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Pete L. Clark
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I think so, yes.

  1. If $V_{/\overline{\mathbb{Q}}}$ be a smooth projective variety. Then on Neron-Severi groups we have $NS(V) = NS(V_{/\mathbb{C}})$. I believe this follows from the existence of the Picard scheme: the component group of a group scheme does not change under extension from one algebraically closed field to another.

  2. Let $S_{\overline{\mathbb{Q}}}$ be a smooth, projective surface. By Castelnuovo's Criterion, minimal models are obtained precisely by a finite sequence of contractions of $-1$-curves. By part 1), every $(-1)$-curve in the Neron-Severi group of $S_{\mathbb{C}}$ is, up to algebraic equivalence, defined over $\overline{\mathbb{Q}}$, so the contractions can be done over $\mathbb{\overline{Q}}$.

Added: In the above there is an implicit assumption that the isomorphism class of the blowdown depends only on the Neron-Severi class of the $(-1)$-curve. I hope that's true; at the moment, I don't quite remember. Alternatively, the above argument certainly produces at least one $\mathbb{C}$-minimal model which is defined over $\overline{\mathbb{Q}}$. If the surface is irrational, minimal models are unique. If the surface is rational, then the minimal models are all known: see e.g. here. They can all be defined over $\overline{\mathbb{Q}}$.

I think so, yes.

  1. If $V_{/\overline{\mathbb{Q}}}$ be a smooth projective variety. Then on Neron-Severi groups we have $NS(V) = NS(V_{/\mathbb{C}})$. I believe this follows from the existence of the Picard scheme: the component group of a group scheme does not change under extension from one algebraically closed field to another.

  2. Let $S_{\overline{\mathbb{Q}}}$ be a smooth, projective surface. By Castelnuovo's Criterion, minimal models are obtained precisely by a finite sequence of contractions of $-1$-curves. By part 1), every $(-1)$-curve in the Neron-Severi group of $S_{\mathbb{C}}$ is, up to algebraic equivalence, defined over $\overline{\mathbb{Q}}$, so the contractions can be done over $\mathbb{\overline{Q}}$.

I think so, yes.

  1. If $V_{/\overline{\mathbb{Q}}}$ be a smooth projective variety. Then on Neron-Severi groups we have $NS(V) = NS(V_{/\mathbb{C}})$. I believe this follows from the existence of the Picard scheme: the component group of a group scheme does not change under extension from one algebraically closed field to another.

  2. Let $S_{\overline{\mathbb{Q}}}$ be a smooth, projective surface. By Castelnuovo's Criterion, minimal models are obtained precisely by a finite sequence of contractions of $-1$-curves. By part 1), every $(-1)$-curve in the Neron-Severi group of $S_{\mathbb{C}}$ is, up to algebraic equivalence, defined over $\overline{\mathbb{Q}}$, so the contractions can be done over $\mathbb{\overline{Q}}$.

Added: In the above there is an implicit assumption that the isomorphism class of the blowdown depends only on the Neron-Severi class of the $(-1)$-curve. I hope that's true; at the moment, I don't quite remember. Alternatively, the above argument certainly produces at least one $\mathbb{C}$-minimal model which is defined over $\overline{\mathbb{Q}}$. If the surface is irrational, minimal models are unique. If the surface is rational, then the minimal models are all known: see e.g. here. They can all be defined over $\overline{\mathbb{Q}}$.

Source Link
Pete L. Clark
  • 65.4k
  • 12
  • 241
  • 381

I think so, yes.

  1. If $V_{/\overline{\mathbb{Q}}}$ be a smooth projective variety. Then on Neron-Severi groups we have $NS(V) = NS(V_{/\mathbb{C}})$. I believe this follows from the existence of the Picard scheme: the component group of a group scheme does not change under extension from one algebraically closed field to another.

  2. Let $S_{\overline{\mathbb{Q}}}$ be a smooth, projective surface. By Castelnuovo's Criterion, minimal models are obtained precisely by a finite sequence of contractions of $-1$-curves. By part 1), every $(-1)$-curve in the Neron-Severi group of $S_{\mathbb{C}}$ is, up to algebraic equivalence, defined over $\overline{\mathbb{Q}}$, so the contractions can be done over $\mathbb{\overline{Q}}$.