My question is about the proof of Example 8 in section 9.2 of the book "Neron models." There we have a semistable curve $X$ over an algebraically closed field $K$ and we let $\pi\colon \widetilde{X} \rightarrow X$ be its normalization. We have established the surjectivity of the natural pullback map $\pi^*: \mathrm{Pic}^0_{X/K} \rightarrow \mathrm{Pic}^0_{\widetilde{X}/K}$ of $K$-group schemes (the target being an abelian variety) and we would like to investigate the kernel $T := \mathrm{Ker}(\pi^*)$. In particular, we would like to show that $T$ is a torus. My question is: why is $T$ a torus?

The argument given for this in loc. cit. seems to be that $T$ being a torus somehow follows from an exact sequence of the form $$1 \rightarrow K^* \rightarrow \prod_{i = 1}^r K^* \rightarrow \prod_{j = 1}^N K^* \rightarrow \mathrm{Pic}(X) \rightarrow \mathrm{Pic}(\widetilde{X}) \rightarrow 1. $$ I understand the proof of this sequence, but this sequence seems to be entirely about abelian groups (at least with the level of justification given in the cited proof), so I fail to understand how it manages to imply the structure of $T$ as an algebraic group.

I would be very grateful if someone could spell out the argument or provide any clarifying comments.