinates $[s,t,u]$ on $\mathbb{P}^2$ such that $L$ equals $\text{Zero}(u)$, such that $p$ equals $[1,0,0]$ and such that $q$ equals $[0,1,0]$. Then $\tau=t/s$ is a local coordinate on $X$ near $p$, and $\sigma=s/t$ is a local coordinate on $X$ near $q$. Since $L$ is not an inflection line near $p$ nor $q$, locally near $p$, $u/s = \tau^2 a$ as regular sections of $\mathcal{O}_X$, where $a\in \mathcal{O}_{X,p}$ is invertible. Similarly, $u/t=\sigma^2 b$, where $b\in \mathcal{O}_{X,q}$ is invertible. Of course the completion $\widehat{\mathcal{O}}_{X,p}$ is isomorphic to $k[[\tau]]$, and similarly $\widehat{\mathcal{O}}_{X,q}$ is isomorphic to $k[[\sigma]]$. Thus, it makes sense to consider $a$ as a power series $a(\tau) = a_0 + a_1\tau + a_2\tau^2 +\dots$, and similarly for $b$.

Near $p$, on an open neighborhood where $\tau$ is a coordinate and where $a$ is invertible, the equation of the tangent line at $\tau_0$ is $$(u-\tau_0^2 a(\tau_0)s) - (2\tau_0 a(\tau_0) + \tau_0^2 a'(\tau_0))(t-\tau_0 s) =$$ $$\tau_0^2( a(\tau_0) - \tau_0 a'(\tau_0))s + \tau_0(-2a(\tau_0) - \tau_0a'(\tau_0))t + u.$$ In particular, if we represent the universal line in $\mathbb{P}^2$ as $$As + Bt + Cu = 0,$$ so that the homogeneous coordinates in the dual projective space are $[A,B,C]$, then the tangent line to the dual curve of $X$ near $p$ is given by $\text{Zero}(A)$. Now, by symmetry, do the same computation for $q$. You should have that the tangent line near $q$ is given by $\text{Zero}(B)$. Thus, if $L$ is not an inflection line to $X$, then even if it happens that two points $p$ and $q$ both map to $[L]$ under the dual curve morphism, nonetheless, the corresponding branches of the dual curve at $[L]$ have distinct tangent lines.

Notice, this goes wrong in characteristic $2$, because then $-2a(\tau_0)$ is zero. Also, in every positive characteristic $p$, there does exist a "funny curve" such that every tangent line to $X$ is an inflection line. However, in characteristics $p\neq 2$, so long as $X$ is not a "funny curve", then the analysis above is correct.

• Thank you for your answer. I have some questions though. You write "The equation of the tangent line at $\tau_0$is..." - are you using the analytic topology to get this to be true? I think of $X$ as being given by a homogenous polynomial $F(x,y,z)$ and don't see how this follows, so would be very grateful for some more exposition or detail on how you reason there. – user44591 Nov 29 '16 at 21:08
• You do not need to use the analytic topology. You can pass from the local ring $\mathcal{O}_{X,p}$ to the completion $\widehat{\mathcal{O}}_{X,p}$, where the germ of $\tau$ gives an isomorphism of the completion with $k[[\tau]]$. Now, using this isomorphism, think of $a$ as a power series in $\tau$ and do the computation in the power series ring. Working with the completion of the local ring is similar to working analytically, but the completion of the local ring is defined purely algebraically. – Jason Starr Nov 29 '16 at 22:20
• Sorry for yet another comment. I should be more precise with what I have some questions about. You write "Near p, on an oben neighborhood where $\tau$ is a coordinate and $a$ is invertible, the equation of the tangent line at $\tau_0$ is.." 1. With open, do you mean open in the Zariski topology? 3. Why is the equation of the tangent line in such a neighborhood given by this equation? This is probably elementary, but I'm thinking of $X$ as given by say the zero set of $F(x,y,z)$ (cont.) – user44591 Nov 30 '16 at 13:18
• and according to this, I would say that the tangent line at the point is given by the sum of the partial derivatives of $F.$ I don't see why this gives the same tangent line as yours. – user44591 Nov 30 '16 at 13:19
• By "coordinate on an open subset", I mean a Zariski open subset $U\subset X$ and a regular morphism $\tau:U\to \mathbb{A}^1$ that is everywhere smooth (equivalently, unramified, equiv. 'etale). For such a morphism, for every $x_0\in U$, for the constant $\tau_0=\tau(x_0)$, $\tau - \tau_0$ is an element of $\mathfrak{m}_{X,x_0}$ that gives an isomorphism of $\widehat{\mathcal{O}}_{X,x_0}$ with $k[[ \tau - \tau_0 ]]$. The tangent line at $x_0$ is the unique linear homogeneous equation whose restriction to $X$ vanishes to order $2$ at $x_0$, i.e., it is a unit times $(\tau-\tau_0)^2$. – Jason Starr Nov 30 '16 at 13:38