TITLE: Detail regarding tangent spaces and dual varieties from Harris's Algebraic Geometry: A First Course
QUESTION [5 upvotes]: In Harris's Algebraic Geometry: A First Course, Example 16.20, the author shows that the dual of the dual variety $X^{*}$ is the original variety $X$. I think in chapter 15, Harris mentions that he'll be working in characteristic 0 in the subsequent chapters.
I might have missed more, but, if we assume $X_{sm}\subset X$ and $X_{sm}^{*}\subset X$ are dense, the only place I can see where he used characteristic zero is where he let $\Phi$ be the closure of $\tilde{\Phi}:=\{(p,H):p\in X_{sm}, H\supset T_pX\}$ in $X\times X^{*}$, and claimed using Proposition 14.4 that the projections $\pi_1: \Phi\rightarrow X$ and $\pi_2: \Phi\rightarrow X^{*}$ are surjective on tangent spaces generically.
1) Is it true that the maps $\pi_1$ and $\pi_2$ are surjective on tangent spaces generically in general characteristic? Since the maps are projecting away variables, maybe the pathologies related to characteristic $p$ wouldn't happen.  
2) In general characteristic, if $X_{sm}$ is a dense open set in $X$, is $(X^{*})^{*}=X$? Can something weird like $X_{sm}\subset X$ is dense, but $X_{sm}^{*}\subset X^{*}$ is empty happen? 
3) Finally, is there a good reference for the proof that $(X^{*})^{*}=X$ if I'm just interested in plane curves (but a completely algebraic proof, not one where you parametrize the curve over $\mathbb{C}$)?

REPLY [2 votes]: Theorem 8.13 of this text (it is in Spanish) shows that $C=C^{**}$ for a plane algebraic curve $C\subseteq\mathbb{P}^{2}$. The idea is the next one:
Let $a\in C$ be a smooth point. Then, $T_{a}C\in C^{*}$ as a point in $(\mathbb{P}^{2})^{*}$. Let's consider the set of lines of $(\mathbb{P}^{2})^{*}$ that passes through $T_{a}C\in(\mathbb{P}^{2})^{*}$. Taking a look at the multiplicity of intersection of each one with $C^{*}$ at $T_{a}C$, you can see that $T_{T_{a}C}C^{*}=\{L\in(\mathbb{P}^{2})^{*}:p\in L\}$, that is a line of $(\mathbb{P}^{2})^{*}$, i.e. a point of $(\mathbb{P}^{2})^{**}$. In fact, with the identification of $(\mathbb{P}^{2})^{**}$ with $\mathbb{P}^{2}$, we have that $p=\{L\in(\mathbb{P}^{2})^{*}:p\in L\}$. Accordingly, $p\in C^{**}$. This easily implies $C\subseteq C^{**}$, and therefore $C=C^{**}$ since they are irreducible projective sets of the same dimension.