We are going to prove the stronger assertion that

$$0\to\mathbf{R}\mathcal{X}_{0}\to\Div_{0}(\mathcal{X})_{\mathbf{R}}\to N^{1}(\mathcal{X}/S)\to N^{1}(\mathcal{X}_{K}/K)\to 0$$

is exact for every regular model $\mathcal{X}$ of $X$. We first prove the exactness at $\Div_{0}(\mathcal{X})_{\mathbf{R}}$. Let $\mathcal{X}_{0}=\sum_{i\in I}b_{i}E_{i}$ be the irreducible decomposition of the special fiber. We claim that $\mathcal{X}_{0}$ is connected. Since $X\simeq\mathcal{X}_{K}^{\mathrm{an}}$ is connected by assumption, the GAGA principle implies that $\mathcal{X}_{K}$ is also connected. If $\mathcal{X}_{0}$ were disconnected then $H^{0}(\mathcal{X},\mathcal{O}_{\mathcal{X}})$ would split as a product by the Grothendieck-Zariski theorem on formal functions [Har77, Theorem 11.1], which would contradict the connectedness of $\mathcal{X}_{K}$. Since $\mathcal{X}$ is regular each $E_{i}$ is Cartier. Pick any ample divisor $\mathcal{A}$ on $\mathcal{X}$ and define a quadratic form $q$ on $\mathbf{R}^{I}$ by setting

$$q(a):=-\left(\sum_{i}a_{i}E_{i}\right)^{2}\cdot\mathcal{A}^{\dim X-1}.$$

We have $q_{ij}\leq 0$ for $i\neq j$, and the matrix $(q_{ij})$ is indecomposable since $\mathcal{X}_{0}$ is connected. By [BPV, Lemma 2.10] it follows that $b$ spans the kernel of $q$. Now let $D=\sum_{i}a_{i}E_{i}$ be a vertical $\mathbf{R}$-divisor whose numerical class on $\mathcal{X}_{0}$ is $0$. It follows that $a$ belongs to the kernel of $q$, hence is proportional to $b$, which precisely means that $D\in\mathbf{R}\mathcal{X}_{0}$ as desired.

Let us now turn to exactness at $N^{1}(\mathcal{X}/S)$, which amounts to the following assertion: every numerically trivial $L\in\Pic(\mathcal{X}_{K})$ admits a numerically trivial extension $\mathcal{L}\in\Pic(\mathcal{X})_{\mathbf{Q}}$.

Arguing as in [Kün96, Lemma 8.1], assume first that $X$ is one-dimensional. Let $\mathcal{L}\in\Pic(\mathcal{X})_{\mathbf{Q}}$ be an arbitrary extension of $L$ to the regular model $\mathcal{X}$. In the notation above we have $\sum_{i}b_{i}(\mathcal{L}\cdot E_{i})=0$ since $\mathcal{L}$ is numerically trivial on the generic fiber $X$. Since $b=(b_{i})_{i\in I}$ spans the kernel of the intersection matrix $(E_{i}\cdot E_{j})$, we may thus find $a\in\mathbf{Q}^{I}$ such that $\sum_{i}a_{i}E_{i}\cdot E_{j}=\mathcal{L}\cdot E_{j}$ for $j\in I$, which shows that $\mathcal{L}-\sum_{i}a_{i}E_{i}$ is a numerically trivial extension of $L$ to $\mathcal{X}$.

We now consider the general case, again following [Kün96, Lemma 8.1]. Given any $S$-scheme $Y$ we write $\mathcal{X}_{Y}:=\mathcal{X}\times_{S}Y$. Since $L$ is numerically trivial on $\mathcal{X}_{K}$, there exists a finite extension $K^{\prime}/K$ such that the pull-back of $L$ to $\mathcal{X}_{K^{\prime}}$ is algebraically equivalent to $0$ [Mat57]. This implies that there exists a smooth projective $K^{\prime}$-curve $T$, a numerically trivial $\mathbf{Q}$-line bundle $M$ on $T$ and a (Cartier) divisor $D$ on $\mathcal{X}_{T}$ such that

$$L=q_{*}\left(p^{*}M\cdot D\right)$$

in $\Pic(\mathcal{X}_{K})_{\mathbf{Q}}$, where $p:\mathcal{X}_{T}\to T$ and $q:\mathcal{X}_{T}\to\mathcal{X}_{K}$ are the natural morphisms. Now let $\mathcal{T}$ be a regular model of $T$ over the integral closure $S^{\prime}$ of $S$ in $K^{\prime}$, and consider the commutative diagram

(4.3)

where we also use for simplicity $p$ and $q$ to denote the natural projections $\mathcal{X}_{\mathcal{T}}\to\mathcal{T}$ and $\mathcal{X}_{\mathcal{T}}\to\mathcal{X}$. By the one-dimensional case, $M$ extends to a numerically trivial $\mathbf{Q}$-line bundle $\mathcal{M}\in\Pic(\mathcal{T})_{\mathbf{Q}}$. Let also $\mathcal{D}$ be the closure of $D$ in $\mathcal{X}_{\mathcal{T}}$, which is a priori merely a Weil divisor. We may then set

$$\mathcal{L}:=q_{*}\left(p^{*}\mathcal{M}\cdot\mathcal{D}\right).$$

Note that $\mathcal{L}$ belongs to $\CH^{1}(\mathcal{X})_{\mathbf{Q}}=\Pic(\mathcal{X})_{\mathbf{Q}}$ since $\mathcal{X}$ is regular. It is clear that $\mathcal{L}$ extends $L$, and it remains to show that $\deg(\mathcal{L}\cdot C)=0$ for each vertical projective curve $C$ on $\mathcal{X}$. Since $\mathcal{X}$ are regular, $\CH(\mathcal{X})_{\mathbf{Q}}$ is a graded commutative algebra with respect to cup-product, by [GS87, §8.3]. As in [GS92, §2.3] one can then define the cap-product $\alpha\cdot_{q}\beta$ of $\alpha\in\CH(\mathcal{X})_{\mathbf{Q}}$ and $\beta\in\CH_{*}(\mathcal{X}_{\mathcal{T}})_{\mathbf{Q}}$, which turns $\CH(\mathcal{X}_{\mathcal{T}})_{\mathbf{Q}}$ into a graded $\CH(\mathcal{X})_{\mathbf{Q}}$-module such that both $q_{*}:\CH(\mathcal{X}_{\mathcal{T}})_{\mathbf{Q}}\to\CH(\mathcal{X})_{\mathbf{Q}}$ and multiplication with $\beta^{\prime}\in\Pic(\mathcal{X}_{\mathcal{T}})_{\mathbf{Q}}$ are maps of $\CH(\mathcal{X})_{\mathbf{Q}}$-modules. Applying this with $\beta^{\prime}=p^{*}\mathcal{M}\in\Pic(\mathcal{X}_{\mathcal{T}})_{\mathbf{Q}}$, which is numerically trivial on the special fiber of $\mathcal{X}_{\mathcal{T}}$, we get

$$\deg\left(C\cdot\mathcal{L}\right)=\deg\left(C\cdot_{q}\left(\beta^{\prime}\cdot\mathcal{D}\right)\right)=\deg\left(\beta^{\prime}\cdot\left(C\cdot_{q}\mathcal{D}\right)\right)=0.$$

Finally the surjectivity of $N^{1}(\mathcal{X}/S)\to N^{1}(\mathcal{X}_{K}/K)$ is clear since $N^{1}(\mathcal{X}_{K}/K)$ is spanned by classes of Cartier divisors on $X$, the closures in $\mathcal{X}$ of which are also Cartier since $\mathcal{X}$ is regular. ∎