# Is this $S$-birational map an open immersion on its domain of definition?

My question is about a claim on the bottom of p. 121 of the book "Neron models" by Bosch, Lutkebohmert, and Raynaud, so I will freely use the general terminology recalled in this book, but will introduce the relevant notation here. All the references are to the book mentioned above.

Let $S$ be a noetherian, normal, strictly henselian local domain (most of these assumptions are likely to be irrelevant for the actual question, but I'll keep them), let $X$ be a smooth, separated, finite type $S$-scheme with $X \rightarrow S$ surjective, and let $m\colon X\times_S X\ --> X$ be an $S$-birational group law, which is assumed to be strict in the sense that there is an open subscheme $U \subset X\times_S X$ that is $X$-dense with respect to both projections and such that $m$ is defined on $U$, as are the universal right and left translations $$X \times_S X\ --> X\times_S X, \quad (x, y) \mapsto (xy, y) \\ X \times_S X\ --> X\times_S X, \quad (x, y) \mapsto (x, xy),$$ which are moreover required to be open immersions on $U$.

All products being over $S$, let $\Gamma$ be the schematic image in $X^3$ of the graph of $m\colon U \rightarrow X$. As is proved in 5.3/3 each projection $q_{ij}\colon X^3 \rightarrow X^2$ is an open immersion on $\Gamma$ with the image that is $X$-dense with respect to both projections. In particular, it follows that for every $S$-scheme $T$ and all $a, c \in X(T)$ there is at most one $x \in X(T)$ with $ax = c$. Suggestively denoting such $x$ by $a^{-1} c$ (if it exists), it also follows that $$f = q_{23} \circ q_{13}^{-1}\colon X\times_S X\ --> X\times_S X, \quad (a, c) \mapsto (a^{-1}c, c)$$ is an $S$-birational map whose domain of definition and image are $X$-dense with respect to both projections.

My question is: is $f$ an open immersion on its domain of definition, as is claimed on p. 121?

The apparent difficulty that is causing my trouble in seeing this is that even though I understand $f$ on $q_{13}(\Gamma)$, it could conceivably happen that the domain of definition of $f$ is larger, to the effect that $f$ looses direct touch with $m$. To get the conclusion from Zariski's main theorem, it would suffice to argue that $f$ is injective (on $T$-points for every $T$), but the injectivity is not clear to me either (same difficulty).

What I am asking about is used in the proof of Lemma 6 on p. 126 although the full strength of the claim is seemingly not needed there. In particular, the said proof (which seems absolutely crucial for the overall goal of the book) may be salvaged even if my question turns out to have a negative answer.

• Probably not. In my copy of that book I made a margin note to myself that this open immersion claim seemed doubtful, but that it is at least true on $\Gamma$ inside $X^2$ (an $X$-dense open subscheme via Lemma 3 on p. 121), and correspondingly when the same probable error occurs on page 126 one only needs the open immersion property on such a smaller domain, so this issue does not affect the intended main result of interest. Lots of books have small errors. – user74230 Nov 26 '14 at 22:03
• Thanks, this matches my observations. I guess one knows though that the claim must be true a posteriori, because the goal of all this is that $X$ may be enlarged to an $S$-group scheme $\overline{X}$, for which the corresponding map is an open immersion (even an isomorphism). I wonder if we're overlooking some argument that would prove the open immersion claim directly... – Question Mark Nov 26 '14 at 22:34
• Life is finite. Since the weaker claim suffices for the proof, and a direct fix isn't apparent after some thought, best to just move on and not worry about it (since the whole discussion is in the middle of something rather technical anyway). – user74230 Nov 27 '14 at 1:08