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Let $X \subset \mathbb{P}^3$ be a smooth surface of degree $d>1$ and consider the Gauss map $X \to \mathbb{P}^{3*}$, which sends a point of $X$ to its tangent plane. To see that the image $X^*$ of $X$ is a surface, I would like to prove that the Gauss map cannot be constant along a curve. How can I do this?

Also, what is the degree of $X^*$ in terms of $d$? We can assume that a general tangent plane to $X$ is tangent at only one point (e.g. in characteristic zero).

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Suppose $X$ is cut out by a polynomial $P$ of degree $d$. The Gauss map can be identified with the map which sends a vector $x$ to the partials $\{ \frac{\partial P}{\partial x_i} \}$; in particular, it is a map of degree $d-1$ (on the open subset $U$ where it is defined), i.e. $U \to (\mathbb{P}^3)^*$ pulls $\mathcal{O}(1)$ back to $\mathcal{O}(d-1)$ on $U$. But a map of degree $d-1$ cannot contract any curve because $\mathcal{O}(d-1)$ is ample and not trivial.

Concretely: if the map contracted a curve, one could choose a nontrivial linear form $\ell$ such that the map $x \mapsto \ell \circ \{ \frac{\partial P}{\partial x_i} \}$ has no zeros on a curve (take a linear form that doesn't vanish on the image of the contracted curve); this is a contradiction.

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