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Hironaka's example of a proper non-projective 3-fold has a $\mathbb Z/2$ action which is free (actually, there's one fixed point, but you can just throw it out) whose quotient is not a scheme. (see an appendix of Hartshorne (page 443) or Shafarevich (page 75))

In general, the quotient of a scheme by a free action of a group is an algebraic space. If this group is finite, this is just because the action induces an etale equivalence relation (this is where you use freeness), and any quotient of a scheme by an etale equivalence relation is an algebraic space. For infinite groups, the quotient will be an algebraic stack fibered in sets, so it will be an algebraic space (c.f. this question).

You are then reduced to the question, when is an algebraic space a scheme?

Here's one situation where you get a scheme quotient. If you have a quasi-projective variety $X$ with an action of a connected reductive group $G$, then there is a geometric quotient $X/G$ if there exists a line bundle $L$ (with $G$-action compatible with the $G$-action on $X$) such that every point of $X$ is stable with respect to $L$. That is, if for every point $x\in X$, there is an invariant section $s$ of some tensor power of $L$ such that $X_s$ (the non-vanishing locus of $s$) is an open affine neighborhood of $x$ in which every $G$-orbit is closed. This is Theorem 1.10 in Geometric Invariant Theory. I don't know if knowing that the action of $G$ is free helps in finding such an $L$.

Hironaka's example of a proper non-projective 3-fold has a $\mathbb Z/2$ action which is free (actually, there's one fixed point, but you can just throw it out) whose quotient is not a scheme. (see an appendix of Hartshorne (page 443) or Shafarevich (page 75))

In general, the quotient of a scheme by a free action of a group is an algebraic space. If this group is finite, this is just because the action induces an etale equivalence relation (this is where you use freeness), and any quotient of a scheme by an etale equivalence relation is an algebraic space. For infinite groups, the quotient will be an algebraic stack fibered in sets, so it will be an algebraic space (c.f. this question).

You are then reduced to the question, when is an algebraic space a scheme?

Here's one situation where you get a scheme quotient. If you have a quasi-projective variety $X$ with an action of a connected reductive group $G$, then there is a geometric quotient $X/G$ if there exists a line bundle $L$ (with $G$-action compatible with the $G$-action on $X$) such that every point of $X$ is stable with respect to $L$. That is, if for every point $x\in X$, there is an invariant section $s$ of some tensor power of $L$ such that $X_s$ (the non-vanishing locus of $s$) is an open affine neighborhood of $x$ in which every $G$-orbit is closed. This is Theorem 1.10 in Geometric Invariant Theory. I don't know if knowing that the action of $G$ is free helps in finding such an $L$.

Hironaka's example of a proper non-projective has a Z/2 action which is free (actually, there's one fixed point, but you can just throw it out) whose quotient is not a scheme. (see an appendix of Hartshorne (page 443) or Shafarevich (page 75))

In general, the quotient of a scheme by a free action of a finite group is an algebraic space. This is just because the action induces an etale equivalence relation (this is where you use freeness), and any quotient of a scheme by an etale equivalence relation is an algebraic space.

If you have a quasi-projective variety X with an action of a connected reductive group G, then there is a geometric quotient X/G if there exists a line bundle L (with G-action compatible with the G-action on X) such that every point of X is stable with respect to L. That is, if for every point x∈X, there is an invariant section f of some tensor power of L such that X_f (the non-vanishing locus of f) is an open affine neighborhood of x in which every G-orbit is closed. This is Theorem 1.10 in Geometric Invariant Theory. I don't know how much it helps to know that the action of G is free.

Hironaka's example of a proper non-projective 3-fold has a $\mathbb Z/2$ action which is free (actually, there's one fixed point, but you can just throw it out) whose quotient is not a scheme. (see an appendix of Hartshorne (page 443) or Shafarevich (page 75))

In general, the quotient of a scheme by a free action of a group is an algebraic space. If this group is finite, this is just because the action induces an etale equivalence relation (this is where you use freeness), and any quotient of a scheme by an etale equivalence relation is an algebraic space. For infinite groups, the quotient will be an algebraic stack fibered in sets, so it will be an algebraic space (c.f. this question).

You are then reduced to the question, when is an algebraic space a scheme?

Here's one situation where you get a scheme quotient. If you have a quasi-projective variety $X$ with an action of a connected reductive group $G$, then there is a geometric quotient $X/G$ if there exists a line bundle $L$ (with $G$-action compatible with the $G$-action on $X$) such that every point of $X$ is stable with respect to $L$. That is, if for every point $x\in X$, there is an invariant section $s$ of some tensor power of $L$ such that $X_s$ (the non-vanishing locus of $s$) is an open affine neighborhood of $x$ in which every $G$-orbit is closed. This is Theorem 1.10 in Geometric Invariant Theory. I don't know if knowing that the action of $G$ is free helps in finding such an $L$.

Hironaka's standard example of a proper non-projective (pictured in an appendix of [Hartshorne][2]; page 443, unfortunately not on Google books) has a Z/2 action which is free (actually, there's one fixed point, but you can just throw it out) whose quotient is not a scheme.

In general, the quotient of a scheme by a free action of a finite group is an algebraic space. This is just because the action induces an etale equivalence relation (this is where you use freeness), and any quotient of a scheme by an etale equivalence relation is an algebraic space.

If you have a quasi-projective variety X with an action of a connected reductive group G, then there is a geometric quotient X/G if there exists a line bundle L (with G-action compatible with the G-action on X) such that every point of X is stable with respect to L. That is, if for every point x∈X, there is an invariant section f of some tensor power of L such that X_f (the non-vanishing locus of f) is an open affine neighborhood of x in which every G-orbit is closed. I don't know how much it helps to know that the action of G is free.

Hironaka's example of a proper non-projective has a Z/2 action which is free (actually, there's one fixed point, but you can just throw it out) whose quotient is not a scheme. (see an appendix of Hartshorne (page 443) or Shafarevich (page 75))

In general, the quotient of a scheme by a free action of a finite group is an algebraic space. This is just because the action induces an etale equivalence relation (this is where you use freeness), and any quotient of a scheme by an etale equivalence relation is an algebraic space.

If you have a quasi-projective variety X with an action of a connected reductive group G, then there is a geometric quotient X/G if there exists a line bundle L (with G-action compatible with the G-action on X) such that every point of X is stable with respect to L. That is, if for every point x∈X, there is an invariant section f of some tensor power of L such that X_f (the non-vanishing locus of f) is an open affine neighborhood of x in which every G-orbit is closed. This is Theorem 1.10 in Geometric Invariant Theory. I don't know how much it helps to know that the action of G is free.