Let $k$ be a field and $M$ be sharp monoid (with no invertible element) consider the log point $\eta_M=(\operatorname{Spec}(k), M)$. Let $X$ be a fine saturated scheme over $\eta_M$ such that the structure morphism from $X$ to $\eta_M$ is log-smooth.( We can assume $X$ has a global chart $a:X \to \operatorname{Spec}(\mathbb{Z}[P]))$. Does that imply that $X$ is reduced? What if $M= \mathbb{N}$ or identity monoid (i.e the log structure on $\operatorname{Spec}(k)$ is the trivial one)? If no, what is the simplest counterexample?
1 Answer
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TL;DR the answer is no, but yes if(f) the morphism is saturated.
Notation: $\mathbf{A}_P = {\rm Spec}(k[P])$ with the standard log structure.
Basic counterexample. Let $n$ be an integer invertible in $k$, and consider the map $$ f_n\colon \mathbf{A}_P\to \mathbf{A}_P $$ induced by the multiplication by $n$ map $P\to P$. The map $f_n$ is log smooth (or even log etale), and hence so is its base change to the log point $\eta_P = {\rm Spec}(P\to k)$. But the fiber over $\eta_P$ is $$ {\rm Spec}(P\to k[P]/I) $$ where $I$ is the ideal generated by $n\cdot (P\setminus\{0\})$, which is non-reduced if $n>1$ and $P$ is nontrivial.
In the simplest example $P=\mathbf{N}$ you get the standard tamely ramified map $t\mapsto t^n\colon \mathbf{A}^1\to \mathbf{A}^1$.
The assumption that is often used to prevent non-reduced fibers is that the morphism is saturated. You can read about saturated morphisms in Tsuji's manuscript (now published in the Tunisian Journal of Mathematics, see Theorem II.4.2) or in Ogus' forthcoming book.
answered Jun 7, 2018 at 20:04
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$\begingroup$ Thank you very much for the answer. What if the log structure on $k$ is trivial? Shouldn't it be true then? $\endgroup$– ABCCommented Jun 9, 2018 at 18:10
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$\begingroup$ Yes. In this case $X$ looks locally like $\mathbf{A}_P$ for a fine and saturated $P$, which is reduced. $\endgroup$ Commented Jun 11, 2018 at 9:54