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In Lang's Survey on Diophantine Geometry, page 40, he said the following:

Let $F=k(Y)$ be a function field of variety $Y$ over the constant field $k$ and $X_F$ a non-singular projective variety over $F$. Then we may view $X_F$ as the generic fiber of a family, namely there exists a morphism $\pi : X \rightarrow Y$ such that the generic fiber is $X_F$. Then there exists a non-empty Zariski open $Y_0$ of $Y$ such that $\pi$ is smooth over $Y_0$. For each $y \in Y_{0}(K^a)$ we get a fiber $X_y$.

A rational point $P \in X_F(F)$ corresponds to a rational section $s_P: Y \rightarrow X$ and for $y \in Y_0$ the imbedding $\{ y\} \subset Y$ induces a point $s_P(y) \in X_y(k(y))$. The map $X_F(F) \rightarrow X_y(k(y))$ where $P \mapsto s_P(y)$ is called the specialization map. If $X=A$ is an abelian variety, then the specialization map is a homomorphism.

My questions are:

1. For all $P\in X_F(F)$, all $s_P$ are just rational maps, why are they all defined at $y \in Y_{0}(K^a)$?

2. Why is the specialization map a homomorphism when $X=A$ is an abelian variety?

Thank you.

Edited: So we have the answer (please refer to responses below) for 1 and 2 if $X$ is an abelian scheme over a smooth scheme $Y_0$. So is it true that given any abelian variety $X_F$, there exists an abelian scheme $\pi : X \rightarrow Y_0$ over $Y_0$ for some non-empty Zariski open smooth scheme $Y_0 \subset Y$, such that the generic fiber is $X_F$?

In Lang's Survey on Diophantine Geometry, page 40, he said the following:

Let $F=k(Y)$ be a function field of variety $Y$ over the constant field $k$ and $X_F$ a non-singular projective variety over $F$. Then we may view $X_F$ as the generic fiber of a family, namely there exists a morphism $\pi : X \rightarrow Y$ such that the generic fiber is $X_F$. Then there exists a non-empty Zariski open $Y_0$ of $Y$ such that $\pi$ is smooth over $Y_0$. For each $y \in Y_{0}(K^a)$ we get a fiber $X_y$.

A rational point $P \in X_F(F)$ corresponds to a rational section $s_P: Y \rightarrow X$ and for $y \in Y_0$ the imbedding $\{ y\} \subset Y$ induces a point $s_P(y) \in X_y(k(y))$. The map $X_F(F) \rightarrow X_y(k(y))$ where $P \mapsto s_P(y)$ is called the specialization map. If $X=A$ is an abelian variety, then the specialization map is a homomorphism.

My questions are:

1. For all $P\in X_F(F)$, all $s_P$ are just rational maps, why are they all defined at $y \in Y_{0}(K^a)$?

2. Why is the specialization map a homomorphism when $X=A$ is an abelian variety?

Thank you.

Edited: So we have the answer (please refer to responses below) for 1 and 2 if $X$ is an abelian scheme over a smooth scheme $Y_0$. So is it true that given any abelian variety $X_F$, there exists an abelian scheme $\pi : X \rightarrow Y_0$ over $Y_0$ for some non-empty Zariski open smooth scheme $Y_0 \subset Y$, such that the generic fiber is $X_F$? The relevant context for this question is from Serre's Lectures on the Mordell-Weil Theorem, chapter 11, page 152 where he talks about Neron specialization theorem:

Let $k$ be a number field and $A$ an abelian variety over $K=k(T_1, \ldots, T_n)$. Since $K$ is the function field of $\mathbb{P}^n$ and $A$ is defined over $K$, $A$ comes from some abelian scheme $A_U$ over a non-empty open subset $U$ of $\mathbb{P}^n$.

So my question is: why does such abelian scheme $A_U$ exist? Thank you.

In Lang's Survey on Diophantine Geometry, page 40, he said the following:

Let $F=k(Y)$ be a function field of variety $Y$ over the constant field $k$ and $X_F$ a non-singular projective variety over $F$. Then we may view $X_F$ as the generic fiber of a family, namely there exists a morphism $\pi : X \rightarrow Y$ such that the generic fiber is $X_F$. Then there exists a non-empty Zariski open $Y_0$ of $Y$ such that $\pi$ is smooth over $Y_0$. For each $y \in Y_{0}(K^a)$ we get a fiber $X_y$.

A rational point $P \in X_F(F)$ corresponds to a rational section $s_P: Y \rightarrow X$ and for $y \in Y_0$ the imbedding $\{ y\} \subset Y$ induces a point $s_P(y) \in X_y(k(y))$. The map $X_F(F) \rightarrow X_y(k(y))$ where $P \mapsto s_P(y)$ is called the specialization map. If $X=A$ is an abelian variety, then the specialization map is a homomorphism.

My questions are:

1. For all $P\in X_F(F)$, all $s_P$ are just rational maps, why are they all defined at $y \in Y_{0}(K^a)$?

2. Why is the specialization map a homomorphism when $X=A$ is an abelian variety?

Thank you.

In Lang's Survey on Diophantine Geometry, page 40, he said the following:

Let $F=k(Y)$ be a function field of variety $Y$ over the constant field $k$ and $X_F$ a non-singular projective variety over $F$. Then we may view $X_F$ as the generic fiber of a family, namely there exists a morphism $\pi : X \rightarrow Y$ such that the generic fiber is $X_F$. Then there exists a non-empty Zariski open $Y_0$ of $Y$ such that $\pi$ is smooth over $Y_0$. For each $y \in Y_{0}(K^a)$ we get a fiber $X_y$.

A rational point $P \in X_F(F)$ corresponds to a rational section $s_P: Y \rightarrow X$ and for $y \in Y_0$ the imbedding $\{ y\} \subset Y$ induces a point $s_P(y) \in X_y(k(y))$. The map $X_F(F) \rightarrow X_y(k(y))$ where $P \mapsto s_P(y)$ is called the specialization map. If $X=A$ is an abelian variety, then the specialization map is a homomorphism.

My questions are:

1. For all $P\in X_F(F)$, all $s_P$ are just rational maps, why are they all defined at $y \in Y_{0}(K^a)$?

2. Why is the specialization map a homomorphism when $X=A$ is an abelian variety?

Thank you.

Edited: So we have the answer (please refer to responses below) for 1 and 2 if $X$ is an abelian scheme over a smooth scheme $Y_0$. So is it true that given any abelian variety $X_F$, there exists an abelian scheme $\pi : X \rightarrow Y_0$ over $Y_0$ for some non-empty Zariski open smooth scheme $Y_0 \subset Y$, such that the generic fiber is $X_F$?

In Lang's Survey on Diophantine Geometry, page 40, he said the following:

Let $F=k(Y)$ be a function field of variety $Y$ over the constant field $k$ and $X_F$ a non-singular projective variety over $F$. Then we nay view $X_F$ as the generic fiber of a family, namely there exists a morphism $\pi : X \rightarrow Y$ such that the generic fiber is $X_F$. Then there exists a non-empty Zariski open $Y_0$ of $Y$ such that $\pi$ is smooth over $Y_0$. For each $y \in Y_{0}(K^a)$ we get a fiber $X_y$.

A rational point $P \in X_F(F)$ corresponds to a rational section $s_P: Y \rightarrow X$ and for $y \in Y_0$ the imbedding $\{ y\} \subset Y$ induces a point $s_P(y) \in X_y(k(y))$. The map $X_F(F) \rightarrow X_y(k(y))$ where $P \mapsto s_P(y)$ is called the specialization map. If $X=A$ is an abelian variety, then the specialization map is a homomorphism.

My questions are:

1. For all $P\in X_F(F)$, all $s_P$ are just rational maps, why are they all defined at $y \in Y_{0}(K^a)$?

2. Why is the specialization map a homomorphism when $X=A$ is an abelian variety?

Thank you.

In Lang's Survey on Diophantine Geometry, page 40, he said the following:

Let $F=k(Y)$ be a function field of variety $Y$ over the constant field $k$ and $X_F$ a non-singular projective variety over $F$. Then we may view $X_F$ as the generic fiber of a family, namely there exists a morphism $\pi : X \rightarrow Y$ such that the generic fiber is $X_F$. Then there exists a non-empty Zariski open $Y_0$ of $Y$ such that $\pi$ is smooth over $Y_0$. For each $y \in Y_{0}(K^a)$ we get a fiber $X_y$.

A rational point $P \in X_F(F)$ corresponds to a rational section $s_P: Y \rightarrow X$ and for $y \in Y_0$ the imbedding $\{ y\} \subset Y$ induces a point $s_P(y) \in X_y(k(y))$. The map $X_F(F) \rightarrow X_y(k(y))$ where $P \mapsto s_P(y)$ is called the specialization map. If $X=A$ is an abelian variety, then the specialization map is a homomorphism.

My questions are:

1. For all $P\in X_F(F)$, all $s_P$ are just rational maps, why are they all defined at $y \in Y_{0}(K^a)$?

2. Why is the specialization map a homomorphism when $X=A$ is an abelian variety?

Thank you.