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Let $X$ be a smooth hypersurface in $\mathbb{P}^{n}$ of degree $\geq 2$ and denote by $\check{\mathbb{P}}^{n}$ the dual projective space. I am trying to show that the image of $X$ under the Gauss map : \begin{alignat*}{5} \gamma : \: & X \: & \rightarrow \: & \check{\mathbb{P}}^{n} \: & \\ \: & P \: & \mapsto \: & [\frac{\partial F}{\partial x_0}(P), \: & \cdots, \frac{\partial F}{\partial x_n}(P)] \end{alignat*} is still a projective hypersurface (non necessarily smooth). It seems that my reasoning is wrong but I don't understand why, neither how to solve the problem if my solution is indeed not correct. A hint I received is to show that the fibers are zero-dimensional, I think that then the result will be direct since for any surjective morphism of quasi-projective varieties $\phi : X \rightarrow Y$, given $y \in Y$ we know that $\dim(\phi^{-1}(y)) = \dim(X)- \dim(Y)$.

So I only have to show that the fibers are zero dimensional. A "point" on $\gamma(X)$ is a tangent space $T$ to $X$ at a given point $P\in X$. So assume that there is a fibre of strictly positive dimension. Then it means that $\gamma$ would be constant on a curve $C$ contained in $X$. But then this means that the tangent space to each point of the curve is the same. Intuitively I would say that this will contradict the smoothness of $X$ but I don't succeed in showing it properly.

Any suggestion would be highly appreciated!

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Suppose the map is not finite. Then $\gamma$ is constant on a complete curve $C\subset X$. The partials $\partial_i F$ restricted to $C$ are then scalar multiples of each other, so they have a common zero. But that is absurd.

Note that this works because $d\ge 2$ means the partials are non-constant polynomials. When $d=1$, the image of the Gauss map is a point, and the partials are all constant.

Jake Levinson mentioned the following details in the comments.

[F]or morphisms between projective varieties (such as the Gauss map), "finite" is equivalent to "finite fibers", so "not finite" means some fiber is positive-dimensional, hence contains a curve. The curve is complete because it is a closed subscheme of a projective variety (and projective varieties are complete).

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  • $\begingroup$ Where are you using $d\geq 2$? $\endgroup$ Nov 5, 2015 at 23:44
  • $\begingroup$ @JakeLevinson If $d=1$, the map is to a point, because the hypersurface is a hyperplane. $\endgroup$
    – Potato
    Nov 5, 2015 at 23:50
  • $\begingroup$ Indeed, and that is not absurd... $\endgroup$ Nov 5, 2015 at 23:52
  • $\begingroup$ @JakeLevinson Then I think I don't understand your question. The problem asks about $d\ge 2$, so the partials are non-constant polynomials. $\endgroup$
    – Potato
    Nov 5, 2015 at 23:59
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    $\begingroup$ @Marianne: for morphisms between projective varieties (such as the Gauss map), "finite" is equivalent to "finite fibers", so "not finite" means some fiber is positive-dimensional, hence contains a curve. The curve is complete because it is a closed subscheme of a projective variety (and projective varieties are complete). $\endgroup$ Nov 6, 2015 at 0:11

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