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Let $n\geq 2$ and $\mathbb{P}^n(\mathbf{C})$ be the complexe projective space of dimension $n$. Let $H\subseteq \mathbb{P}^n(\mathbf{C})$ be a hypersurface of degree $d$ where the coordinates in $\mathbb{P}^n(\mathbf{C})$ are chosen to be $x=[x_1,\ldots,x_{n+1}]$. Since $H$ is a hypersurface of degree $d$, it is defined by the zero locus of a homogeneous polynomial $F=F(x_1,\ldots,x_{n+1})$ of degree $d$. Similarly, let $H'\subseteq \mathbb{P}^n(\mathbf{C})$ be a second hypersurface of degree $d$ where the coordinates in $\mathbb{P}^n(\mathbf{C})$ are chosen to be $x'=[x_1',\ldots,x_{n+1}']$ and the defining homogeneous polynomial of $H'$ is $F'=F'(x_1',\ldots,x_{n+1}')$.

Let $\phi:H\rightarrow H'$ be a biregular map. So $\phi=[\phi_1,\ldots,\phi_{n+1}]$ where the $\phi_i$'s may be assumed to be homogeneous polynomials of degree $e\geq 1$ such that $\gcd(\phi_1,\ldots,\phi_{n+1})=1$. By convention the trivial polynomial $0$ is consider to be homogeneous in any degree.

Question: Do we necessarily have $e=1$ ?

Let me give an interesting special case where the question has a positive answer.

Proposition: Assume that $P_0=[0,0,\ldots,0,1]$ is a point, both on $H$ and $H'$, such that $\phi(P_0)=P_0$. Assume furthermore that if $x'\in H'$ is such that $x_1'=0$ then necessarily $x_2'=0$. Then $e=1$.

Proof Let us suppose that $e\geq 2$. We will reach a contradiction by showing that the variety set $\phi^{-1}(P_0)$ has a too large total multiplicity. Notice that $\phi^{-1}(P_0)$ correspond to the zero locus of the following set of polynomials in the variables $x_1,\ldots,x_{n+1}$: $$ F=0, \phi_1=0,\phi_3=0,\ldots,\phi_{n+1}=0. $$ This variety, say $V\subseteq H$, must be zero dimensional becaue $\phi$ is a finite to one map. Therefore, by Bezout theorem, this intersection set, counting multiplicities, has size $de^{n-1}$. Since $\phi$ is injective and $\phi(P_0)=P_0$, it follows that this set consists of a single point, namely $P_0$, which has multiplicity $de^{n-1}$. On the other hand, the subvariety $V\subseteq H$ may also be characterized as being the zero locus of $$ F=0,x_1=0,x_2=0,\ldots, x_{n}=0. $$ Again by Bezout theorem, the multiplicity of $P_0$ must be $d$. But $d<de^{n-1}$. Contradiction.

Note that the result of the previous proposition covers the case of an elliptic curve written in Weierstrass normal form.

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    $\begingroup$ What you write is obviously wrong when $n=2$ and $d=2$ or $d=3$. One basic problem is your hypothesis that every morphism $\phi$ is given as the restriction to $H$ of an $(n+1)$-tuple of homogeneous polynomials. $\endgroup$ – Jason Starr Sep 19 '15 at 15:04
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    $\begingroup$ Maybe the only counterexample is when $n$ equals $2$ and $d$ equals $3$ ($d=2$ does not seem to be a counterexample). $\endgroup$ – Jason Starr Sep 19 '15 at 16:29
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Since $\phi$ is biregular, it induces an isomorphism $\mathrm{Pic}(H')\stackrel{\sim}{\rightarrow }\mathrm{Pic}(H)$, which maps $\mathcal{O}_{H'}(1)$ to $\mathcal{O}_H(e)$. If $e>1$ this implies that $\mathcal{O}_{H'}(1)$ is divisible in $\mathrm{Pic}(H')$, which is impossible if your hypersurfaces have reasonable singularities, in particular if they are smooth.

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  • $\begingroup$ Thanks abx for the simple argument. At least it covers the generic case, namely smooth hypersurfaces. $\endgroup$ – Hugo Chapdelaine Sep 20 '15 at 12:54
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This seems to be true for $n\geq 4$ and if the hypersurfaces are assumed to be smooth, since then Picard groups are generated by $\mathcal{O}(1)$, by Lefschetz theorem.

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