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Assume $R$ is a discrete valuation ring with uniformizing parameter $t$, i.e. $\mathfrak{m}_R=(t)$. We denote $\widehat{R}$ the completion of $R$ with respect to $(t)$. Let $Y$ be a flat locally Noetherian $R$-scheme and denote $\widehat{Y}:= Y \times_{\text{Spec } R} \text{Spec } \widehat{R}$. denote by $p:\hat{Y} \to Spec(\hat{R})$ the canonical structure morphism. An ideal sheaf $\mathcal{I} \subset \mathcal{O}_{\widehat{Y}}$ gives rise to a closed subscheme $V(\mathcal{I}) \subset \widehat{Y}$. Assume that $V(\mathcal{I})$ is contained in $V(t)$. Where $V(t)$ is by abusing of notation nothing but the closed fiber $p^{-1}(\mathfrak{m}_R) $.

Let $\pi : \widehat{Y} \to Y$ be the canonical morphism. As $Y$ is flat over $R$, the canonical homomorphism $\mathcal{O}_Y \to \pi_*\mathcal{O}_{\widehat{Y}} = \mathcal{O}_{Y} \otimes_R \widehat{R}$ stays injective. Thus the definition $\mathcal{I}_0:= \pi_* \mathcal{I} \cap \mathcal{O}_Y$ make sense.

Q: Why does the assumption $V(\mathcal{I}) \subset V(t)$ imply that $\mathcal{I}= \pi^*\mathcal{I}_0$?

My ideas: we can agrue locally or even on stalks. let $x \in \hat{Y}, \pi(x) :=y \in Y$. then by definition of pullback sheaf the stalk is given by

$${(\pi^*\mathcal{I}_0)}_x= {(\mathcal{I}_{0})}_y \otimes_{\mathcal{O}_{Y,y}} \mathcal{O}_{X,x}= {(\pi_* \mathcal{I} \cap \mathcal{O}_Y)}_y \otimes_{\mathcal{O}_{Y,y}} \mathcal{O}_{\hat{Y},x}$$

and we have to compate is with ${(\mathcal{I})}_x$.

we know that $V(\mathcal{I}) \subset V(t)$ and since

$R$ is DVR, so $Spec(\hat{R})= \{\sigma:=(t), \eta \}$, with unique close point $\sigma=(t)$ and generic point $\eta$. denote by $p:\hat{Y} \to Spec(\hat{R})$ the canonical structure map. we have disjunct topological decomposion by fibers $\hat{Y}= \hat{Y}_{\sigma} \cup \hat{Y}_{\eta}=p^{-1}(\sigma) \cup p^{-1}(\eta)$.

since $V(\mathcal{I}) \subset V(t)$ is assumed, I have to solve two problems:

  1. Why ${(\pi^*\mathcal{I}_0)}_x =0$ if $x \in \hat{Y}_{\eta}$?

  2. Can I simplify ${(\pi_* \mathcal{I} \cap \mathcal{O}_Y)}_y \otimes_{\mathcal{O}_{Y,y}} \mathcal{O}_{\hat{Y},x}$ if I know that $x \in \hat{Y}_{\sigma}$. what do I know about $\pi^* \pi_* \mathcal{I}$ when $\pi$ flat (it is flat, because $R \to \hat{R}$ is flat!)

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  • $\begingroup$ You probably want to assume that $\mathcal{I}$ is quasi-coherent. Given this, such an $\mathcal{I}$ is determined by its corresponding closed subscheme $Z \subset \widehat{Y}$, which by your assumption is a closed subscheme of some $\widehat{Y}/t^n \cong Y/t^n$. The ideal $\mathcal{I}_0$ is also quasi-coherent and its corresponding closed subscheme $Z' \subset Y$ is the schematic image of $Z$ in $Y$ (basically by definition, or because the morphism $\widehat{Y} \rightarrow Y$ is qcqs). By general definition, the schematic image $Z'$ is $Z$ itself viewed inside $Y/t^n$, hence the result. $\endgroup$
    – Kestutis Cesnavicius
    Commented Dec 7, 2019 at 22:08
  • $\begingroup$ @KestutisCesnavicius:the last step I not understand: firstly, we indeed assume $\mathcal{I}$ be qc and thus it is uniquely determined by closed subscheme $Z = V(I) \subset \widehat{Y}$. futhermore by qcqs as you said we can explicitely determine the qc ideal sheaf which determine the schematic image $Z'$, namely it's explicitly kernel $K$ of $\mathcal{O}_{Y} \to g_*\mathcal{O}_Z$ where $g$ is the composition $Z \subset \hat{Y} \to Y$. direct calculation gives indeed $K=\mathcal{I}_0= \pi_* \mathcal{I} \cap \mathcal{O}_Y$. how $\endgroup$
    – user267839
    Commented Dec 7, 2019 at 23:21
  • $\begingroup$ your last step, that "... by general definition, the schematic image $Z′=V(I_0)$ is $Z$ itself viewed inside $Y/t^n$..." translates to $\mathcal{I}= \pi^*\mathcal{I}_0$ I not understand. by assumption $Z$ is contained in $V(t) =\widehat{Y}/t^n$ and as $\widehat{Y}/t^n \cong Y/t^n$ and keeping in mind that this iso is given by restriction of $\pi$, we conclude that direct image is contained $Z' \subset Y/t^n$. that's fine. why does this already imply $\mathcal{I}= \pi^*\mathcal{I}_0$? $\endgroup$
    – user267839
    Commented Dec 7, 2019 at 23:21
  • $\begingroup$ Essentially, I need to know if and why following is true: Let $f: X \to Y$ morphism with "nice enough properties (like qcqs and and and...) between nice enough schemes as in our case" and $V \subset Y$ is reduced closed subscheme defined by ideal sheaf $J$. then $f^{-1}(V)$ is defiend by pulled back ideal $f^*J$. is this true & why? if we know it, then we are done, since $\widehat{Y}/t^n \cong Y/t^n$ is induced by $\pi$ and the whole story between $Z$ and $Z'$ take place inside $\widehat{Y}/t^n$ resp. $ Y/t^n$. $\endgroup$
    – user267839
    Commented Dec 8, 2019 at 3:23
  • $\begingroup$ By the very definition of schematic image, it is the smallest closed subscheme of $Y$ through which $Z$ factors, so it is just $Z$ viewed inside $Y$ instead of $\widehat{Y}$. To conclude, I indeed use that pulling back quasi-coherent sheaves of ideals and their associated closed subschemes are compatible; this basically amounts to the right exactness of the tensor product. $\endgroup$
    – Kestutis Cesnavicius
    Commented Dec 8, 2019 at 13:45

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