natural inclusion $ε(R)→E(K)$,$x→x$, regarding Neron model of elliptic curve Let $E$ be an elliptic curve over field $K$, and $ε$ be an Neron model of $E/K$.
I heard there is natural inclusion
$ε(R)→E(K)$, $x→x$.
I understand $ε(R)＝${$R$-morphisms, Spec$R→ε$}(sections of $R$-scheme $ε$) , $E(K)＝${$K$-morphisms, Spec$K→E$}($K$-rational point), but I cannot understand why the titled map is inclusion.
Could you give me explanation why $S$-morphisms, Spec$R→ε$ is regards as $K$-morphisms, Spec$K→E$ ?
(In the situation $E$ and $ε$ are defined by the same equation・・・①,$ε(R)$ and $E(R)$ corresponds to $R$($K$)-rational point of $E$, so the inclusion is obvious, but in general ① is not supposed)
Thank you in advance.
 A: You have that $\varepsilon$ is a Neron model of $E/K$. This means that $\varepsilon_K = E$ as schemes over $\operatorname{Spec} K$ (Note that $\varepsilon_K$ is the generic fiber of $\varepsilon \to \operatorname{Spec} R$).
Therefore, by base change, an $R$-valued point of $\varepsilon$ gives a morphism $$\operatorname{Spec} K = (\operatorname{Spec} R)_K \to \varepsilon_K = E,$$
i.e. a $K$-valued point of $E$. This is the map $\varepsilon(R) \to E(K)$.
For the term inclusion to make sense, you would at least want this map to be injective. As a consequence of the Neron mapping property, a $K$-valued point of $E$ is determined by a unique $R$-valued point of $\varepsilon$. In particular, this means that the map $\varepsilon(R) \to E(K)$ is bijective.
To address whether you can view $\varepsilon(R) \to E(K)$ as an inclusion without using the mapping property, note that if $r, s$ are two $R$-valued points of $\varepsilon$ whose base changes to $\operatorname{Spec} K$ are equal, i.e. $r_K = s_K$, then by definition of the base change you can show that the compositions $$
\operatorname{Spec} K \to \operatorname{Spec} R \xrightarrow{r} \varepsilon, \\
\operatorname{Spec} K \to \operatorname{Spec} R \xrightarrow{s} \varepsilon \\
$$ are equal. This is to say, $r$ and $s$ agree on the generic point $\xi$ of $\operatorname{Spec} R$.
Using @Alex's comment, since $\varepsilon$ is a separated scheme over $\operatorname{Spec} R$, the diagonal morphism $\Delta : \varepsilon \to \varepsilon \times_{\operatorname{Spec} R} \varepsilon$ is a closed immersion. Hence the locus of agreement of $r, s$ (the pullback of $\Delta$ along $(r, s) : \operatorname{Spec} R \to \varepsilon \times_{\operatorname{Spec} R} \varepsilon$) is a closed immersion $i : V \to \operatorname{Spec} R$. In particular $ri = si$, so $r$ and $s$ agree on a closed subscheme of $\operatorname{Spec} R$. Since $\operatorname{Spec} R$ is reduced and irreducible and $r$ and $s$ agree on the generic point, they must agree on $\operatorname{Spec} R$ and you have $r = s$.
