Is it true that the dual of a smooth hypersurface $X$ of $\mathbb{P}^n$ of degree $d\ge 2$ is a hypersurface? If yes, could you give me a simple proof ? Or a reference?
Note that in this case, the dual is birational to $X$.
Is it true that the dual of a smooth hypersurface $X$ of $\mathbb{P}^n$ of degree $d\ge 2$ is a hypersurface? If yes, could you give me a simple proof ? Or a reference? Note that in this case, the dual is birational to $X$. 


Yes, the dual is a hypersurface. Here is a simple proof when $n\geq 4$. First of all, by the Lefschetz hyperplane theorem, the restriction homomorphism, $$ r : \text{Pic}(\mathbb{P}^n) \to \text{Pic}(X),$$ is an isomorphism. In particular, every invertible sheaf on $X$ is either ample, trivial, or antiample. Denote the Gauss map on $X$ by, $$ G : X \to (\mathbb{P}^n)^\vee, \ \ x \mapsto \mathbb{T}_xX, $$ where $\mathbb{T}_x X$ denotes the tangent hyperplane to $X$ at $x$. The invertible sheaf $G^*\mathcal{O}(1)$ is either ample, trivial or antiample. If $G$ is nonconstant, then $G^*\mathcal{O}(1)$ has positive degree Hilbert polynomial, hence $G^*\mathcal{O}(1)$ is neither trivial nor antiample. Therefore $G^*\mathcal{O}(1)$ is ample. In particular, $G$ can have no positive dimensional fiber, since this would give a subvariety of $X$ on which $G^*\mathcal{O}(1)$ is not ample. Therefore $G$ is finite, and hence $\text{dim}(G(X))$ equals $\text{dim}(X)$. Finally, if $G$ were constant with image $[H]$, for $H$ a fixed hyperplane in $\mathbb{P}^n$, then for every $x\in X$, the tangent hyperplane to $\mathbb{T}_x X$ equals $H$. In particular, for every $x\in X$, $x$ is contained in $H$. Thus $X$ is contained in $H$, contradicting that $X$ is a smooth hypersurface of degree $d>1$. Edit. I just want to add, I heard this argument long ago, but I do not know who first proved this result. Fyodor Zak has done important work on the Gauss map, so you might consult his work to try and find a reference. Second edit. The OP asks about $n=3$. One could probably adapt the "fast" argument above using NoetherLefschetz, but it is probably more honest just to compute $G^*\mathcal{O}(1)$. Indeed, the morphism $G$ arises from the short exact sequence sequence, $$ 0 \to \mathcal{O}_{\mathbb{P}^n}(d)_X \xrightarrow{u} \Omega_{\mathbb{P}^n/k}_X \to \Omega_{X/k} \to 0,$$ and the Euler sequence (restricted to $X$), $$ 0 \to \Omega_{\mathbb{P}^n/k}_X \xrightarrow{v} \mathcal{O}_{\mathbb{P}^n}(1)^{\oplus(n+1)}_X \to \mathcal{O}_{\mathbb{P}^n}_X \to 0. $$ The adjoint of $v\circ u$ is an invertible quotient, $$ \mathcal{O}_{\mathbb{P}^n}(1)^{\oplus(n+1)}_X \to \mathcal{O}_{\mathbb{P}^n}(d)_X, $$ which is equivalent to an invertible quotient (after twisting), $$\mathcal{O}_{\mathbb{P}^n}^{\oplus(n+1)}_X \to \mathcal{O}_{\mathbb{P}^n}(d1)_X. $$ This invertible quotient defines the morphism $G:X\to (\mathbb{P}^n)^\vee$. Thus, $G^*\mathcal{O}(1)$ is isomorphic to $\mathcal{O}_{\mathbb{P}^n}(d1)_X$. Thus, if $d>1$, then $G^*\mathcal{O}(1)$ is ample. Hence $G$ is finite. 


A reference for this statement is Dolgachev's beautiful book Classical Algebraic Geometry, Chapter 1. 

