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Olivier Benoist
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The answer is negative. Let me give a counter-example by modifying the example of a singular complex algebraic surface that is not a scheme given by Knutson in [Algebraic Spaces, p.21-22].

Let me work over $\mathbb{C}$. Consider the pencil in $\mathbb{P}^2$ generated by two smooth cubic curves $C$ and $C'$ intersecting transversally in nine points $P_1,\dots, P_9$. Let us choose $P_1$ as the origin of the group law on $C$. Blowing up these nine points, we get a morphism $X\to\mathbb{P}^1$ whose fiber over $0$ is the elliptic curve $X_0=C$. Choose, as usual, an inflection point $O\in C$ as the origin of the group law on $C$.

Let $\hat{X}$ be the blow-up of $X$ in a tenth point $Q\in X_0$$Q\in X_0=C$ (chosen generic, so that no multiple of $Q$ is not in the subgroup of $C$ generated by $P_1,\dots, P_9$ : here, we use the uncountability of the base field). At this point, the strict transform of $X_0$ in $\hat{X}$ has negative self-intersection and may be contracted to a point $y$ in a surface $Y\to\mathbb{P}^1$. Let $T$ be the local ring of $\mathbb{P}^1$ at $0$. I claim that $Y_T\to T$ is the counter-example we are looking for.

First, the fibers of $Y_T\to T$ (and their infinitesimal neighbourhoods) are schemes because they are one-dimensional [Algebraic Spaces V 4.9]. However, if $Y_T$ were a scheme, it would be possible to find a curve $D$ in $Y$ intersecting the tenth exceptional divisor, but not containing $y$. Its strict transform in $\mathbb{P}^2$ would be a plane curve meeting $C$ only at $P_1,\dots, P_9, Q$. By the choice of $Q$, this plane curve would meet $C$ only at $P_1,\dots, P_9$. This contradicts the fact that $D$ intersects the tenth exceptional divisor.

The answer is negative. Let me give a counter-example by modifying the example of a singular complex algebraic surface that is not a scheme given by Knutson in [Algebraic Spaces, p.21-22].

Let me work over $\mathbb{C}$. Consider the pencil in $\mathbb{P}^2$ generated by two smooth cubic curves $C$ and $C'$ intersecting transversally in nine points $P_1,\dots, P_9$. Let us choose $P_1$ as the origin of the group law on $C$. Blowing up these nine points, we get a morphism $X\to\mathbb{P}^1$ whose fiber over $0$ is the elliptic curve $X_0=C$.

Let $\hat{X}$ be the blow-up of $X$ in a tenth point $Q\in X_0$ (chosen generic, so that $Q$ is not in the subgroup of $C$ generated by $P_1,\dots, P_9$). At this point, the strict transform of $X_0$ in $\hat{X}$ has negative self-intersection and may be contracted to a point $y$ in a surface $Y\to\mathbb{P}^1$. Let $T$ be the local ring of $\mathbb{P}^1$ at $0$. I claim that $Y_T\to T$ is the counter-example we are looking for.

First, the fibers of $Y_T\to T$ (and their infinitesimal neighbourhoods) are schemes because they are one-dimensional [Algebraic Spaces V 4.9]. However, if $Y_T$ were a scheme, it would be possible to find a curve $D$ in $Y$ intersecting the tenth exceptional divisor, but not containing $y$. Its strict transform in $\mathbb{P}^2$ would be a plane curve meeting $C$ only at $P_1,\dots, P_9, Q$. By the choice of $Q$, this plane curve would meet $C$ only at $P_1,\dots, P_9$. This contradicts the fact that $D$ intersects the tenth exceptional divisor.

The answer is negative. Let me give a counter-example by modifying the example of a singular complex algebraic surface that is not a scheme given by Knutson in [Algebraic Spaces, p.21-22].

Let me work over $\mathbb{C}$. Consider the pencil in $\mathbb{P}^2$ generated by two smooth cubic curves $C$ and $C'$ intersecting transversally in nine points $P_1,\dots, P_9$. Blowing up these nine points, we get a morphism $X\to\mathbb{P}^1$ whose fiber over $0$ is the elliptic curve $X_0=C$. Choose, as usual, an inflection point $O\in C$ as the origin of the group law on $C$.

Let $\hat{X}$ be the blow-up of $X$ in a tenth point $Q\in X_0=C$ (chosen generic, so that no multiple of $Q$ is in the subgroup of $C$ generated by $P_1,\dots, P_9$ : here, we use the uncountability of the base field). At this point, the strict transform of $X_0$ in $\hat{X}$ has negative self-intersection and may be contracted to a point $y$ in a surface $Y\to\mathbb{P}^1$. Let $T$ be the local ring of $\mathbb{P}^1$ at $0$. I claim that $Y_T\to T$ is the counter-example we are looking for.

First, the fibers of $Y_T\to T$ (and their infinitesimal neighbourhoods) are schemes because they are one-dimensional [Algebraic Spaces V 4.9]. However, if $Y_T$ were a scheme, it would be possible to find a curve $D$ in $Y$ intersecting the tenth exceptional divisor, but not containing $y$. Its strict transform in $\mathbb{P}^2$ would be a plane curve meeting $C$ only at $P_1,\dots, P_9, Q$. By the choice of $Q$, this plane curve would meet $C$ only at $P_1,\dots, P_9$. This contradicts the fact that $D$ intersects the tenth exceptional divisor.

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Olivier Benoist
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The answer is negative. Let me give a counter-example by modifying the example of a singular complex algebraic spacesurface that is not a scheme given by Knutson in [algebraic spaces[Algebraic Spaces, p.21-22].

Let me work over $\mathbb{C}$. Consider the pencil in $\mathbb{P}^2$ generated by two smooth cubic curves $C$ and $C'$ intersecting transversally in nine points $P_1,\dots, P_9$. Let us choose $P_1$ as the origin of the group law on $C$. Blowing up these nine points, we get a morphism $X\to\mathbb{P}^1$ whose fiber over $0$ is anthe elliptic curve $C=X_0$. The points $P_1,\dots P_9$ induce sections, let us choose the first one as the origin of the group law on $X_0$$X_0=C$.

Blow upLet $\hat{X}$ be the blow-up of $X$ in a tenth point $Q\in X_0$ chosen(chosen generic, so that $Q$ is not in the subgroup of $C$ generated by the images of the nine sections$P_1,\dots, P_9$). At this point, the strict transform of $X_0$ in $\hat{X}$ has negative self-intersection and may be contracted to a point $y$ in a surface $Y\to\mathbb{P}^1$. Let $T$ be the local ring of $\mathbb{P}^1$ at $0$. I claim that $Y_T\to T$ is athe counter-example we are looking for.

First, the fibers of $Y_T\to T$ (and their infinitesimal neighbourhoods) are schemes bybecause they are one-dimensional [Algebraic spacesSpaces V 4.9]. However, if $Y_T$ were a scheme, it would be possible to find a curve $D$ in $Y$ intersecting the tenth exceptional divisor, but not meetingcontaining $y$. Its imagestrict transform in $\mathbb{P}^2$ would be a plane curve meeting $C$ only at $P_1,\dots, P_9, Q$. By the choice of $Q$, it meetsthis plane curve would meet $C$ only at $P_1,\dots, P_9$. This contradicts the fact that its stric transform in $Y$ should intersect$D$ intersects the tenth exceptional divisor.

The answer is negative. Let me give a counter-example by modifying the example of singular complex algebraic space that is not a scheme given by Knutson in [algebraic spaces, p.21-22].

Let me work over $\mathbb{C}$. Consider the pencil in $\mathbb{P}^2$ generated by two cubic curves $C$ and $C'$ intersecting transversally in nine points $P_1,\dots, P_9$. Blowing up these nine points, we get a morphism $X\to\mathbb{P}^1$ whose fiber over $0$ is an elliptic curve $C=X_0$. The points $P_1,\dots P_9$ induce sections, let us choose the first one as the origin of the group law on $X_0$.

Blow up $X$ in a tenth point $Q\in X_0$ chosen generic, so that $Q$ is not in the subgroup generated by the images of the nine sections. At this point, the strict transform of $X_0$ has negative self-intersection and may be contracted to a point $y$ in a surface $Y\to\mathbb{P}^1$. Let $T$ be the local ring of $\mathbb{P}^1$ at $0$. I claim that $Y_T\to T$ is a counter-example.

First, the fibers (and their infinitesimal neighbourhoods) are schemes by [Algebraic spaces V 4.9]. However, if $Y_T$ were a scheme, it would be possible to find a curve in $Y$ intersecting the tenth exceptional divisor, but not meeting $y$. Its image in $\mathbb{P}^2$ would be a plane curve meeting $C$ only at $P_1,\dots, P_9, Q$. By the choice of $Q$, it meets $C$ only at $P_1,\dots, P_9$. This contradicts the fact that its stric transform in $Y$ should intersect the tenth exceptional divisor.

The answer is negative. Let me give a counter-example by modifying the example of a singular complex algebraic surface that is not a scheme given by Knutson in [Algebraic Spaces, p.21-22].

Let me work over $\mathbb{C}$. Consider the pencil in $\mathbb{P}^2$ generated by two smooth cubic curves $C$ and $C'$ intersecting transversally in nine points $P_1,\dots, P_9$. Let us choose $P_1$ as the origin of the group law on $C$. Blowing up these nine points, we get a morphism $X\to\mathbb{P}^1$ whose fiber over $0$ is the elliptic curve $X_0=C$.

Let $\hat{X}$ be the blow-up of $X$ in a tenth point $Q\in X_0$ (chosen generic, so that $Q$ is not in the subgroup of $C$ generated by $P_1,\dots, P_9$). At this point, the strict transform of $X_0$ in $\hat{X}$ has negative self-intersection and may be contracted to a point $y$ in a surface $Y\to\mathbb{P}^1$. Let $T$ be the local ring of $\mathbb{P}^1$ at $0$. I claim that $Y_T\to T$ is the counter-example we are looking for.

First, the fibers of $Y_T\to T$ (and their infinitesimal neighbourhoods) are schemes because they are one-dimensional [Algebraic Spaces V 4.9]. However, if $Y_T$ were a scheme, it would be possible to find a curve $D$ in $Y$ intersecting the tenth exceptional divisor, but not containing $y$. Its strict transform in $\mathbb{P}^2$ would be a plane curve meeting $C$ only at $P_1,\dots, P_9, Q$. By the choice of $Q$, this plane curve would meet $C$ only at $P_1,\dots, P_9$. This contradicts the fact that $D$ intersects the tenth exceptional divisor.

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Olivier Benoist
  • 6.5k
  • 2
  • 38
  • 55

The answer is negative. Let me give a counter-example by modifying the example of singular complex algebraic space that is not a scheme given by Knutson in [algebraic spaces, p.21-22].

Let me work over $\mathbb{C}$. Consider the pencil in $\mathbb{P}^2$ generated by two cubic curves $C$ and $C'$ intersecting transversally in nine points $P_1,\dots, P_9$. Blowing up these nine points, we get a morphism $X\to\mathbb{P}^1$ whose fiber over $0$ is an elliptic curve $C=X_0$. The points $P_1,\dots P_9$ induce sections, let us choose the first one as the origin of the group law on $X_0$.

Blow up $X$ in a tenth point $Q\in X_0$ chosen generic, so that $Q$ is not in the subgroup generated by the images of the nine sections. At this point, the strict transform of $X_0$ has negative self-intersection and may be contracted to a point $y$ in a surface $Y\to\mathbb{P}^1$. Let $T$ be the local ring of $\mathbb{P}^1$ at $0$. I claim that $Y_T\to T$ is a counter-example.

First, the fibers (and their infinitesimal neighbourhoods) are schemes by [Algebraic spaces V 4.9]. However, if $Y_T$ were a scheme, it would be possible to find a curve in $Y$ intersecting the tenth exceptional divisor, but not meeting $y$. Its image in $\mathbb{P}^2$ would be a plane curve meeting $C$ only at $P_1,\dots, P_9, Q$. By the choice of $Q$, it meets $C$ only at $P_1,\dots, P_9$. This contradicts the fact that its stric transform in $Y$ should intersect the tenth exceptional divisor.