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Up to passing to the Galois closure $\bar{X} \to X$ and using Riemann Extension Theorem (ensuring that $\bar{X}$ is algebraic) we may assume that $X \to Y$ is a Galois cover, induced by the action of a finite group $G$. Then, quoting M. Roth answer to thisthis MathOverflow question

In the case of a quotient of a scheme $X$ by a finite group $G$, a necessary and sufficient condition for the quotient scheme $X/G$ to exist is that the orbit of every point of $X$ be contained in an affine open subset of $X$. This is proved in SGA I, Exposé V, Proposition 1.8.

The condition is in particular satisfied when $X$ is a projective scheme, because the orbit is formed by a finite number of points and we can take the affine subset of $X$ given by the complement of a hyperplane section avoiding all of them (note that the property of being projective is preserved in the Galois closure, since the pullback of an ample divisor by a finite map is again ample).

Up to passing to the Galois closure $\bar{X} \to X$ and using Riemann Extension Theorem (ensuring that $\bar{X}$ is algebraic) we may assume that $X \to Y$ is a Galois cover, induced by the action of a finite group $G$. Then, quoting M. Roth answer to this MathOverflow question

In the case of a quotient of a scheme $X$ by a finite group $G$, a necessary and sufficient condition for the quotient scheme $X/G$ to exist is that the orbit of every point of $X$ be contained in an affine open subset of $X$. This is proved in SGA I, Exposé V, Proposition 1.8.

The condition is in particular satisfied when $X$ is a projective scheme, because the orbit is formed by a finite number of points and we can take the affine subset of $X$ given by the complement of a hyperplane section avoiding all of them (note that the property of being projective is preserved in the Galois closure, since the pullback of an ample divisor by a finite map is again ample).

Up to passing to the Galois closure $\bar{X} \to X$ and using Riemann Extension Theorem (ensuring that $\bar{X}$ is algebraic) we may assume that $X \to Y$ is a Galois cover, induced by the action of a finite group $G$. Then, quoting M. Roth answer to this MathOverflow question

In the case of a quotient of a scheme $X$ by a finite group $G$, a necessary and sufficient condition for the quotient scheme $X/G$ to exist is that the orbit of every point of $X$ be contained in an affine open subset of $X$. This is proved in SGA I, Exposé V, Proposition 1.8.

The condition is in particular satisfied when $X$ is a projective scheme, because the orbit is formed by a finite number of points and we can take the affine subset of $X$ given by the complement of a hyperplane section avoiding all of them (note that the property of being projective is preserved in the Galois closure, since the pullback of an ample divisor by a finite map is again ample).

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Francesco Polizzi
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Up to passing to the Galois closure $\bar{X} \to X$ and using Riemann Extension Theorem (ensuring that $\bar{X}$ is algebraic) we may assume that $X \to Y$ is a Galois cover, induced by the action of a finite group $G$. Then, quoting M. Roth answer to this MathOverflow question

In the case of a quotient of a scheme $X$ by a finite group $G$, a necessary and sufficient condition for the quotient scheme $X/G$ to exist is that the orbit of every point of $X$ be contained in an affine open subset of $X$. This is proved in SGA I, Exposé V, Proposition 1.8.

The condition is in particular satisfied when $X$ is a projectiveprojective scheme, because the orbit is formed by a finite number of points and we can take the affine subset of $X$ given by the complement of a hyperplane section avoiding all of them (note that the property of being projective is preserved in the Galois closure, since the pullback of an ample divisor by a finite map is again ample).

Up to passing to the Galois closure $\bar{X} \to X$ and using Riemann Extension Theorem (ensuring that $\bar{X}$ is algebraic) we may assume that $X \to Y$ is a Galois cover, induced by the action of a finite group $G$. Then, quoting M. Roth answer to this MathOverflow question

In the case of a quotient of a scheme $X$ by a finite group $G$, a necessary and sufficient condition for the quotient scheme $X/G$ to exist is that the orbit of every point of $X$ be contained in an affine open subset of $X$. This is proved in SGA I, Exposé V, Proposition 1.8.

The condition is in particular satisfied when $X$ is a projective scheme, because the orbit is formed by a finite number of points and we can take the affine subset of $X$ given by the complement of a hyperplane section avoiding all of them (note that the property of being projective is preserved in the Galois closure, since the pullback of an ample divisor by a finite map is again ample).

Up to passing to the Galois closure $\bar{X} \to X$ and using Riemann Extension Theorem (ensuring that $\bar{X}$ is algebraic) we may assume that $X \to Y$ is a Galois cover, induced by the action of a finite group $G$. Then, quoting M. Roth answer to this MathOverflow question

In the case of a quotient of a scheme $X$ by a finite group $G$, a necessary and sufficient condition for the quotient scheme $X/G$ to exist is that the orbit of every point of $X$ be contained in an affine open subset of $X$. This is proved in SGA I, Exposé V, Proposition 1.8.

The condition is in particular satisfied when $X$ is a projective scheme, because the orbit is formed by a finite number of points and we can take the affine subset of $X$ given by the complement of a hyperplane section avoiding all of them (note that the property of being projective is preserved in the Galois closure, since the pullback of an ample divisor by a finite map is again ample).

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Francesco Polizzi
  • 66.3k
  • 5
  • 180
  • 283

Up to passing to the Galois closure $\bar{X} \to X$ and using Riemann Extension Theorem (ensuring that $\bar{X}$ is algebraic) we may assume that $X \to Y$ is a Galois cover, induced by the action of a finite group $G$. Then, quoting M. Roth answer to this MathOverflow question

In the case of a quotient of a scheme $X$ by a finite group $G$, a necessary and sufficient condition for the quotient scheme $X/G$ to exist is that the orbit of every point of $X$ be contained in an affine open subset of $X$. This is proved in SGA I, Exposé V, Proposition 1.8.

The condition is in particular satisfied when $X$ is a (quasi)-projectiveprojective scheme, because the orbit is formed by a finite number of points and we can take the affine subset of $X$ given by the complement of a hyperplane section avoiding all of them (note that the property of being (quasi)-projectiveprojective is preserved in the Galois closure, since the pullback of an ample divisor by a finite map is again ample).

Up to passing to the Galois closure $\bar{X} \to X$ and using Riemann Extension Theorem (ensuring that $\bar{X}$ is algebraic) we may assume that $X \to Y$ is a Galois cover, induced by the action of a finite group $G$. Then, quoting M. Roth answer to this MathOverflow question

In the case of a quotient of a scheme $X$ by a finite group $G$, a necessary and sufficient condition for the quotient scheme $X/G$ to exist is that the orbit of every point of $X$ be contained in an affine open subset of $X$. This is proved in SGA I, Exposé V, Proposition 1.8.

The condition is in particular satisfied when $X$ is a (quasi)-projective scheme, because the orbit is formed by a finite number of points and we can take the affine subset of $X$ given by the complement of a hyperplane section avoiding all of them (note that the property of being (quasi)-projective is preserved in the Galois closure, since the pullback of an ample divisor by a finite map is again ample).

Up to passing to the Galois closure $\bar{X} \to X$ and using Riemann Extension Theorem (ensuring that $\bar{X}$ is algebraic) we may assume that $X \to Y$ is a Galois cover, induced by the action of a finite group $G$. Then, quoting M. Roth answer to this MathOverflow question

In the case of a quotient of a scheme $X$ by a finite group $G$, a necessary and sufficient condition for the quotient scheme $X/G$ to exist is that the orbit of every point of $X$ be contained in an affine open subset of $X$. This is proved in SGA I, Exposé V, Proposition 1.8.

The condition is in particular satisfied when $X$ is a projective scheme, because the orbit is formed by a finite number of points and we can take the affine subset of $X$ given by the complement of a hyperplane section avoiding all of them (note that the property of being projective is preserved in the Galois closure, since the pullback of an ample divisor by a finite map is again ample).

added 158 characters in body
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Francesco Polizzi
  • 66.3k
  • 5
  • 180
  • 283
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Francesco Polizzi
  • 66.3k
  • 5
  • 180
  • 283
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