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Angelo
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The stack of curves of genus 0 with at most one node is a quotient stack (see The integral Chow ring of the stack of at most 1-nodal rational curves, but Edidin and Fulghesu), so you are fine in this case.

On the other hand, Andrew Kresch proved that the stack of nodal curves of genus 0, with at most two nodes, is not a quotient stack (Flattening stratification and the stack of partial stabilizations of prestable curves). Given that this stack has only three points, this should gives a counterexample.

The stack of curves of genus 0 with at most one node is a quotient stack (see The integral Chow ring of the stack of at most 1-nodal rational curves, but Edidin and Fulghesu), so you are fine in this case.

On the other hand, Andrew Kresch proved that the stack of nodal curves of genus 0, with at most two nodes, is not a quotient stack (Flattening stratification and the stack of partial stabilizations of prestable curves). Given that this stack has only three points, this should gives a counterexample.

The stack of curves of genus 0 with at most one node is a quotient stack (see The integral Chow ring of the stack of at most 1-nodal rational curves, but Edidin and Fulghesu), so you are fine in this case.

On the other hand, Andrew Kresch proved that the stack of nodal curves of genus 0 with at most two nodes is not a quotient stack (Flattening stratification and the stack of partial stabilizations of prestable curves). Given that this stack has only three points, this should gives a counterexample.

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Angelo
  • 27k
  • 6
  • 92
  • 112

The stack of curves of genus 0 with at most one node is a quotient stack (see The integral Chow ring of the stack of at most 1-nodal rational curves, but Edidin and Fulghesu), so you are fine in this case.

On the other hand, Andrew Kresch proved that the stack of nodal curves of genus 0, with at most two nodes, is not a quotient stack (Flattening stratification and the stack of partial stabilizations of prestable curves). Given that this stack has only three points, this should gives a counterexample.