Let $X$ be a projective variety of dimension n. Then there exists a finite surjective morphism $X \to \mathbf P^n$. Let $d$ be the minimal degree of such a finite surjective morphism.

Let $d^\prime > d$.

Question. Does there exist a finite surjective morphism $X\to \mathbf P^n$ of degree $d^\prime$?

The answer is yes for multiples of $d$, but what about the general case?

Also, by Riemann-Roch the answer is yes for $n=1$.

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    $\begingroup$ In fact, I do not think it is true if $n=1$. If you choose $X$ to be an hyperelliptic curve of genus $g\geq 3$, one has $d=2$, But such a curve cannot be trigonal, which exactly means that you cannot reach $d'=3$. $\endgroup$ – Olivier Benoist Mar 5 '14 at 23:20
  • $\begingroup$ There is a paper in Inventiones (1989) by Paranjape, K. H. and Srinivas, V. Self maps of homogeneous spaces where they prove that finite surjective morphisms between Grassmannians are isomorphisms, $\endgroup$ – P Vanchinathan Mar 6 '14 at 9:13

The answer is no in general. For an easy $2$-dimensional case, take for example $X=\mathbb{P}^1\times \mathbb{P}^1$, in which case $d=2$. Then, only the multiples of $d$ are possible for $d'$. Indeed, take a morphism $X\to \mathbb{P}^2$, and denote by $D$ the divisor of its linear system, which satisfies $D^2=d'$. The Picard group of $X$ is generated by $f_1,f_2$, the two fibres of the two projections, and $D=af_1+bf_2$, so $d'=D^2=2ab$.


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