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Let $X$ be a scheme which is smooth, integral, of finite type and separated over $\mathbb{Z}$, equipped with a smooth morphism $h: X \to \mathbb{A}^1_\mathbb{Z}$. Let $Y \subsetneq X$ be a closed subscheme and $\beta: X' \to X$ the blow-up of $Y$.

Suppose that the restriction of $h$ to $Y$ is not dominant and $Y \otimes \mathbb{Q}$ is non-empty. If I understand things correctly, then this means that $h\mid_Y \otimes \mathbb{Q}$ is constant with some value $\nu$. Is it true that the multiplicity of $\beta^{-1}(Y)$ in the divisor of the function $(h \circ \beta) - \nu$ then has to be equal to $1$?

Moreover, if $Z$ is the intersection of $\beta^{-1}(Y)$ with the union of the other components of the divisor of $(h \circ \beta) - \nu$, is it true that $h \circ \beta$ is smooth at every point of $\beta^{-1}(Y) \setminus Z$?

I'm not sure how I should look at this situation...

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    I don't think you can deduce that $h\otimes \mathbf{Q}$ is constant by just looking at some closed subscheme. Do you mean to say that if $h|_{Y}$ is not dominant, then the morphism $h|_{Y}\otimes \mathbf{Q}$ is constant?2012-05-28
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    Yes. Thank you.2012-05-28
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    I think something is wrong. If $Y$ is contained in some $X_p$ ($p$ prime number), then $Y_{\mathbb Q}$ is empty. What would be $\nu$ ?2012-05-28
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    Mum, you skip some part of the proof. He assumes that $Y_{\mathbb Q}$ ($W_{\mathbb Q}$ with his notation) is non-empty.2012-05-28
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    Sorry about that. There are so many details around...2012-05-28
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    OK but then the question becomes a little bit too technical...2012-05-30

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