$\sigma^{\prime}$.

  * (b)

The commutativity of the diagram follows because $\pi^{\prime}\circ
g=f\circ\pi$. From the proof of Lemma 10.3(d) and (e) and Lemma 10.4(c), there
exists the following commutative diagram of toric morphisms:

${X(\tau^{\prime})\times\mathbb{G}_{m}^{d}}$${X(\sigma^{\prime})\times\mathbb{G}_{m}^{d}}$${X(\tau)}$${X(\sigma)}$${X(\tau^{\prime})}$${X(\sigma^{\prime})}$$\scriptstyle{(\pi_{*}^{\prime}{,}\mathrm{id})}$$\scriptstyle{\mu_{*}}$$\scriptstyle{\mathrm{pr}_{1}}$$\scriptstyle{\lambda_{*}}$$\scriptstyle{\mathrm{pr}_{1}}$$\scriptstyle{\pi_{*}}$$\scriptstyle{g_{*}}$$\scriptstyle{f_{*}}$$\scriptstyle{\pi^{\prime}_{*}}$

,where $d$ denotes $\mathrm{rank}(\ker(f))$ and $\lambda\colon
N^{\prime}\oplus\ker(f)\rightarrow N$ and $\mu\colon
L^{\prime}\oplus\ker(f)\rightarrow L$ are isomorphisms defined by the proof of
Lemma 10.3(d) and (e). In particular, the largest diagram is a Cartesian
diagram, and $\lambda_{*}$ and $\mu_{*}$ are isomorphic. Thus, the diagram in
the statement is a Cartesian diagram.

∎

The following lemma is a basic well-known result about the torus orbit
decomposition of the toric varieties.

######  Lemma 10.6.

[1, Prop. 3.2.6] Let $N$ be a lattice of finite rank, $\Delta$ be a strongly
convex rational polyhedral fan in $N_{\mathbb{R}}$, $M$ be the dual lattice of
$N$, $\sigma$ be a cone in $\Delta$. We recall that a global section ring of
$X(\sigma)$ is $k[\sigma^{\vee}\cap M]$. Let $\mathfrak{p}_{\sigma}$ denote a
prime ideal of $k[\sigma^{\vee}\cap M]$ as follows:

$\mathfrak{p}_{\sigma}=\bigoplus_{\omega\in(\sigma^{\vee}\cap
M)\setminus\sigma^{\perp}}k\chi^{\omega}$

Let $O_{\sigma}$ denote a closed subvariety of $X(\sigma)$ associated with
$\mathfrak{p}$. Through an open immersion $X(\sigma)\hookrightarrow
X(\Delta)$, we regard $O_{\sigma}$ as a locally closed subscheme of
$X(\Delta)$.

Then the following statements follow:

  1. (a)

Let $\tau\in\Delta$ be a cone such that $\sigma\not\subset\tau$. Then
$O_{\sigma}\cap X(\tau)=\emptyset$.

  2. (b)

$X(\sigma)=O_{\sigma}\amalg\bigcup_{\tau\precneqq\sigma}X(\tau)$

  3. (c)

$X(\Delta)=\coprod_{\sigma\in\Delta}O_{\sigma}$

  4. (d)

$\overline{O_{\sigma}}=\coprod_{\sigma\preceq\tau}O_{\tau}$

###### Proof.

We prove the statement from (a) to (d) in order.

  1. (a)

Let $\gamma$ denote $\sigma\cap\tau\in\Delta$. Because
$\sigma\not\subset\tau$, $\gamma\neq\sigma$. Let $\omega\in\sigma^{\vee}$ be
an element such that $\gamma=\sigma\cap\omega^{\perp}$. We remark that
$\omega\notin\sigma^{\perp}$ because $\gamma\neq\sigma$. In particular,
$\chi^{\omega}\in\mathfrak{p}_{\sigma}$. On the other hand,
$\mathfrak{p}_{\sigma}k[\gamma^{\vee}\cap M]=k[\gamma^{\vee}\cap M]$ because
$k[\gamma^{\vee}\cap M]=k[\sigma^{\vee}\cap M]_{\chi^{\omega}}$. Thus,
$O_{\sigma}\times_{X(\sigma)}X(\gamma)=\emptyset$. Therefore, $O_{\sigma}\cap
X(\tau)=\emptyset$ because $X(\gamma)=X(\sigma)\cap X(\tau)$.

  2. (b)

From (a), for any $\tau\precneqq\sigma$, $O_{\sigma}\cap X(\tau)=\emptyset$.
Thus, $O_{\sigma}\cap\bigcup_{\tau\precneqq\sigma}X(\tau)=\emptyset$. It is
enough to show that $X(\sigma)\subset
O_{\sigma}\amalg\bigcup_{\tau\precneqq\sigma}X(\tau)$. Let
$\mathfrak{p}\subset k[\sigma^{\vee}\cap M]$ be a prime ideal. We assume that
$\mathfrak{p}\subset\mathfrak{p}_{\sigma}$. From the definition of
$\mathfrak{p}_{\sigma}$, there exists $\omega\in(\sigma^{\vee}\cap
M)\setminus\sigma^{\perp}$ such that $\chi^{\omega}\notin\mathfrak{p}$. In
particular, $\mathfrak{p}\in D(\chi^{\omega})$ where $D(\chi^{\omega})$ is a
fundamental open subset of an affine scheme $X(\sigma)$ associated with
$\chi^{\omega}\in k[\sigma^{\vee}\cap M]$. Let $\tau$ be a denote a face
$\sigma\cap\omega^{\perp}$ of $\sigma$. Then $D(\chi^{\omega})=X(\tau)$ as an
open subset of $X(\sigma)$. Thus, $\mathfrak{p}\in X(\tau)$. Because
$\omega\not\in\sigma^{\perp}$, $\tau\neq\sigma$. Therefore, $X(\sigma)\subset
O_{\sigma}\amalg\bigcup_{\tau\precneqq\sigma}X(\tau)$.

  3. (c)

First, we show that $O_{\sigma}\cap O_{\tau}=\emptyset$ for any distinct
$\sigma,\tau\in\Delta$. Because $\sigma\neq\tau$, $\sigma\not\subset\tau$ or
$\tau\not\subset\sigma$ holds. We may assume $\sigma\not\subset\tau$. Then
$O_{\sigma}\cap X(\tau)$ from (a). Because $O_{\tau}\subset X(\tau)$,
$O_{\sigma}\cap O_{\tau}=\emptyset$. Next, we show that
$X(\Delta)\subset\bigcup_{\sigma\in\Delta}O_{\sigma}$. Let $x\in X(\Delta)$ be
a point. Then there exists minimal cone $\sigma\in\Delta$ such that $x\in
X(\sigma)$ because $\Delta$ is a fan. If $x\notin O_{\sigma}$, there exists
$\tau\precneqq\sigma$ such that $x\in X(\tau)$. However, it is a contradiction
to the minimality of $\sigma$. Thus, $x\in O_{\sigma}$.

  4. (d)

Let $\tau\in\Delta$ be a cone. First, we assume that $\sigma\not\subset\tau$.
Then $O_{\sigma}\cap X(\tau)=\emptyset$ from (a). Thus,
$\overline{O_{\sigma}}\cap X(\tau)=\emptyset$. Second, we assume that
$\sigma\subset\tau$. Let $\mathfrak{q}$ denote $\mathfrak{p}_{\sigma}\cap
k[M\cap\tau^{\vee}]$. From the definition of $\mathfrak{p}_{\sigma}$,
$\mathfrak{q}=\bigoplus_{\omega\in(\tau^{\vee}\cap
M)\setminus\sigma^{\perp}}k\chi^{\omega}$. Thus,
$\mathfrak{q}\subset\mathfrak{p}_{\tau}$. Hence,
$O_{\tau}\subset\overline{O_{\sigma}}\cap X(\tau)$. Therefore, from (c), the
statement holds.

∎

The following lemma is a basic well-known result about the relation between a
toric morphism and torus orbits.

######  Lemma 10.7.

Let $N$ and $N^{\prime}$ be a lattice, $\Delta$ and $\Delta^{\prime}$ be a
strongly convex rational polyhedral fan in $N_{\mathbb{R}}$ and
$N^{\prime}_{\mathbb{R}}$ respectively. Let $\pi$ be a lattice morphism from
$N^{\prime}$ to $N$ and we assume that $\pi$ is compatible with the fans
$\Delta^{\prime}$ and $\Delta$. Then the following statements hold:

  1. (a)

Let $\tau\in\Delta^{\prime}$ be a cone and $\sigma\in\Delta$ be a minimal cone
such that $\pi_{\mathbb{R}}(\tau)\subset\sigma$. Then
$\pi_{*}(O_{\tau})\subset O_{\sigma}$.

  2. (b)

We use the notation of (a). If $\pi$ is surjective,
$\pi_{*}(O_{\tau})=O_{\sigma}$.

  3. (c)

Let $\sigma\in\Delta$ be a cone and $U$ be an open subscheme
$X(\Delta^{\prime})\times_{X(\Delta)}X(\sigma)$. Then
$U=\bigcup_{\tau\in\Delta^{\prime},\pi_{\mathbb{R}}(\tau)\subset\sigma}X(\tau)$.

###### Proof.

Let $M$ and $M^{\prime}$ be the dual lattices of $N$ and $N^{\prime}$
respectively. Let $\pi^{*}$ denote a lattice morphism $M^{\prime}\rightarrow
M$ or a ring morphism $k[M^{\prime}]\rightarrow k[M]$ induced by $\pi$. We
prove the statements from (a) to (c) in order.

  1. (a)

From the assumption, $\pi$ induces the ring morphism $\pi^{*}\colon
k[\sigma^{\vee}\cap M]\rightarrow k[\tau^{\vee}\cap M^{\prime}]$. Thus, it is
enough to show that
$\mathfrak{p}_{\sigma}\subset(\pi^{*})^{-1}(\mathfrak{p}_{\tau})$. Let
$\omega\in(\sigma^{\vee}\cap M)\setminus\sigma^{\perp}$ be an element and
$v^{\prime}\in\tau^{\circ}\cap N^{\prime}$ be an element. Then
$\pi(v^{\prime})\in\sigma^{\circ}\cap N$ from the minimally of $\sigma$.
Because $\omega\notin\sigma^{\perp}$ and $\pi(v^{\prime})\in\sigma^{\circ}\cap
N$, we can check that $\langle
v^{\prime},\pi^{*}(\omega)\rangle=\langle\pi(v^{\prime}),\omega\rangle>0$.
Thus, $\pi^{*}(\omega)\notin\tau^{\perp}$. Therefore,
$\mathfrak{p}_{\sigma}\subset(\pi^{*})^{-1}(\mathfrak{p}_{\tau})$.

  2. (b)

Because $\pi$ is surjective, $\pi^{*}\colon k[\sigma^{\vee}\cap M]\rightarrow
k[\tau^{\vee}\cap M^{\prime}]$ is injective. Let
$f=\sum_{\omega\in\sigma^{\vee}\cap
M}a_{\omega}\chi^{\omega}\in(\pi^{*})^{-1}(\mathfrak{p}_{\tau})$ be a Laurant
polynomial. From the injectivity of $\pi^{*}$,
$\sum_{\omega\in\sigma^{\vee}\cap M}a_{\omega}\chi^{\pi^{*}(\omega)}$ is a
$T_{N^{\prime}}$-invariant decomposition of $\pi^{*}(f)$. In particular, for
any $\omega\in\sigma^{\vee}\cap M$ such that $a_{\omega}\neq 0$,
$\pi^{*}(\omega)\notin\tau^{\perp}$ because
$\pi^{*}(f)\in\mathfrak{p}_{\tau}$. Let $\omega\in\sigma^{\vee}\cap M$ be an
element such that $a_{\omega}\neq 0$. Then there exists $v^{\prime}\in\tau\cap
N^{\prime}$ such that $\langle v^{\prime},\pi^{*}(\omega)\rangle>0$. Thus,
$\langle\pi(v^{\prime}),\omega\rangle=\langle
v^{\prime},\pi^{*}(\omega)\rangle>0$. Because
$\pi_{\mathbb{R}}(\tau)\subset\sigma$, $\pi(v^{\prime})\in\sigma\cap N$.
Hence, $\omega\notin\sigma^{\perp}$ and
$\chi^{\omega}\in\mathfrak{p}_{\sigma}$. Therefore,
$f\in\mathfrak{p}_{\sigma}$ and
$\mathfrak{p}_{\sigma}=(\pi^{*})^{-1}(\mathfrak{p}_{\tau})$ from (a).

The equation $\mathfrak{p}_{\sigma}=(\pi^{*})^{-1}(\mathfrak{p}_{\tau})$
induces an injective ring morphism $\pi^{*}\colon k[\sigma^{\perp}\cap
M]\rightarrow k[\tau^{\perp}\cap M^{\prime}]$. We can check that this ring
morphism is faithfully flat because lattice morphism
$\pi^{*}|_{M\cap\sigma^{\perp}}\colon M\cap\sigma^{\perp}\rightarrow
M^{\prime}\cap\tau^{\perp}$ is split injection. Thus,
$\pi_{*}(O_{\tau})=O_{\sigma}$.

  3. (c)

Let $\tau\in\Delta^{\prime}$ be a cone. First, we assume that
$\pi_{\mathbb{R}}(\tau)\subset\sigma$. Then from (a),
$\pi_{*}(O_{\tau})\subset O_{\sigma}$ so that $O_{\tau}\subset U$

Second, we assume that $\pi_{\mathbb{R}}(\tau)\not\subset\sigma$. From the
assumption of $\pi$, there exists a minimal cone $\gamma\in\Delta$ such that
$\pi_{\mathbb{R}}(\tau)\subset\gamma$. From the assumption of $\tau$,
$\gamma\not\subset\sigma$. Thus, from Lemma10.6(a), $O_{\gamma}\cap
X(\sigma)=\emptyset$. From (a), $\pi_{*}(O_{\tau})\subset O_{\gamma}$ so that
$O_{\tau}\cap U=\emptyset$.

In conclusion, from Lemma10.6(b) and (c), the following equation holds:

$\displaystyle U$
$\displaystyle=\bigcup_{\pi_{\mathbb{R}}(\tau)\subset\sigma,\tau\in\Delta^{\prime}}O_{\tau}$
$\displaystyle=\bigcup_{\pi_{\mathbb{R}}(\tau)\subset\sigma,\tau\in\Delta^{\prime}}X(\tau)$

∎

The following lemma is a basic well-known result. We used it when we consider
the mock toric structure of the closure of a stratum of a mock toric variety.

######  Proposition 10.8.

[1, Prop.3.2.7] Let $N$ be a lattice of finite rank and $\Delta$ be a strongly
convex rational polyhedral fan in $N_{\mathbb{R}}$. For $\sigma\in\Delta$, let
$\Delta^{\sigma}$ denote the following sub fan of $\Delta$.

$\Delta^{\sigma}=\\{\gamma\in\Delta\mid\exists\tau\in\Delta\ \ \mathrm{s.t.}\
\ \gamma\cup\sigma\subset\tau\\}$

Let $\pi^{\sigma}$ denote the quotient map $N\rightarrow
N/(\langle\sigma\rangle\cap N)$ and $\Delta[\sigma]$ denote the following set:

$\Delta[\sigma]=\\{(\pi^{\sigma})_{\mathbb{R}}(\tau)\mid\sigma\subset\tau\in\Delta\\}$

Then the following statements follow:

  * (a)

The set $\Delta[\sigma]$ is a strongly convex rational polyhedral fan in
$(N/(\langle\sigma\rangle\cap N))_{\mathbb{R}}$.

  * (b)

The morphism $\pi^{\sigma}$ is compatible with the fans $\Delta^{\sigma}$ and
$\Delta[\sigma]$.

  * (c)

A closed immersion $\overline{O_{\sigma}}\hookrightarrow X(\Delta)$ passes
through $X(\Delta^{\sigma})\subset X(\Delta)$.

  * (d)

The following composition is isomorphic:

${\overline{O_{\sigma}}}$${X(\Delta^{\sigma})}$${X(\Delta[\sigma])}$$\scriptstyle{(\pi^{\sigma})_{*}}$

, where the former morphism is a closed immersion, and the latter one is a
toric morphism induced by $\pi^{\sigma}$.

###### Proof.

We prove these statements from (a) to (e) in order.

  * (a)

Let $\tau\in\Delta$ be a cone such that $\sigma\preceq\tau$. Because
$\pi^{\sigma}$ is a lattice morphism, $(\pi^{\sigma})_{\mathbb{R}}(\tau)$ is a
rational polyhedral convex cone. Let $x$ and $y\in\tau$ be elements such that
$(\pi^{\sigma})_{\mathbb{R}}(x)+(\pi^{\sigma})_{\mathbb{R}}(y)=0$. Then there
exists $z$ and $w\in\sigma$ such that $x+y=z-w$ because
$\ker((\pi^{\sigma})_{\mathbb{R}})=\langle\sigma\rangle$. Thus, $x$ and
$y\in\sigma$ because $x,y,w\in\tau$, $\sigma\preceq\tau$, and
$x+y+w\in\sigma$. Therefore, $(\pi^{\sigma})_{\mathbb{R}}(\tau)$ is strongly
convex.

Next, we show that $\Delta[\sigma]$ is a fan. Let $M$ and $M^{\sigma}$ be dual
lattices of $N$ and $N/(\langle\sigma\rangle\cap N)$ respectively, and
$(\pi^{\sigma})^{*}\colon M^{\sigma}\rightarrow M$ be a morphism induced by
$\pi^{\sigma}$. Let $\tau\in\Delta$ be a cone such that $\sigma\preceq\tau$.
Then we can check that for any
$\omega\in(\pi^{\sigma})_{\mathbb{R}}(\tau)^{\vee}$,
$(\pi^{\sigma})^{*}_{\mathbb{R}}(\omega)\in\tau^{\vee}$ and
$(\pi^{\sigma})_{\mathbb{R}}(\tau\cap(\pi^{\sigma})^{*}_{\mathbb{R}}(\omega)^{\perp})=(\pi^{\sigma})_{\mathbb{R}}(\tau)\cap\omega^{\perp}$.
In fact, for any $v\in\tau$, the following equation holds:

$\displaystyle\langle v,(\pi^{\sigma})^{*}_{\mathbb{R}}(\omega)\rangle$
$\displaystyle=\langle(\pi^{\sigma})_{\mathbb{R}}(v),\omega\rangle$

In addition to this, we can check that
$(\pi^{\sigma})^{*}_{\mathbb{R}}(M^{\sigma}_{\mathbb{R}})=\langle\sigma\rangle^{\perp}$.
This implies that for any $\omega\in(\pi^{\sigma})_{\mathbb{R}}(\tau)^{\vee}$,
$\sigma\subset\tau\cap(\pi^{\sigma})^{*}_{\mathbb{R}}(\omega)^{\perp}$. Thus,
$(\pi^{\sigma})_{\mathbb{R}}(\tau\cap(\pi^{\sigma})^{*}_{\mathbb{R}}(\omega)^{\perp})\in\Delta[\sigma]$.
Therefore, all faces of $(\pi^{\sigma})_{\mathbb{R}}(\tau)$ are contained in
$\Delta[\sigma]$.

Thirdly, we show that for any $\tau$ and $\gamma\in\Delta$ such that
$\sigma\subset\tau\cap\gamma$, we have that
$(\pi^{\sigma})_{\mathbb{R}}(\tau)\cap(\pi^{\sigma})_{\mathbb{R}}(\gamma)=(\pi^{\sigma})_{\mathbb{R}}(\tau\cap\gamma)$.
Because one direction is easy, we only show that
$(\pi^{\sigma})_{\mathbb{R}}(\tau)\cap(\pi^{\sigma})_{\mathbb{R}}(\gamma)\subset(\pi^{\sigma})_{\mathbb{R}}(\tau\cap\gamma)$.
Let $x\in\tau$ and $y\in\gamma$ be elements such that
$(\pi^{\sigma})_{\mathbb{R}}(x)=(\pi^{\sigma})_{\mathbb{R}}(y)$. Then there
exist $z$ and $w\in\sigma$ such that $x-y=z-w$. The equation $x+w=y+z$
indicates that $x+w$ and $y+z\in\tau\cap\gamma$ because
$\sigma\subset\tau\cap\gamma$. Thus,
$(\pi^{\sigma})_{\mathbb{R}}(x)=(\pi^{\sigma})_{\mathbb{R}}(x+w)\in(\pi^{\sigma})_{\mathbb{R}}(\tau\cap\gamma)$.
This implies that
$(\pi^{\sigma})_{\mathbb{R}}(\tau)\cap(\pi^{\sigma})_{\mathbb{R}}(\gamma)\subset(\pi^{\sigma})_{\mathbb{R}}(\tau\cap\gamma)$.

Finally, we show that for any $\tau$ and $\gamma\in\Delta$ such that
$\sigma\subset\gamma\subset\tau$, we have that
$(\pi^{\sigma})_{\mathbb{R}}(\gamma)\preceq(\pi^{\sigma})_{\mathbb{R}}(\tau)$.
Let $\omega\in\tau^{\vee}$ be an element such that
$\gamma=\tau\cap\omega^{\perp}$. Because
$\langle\sigma\rangle\subset\langle\gamma\rangle\subset\omega^{\perp}$, there
is an element $\omega^{\prime}\in M^{\sigma}_{\mathbb{R}}$ such that
$(\pi^{\sigma})^{*}_{\mathbb{R}}(\omega^{\prime})=\omega$. Then for any
$v\in\tau$, the following equation holds:

$\displaystyle\langle(\pi^{\sigma})_{\mathbb{R}}(v),\omega^{\prime}\rangle$
$\displaystyle=\langle
v,(\pi^{\sigma})^{*}_{\mathbb{R}}(\omega^{\prime})\rangle$
$\displaystyle=\langle v,\omega\rangle$

Thus $\omega^{\prime}\in(\pi^{\sigma}_{\mathbb{R}}(\tau))^{\vee}$ and
$(\pi^{\sigma})_{\mathbb{R}}(\gamma)=(\pi^{\sigma})_{\mathbb{R}}(\tau)\cap(\omega^{\prime})^{\perp}$.
Therefore,
$(\pi^{\sigma})_{\mathbb{R}}(\gamma)\preceq(\pi^{\sigma})_{\mathbb{R}}(\tau)$.

  * (b)

For $\gamma\in\Delta^{\sigma}$, there exists $\tau\in\Delta$ such that
$\gamma\cup\sigma\subset\tau$. Then
$(\pi^{\sigma})_{\mathbb{R}}(\gamma)\subset(\pi^{\sigma})_{\mathbb{R}}(\tau)$
and $(\pi^{\sigma})_{\mathbb{R}}(\tau)\in\Delta[\sigma]$, so that
$\pi^{\sigma}$ is compatible with the fans $\Delta^{\sigma}$ and
$\Delta[\sigma]$.

  * (c)

From Lemma10.6(d), $\overline{O_{\sigma}}$ is decomposed into torus orbits
$O_{\tau}$ where $\tau\in\Delta$ is a cone which contains $\sigma$. Thus, the
following inclusion holds:

$\overline{O_{\sigma}}\subset\bigcup_{\sigma\subset\tau\in\Delta}X(\tau)=X(\Delta^{\sigma})$

  * (d)

Let $\tau\in\Delta$ be a cone such that $\sigma\subset\tau$. First, we show
that the following diagram is commutative and a Cartesian diagram:

${X(\tau)}$${X((\pi^{\sigma})_{\mathbb{R}}(\tau))}$${X(\Delta^{\sigma})}$${X(\Delta[\sigma])}$$\scriptstyle{(\pi^{\sigma})_{*}}$$\scriptstyle{(\pi^{\sigma})_{*}}$

In fact, for any $\gamma\in\Delta^{\sigma}$, we can show that if
$(\pi^{\sigma})_{\mathbb{R}}(\gamma)\subset(\pi^{\sigma})_{\mathbb{R}}(\tau)$,
then $\gamma\subset\tau$ as follows: Let $x\in\gamma$ and $y\in\tau$ be
elements such that
$(\pi^{\sigma})_{\mathbb{R}}(x)=(\pi^{\sigma})_{\mathbb{R}}(y)$. Then there
exists $z$ and $w\in\sigma$ such that $x-y=z-w$. From the definition of
$\Delta^{\sigma}$, there exists $\beta\in\Delta$ such that
$\sigma\cup\gamma\subset\beta$. The equation $x+w=y+z$ indicates that
$x+w\in\tau$ because $\sigma\subset\tau$. Because $x$ and $w\in\beta$, and
$\beta\cap\tau\preceq\beta$, we have $x$ and $w\in\beta\cap\tau$. In
particular, $x\in\tau$. Thus, the diagram above is a Cartesian diagram from
Lemma10.7(c).

Next, we show that for any $\tau\in\Delta$ such that $\sigma\subset\tau$, the
upper composition from $\overline{O_{\sigma}}\cap X(\tau)$ to
$X((\pi^{\sigma})_{\mathbb{R}}(\tau))$ of the following diagram whose all
squares are Cartesian squares is an isomorphism:

${\overline{O_{\sigma}}\cap
X(\tau)}$${X(\tau)}$${X((\pi^{\sigma})_{\mathbb{R}}(\tau))}$${\overline{O_{\sigma}}}$${X(\Delta^{\sigma})}$${X(\Delta[\sigma])}$$\scriptstyle{(\pi^{\sigma})_{*}}$$\scriptstyle{(\pi^{\sigma})_{*}}$

If this claim holds, then the statement (c) follows because
$X(\Delta[\sigma])$ is covered by
$\\{X((\pi^{\sigma})_{\mathbb{R}}(\tau))\\}_{(\sigma\subset)\ \tau\in\Delta}$.

Once the argument comes to affine toric varieties, we only check it
algebraically. We use the notation in the proof of statement (a), and we can
show that
$(\pi^{\sigma})^{*}_{\mathbb{R}}((\pi^{\sigma})_{\mathbb{R}}(\tau)^{\vee})=\langle\sigma\rangle^{\perp}\cap\tau^{\vee}$.
Moreover, $(\pi^{\sigma})^{*}((\pi^{\sigma})_{\mathbb{R}}(\tau)^{\vee}\cap
M^{\sigma})=\langle\sigma\rangle^{\perp}\cap\tau^{\vee}\cap M$ because
$\pi^{\sigma}(M^{\sigma})=\langle\sigma\rangle^{\perp}\cap M$. By the way, the
ring morphism associated with $(\pi^{\sigma})_{*}$ is a natural inclusion
$k[(\pi^{\sigma})_{\mathbb{R}}(\tau)^{\vee}\cap M^{\sigma}]\hookrightarrow
k[\tau^{\vee}\cap M]$. On the other hand, the ring morphism associated with
the closed immersion $\overline{O_{\sigma}}\cap X(\tau)\rightarrow X(\tau)$ is
$k[\tau^{\vee}\cap M]\rightarrow
k[\tau^{\vee}\cap\langle\sigma\rangle^{\perp}\cap M]$ whose kernel is
generated by $\\{\chi^{\omega}\\}_{\omega\in(\tau^{\vee}\cap
M)\setminus\langle\sigma\rangle^{\perp}}$. Therefore, the morphism
$\overline{O_{\sigma}}\cap X(\tau)\rightarrow
X((\pi^{\sigma})_{\mathbb{R}}(\tau))$ is an isomorphism.

∎

######  Corollary 10.9.

We keep the notation in Proposition 10.8, let $\sigma,\tau,$ and $\gamma$ be a
cone in $\Delta$ such that $\sigma\subset\tau\cap\gamma$. the following
statements follow:

  1. (a)

If
$(\pi^{\sigma})_{\mathbb{R}}(\gamma)\subset(\pi^{\sigma})_{\mathbb{R}}(\tau)$,
then $\gamma\subset\tau$.

  2. (b)

$(\pi^{\sigma})_{\mathbb{R}}(\gamma)\cap(\pi^{\sigma})_{\mathbb{R}}(\tau)=(\pi^{\sigma})_{\mathbb{R}}(\gamma\cap\tau)$.

###### Proof.

Both statements follow from the proof of Proposition 10.8. ∎

The following lemma concerns toric varieties, which have a dominant toric
morphism to $\mathbb{A}^{1}_{k}$.

######  Lemma 10.10.

Let $N$ be a lattice of finite rank and $\Delta$ be a strongly convex rational
polyhedral fan in $(N\oplus\mathbb{Z})_{\mathbb{R}}$. We assume that
$\mathrm{pr}_{2}\colon N\oplus\mathbb{Z}\rightarrow\mathbb{Z}$ is compatible
with the fans $\Delta$ and $\Delta_{!}$. Let $\Delta_{0}$ denote the following
sub fan of $\Delta$:

$\Delta_{0}=\\{\sigma\in\Delta\mid(\mathrm{pr}_{2})_{\mathbb{R}}(\sigma)=\\{0\\}\\}$

Let $\Delta_{(0)}$ denote the following set:

$\Delta_{(0)}=\\{(\mathrm{pr}_{1})_{\mathbb{R}}(\sigma)\mid\sigma\in\Delta_{0}\\}$

Then the following statements follow:

  1. (a)

The set $\Delta_{(0)}$ is a strongly convex rational polyhedral fan in
$N_{\mathbb{R}}$.

  2. (b)

A toric morphism $(\mathrm{pr}_{1})_{*}\times(\mathrm{pr}_{2})_{*}\colon
X(\Delta_{0})\rightarrow X(\Delta_{(0)})\times{\mathbb{G}_{m}^{1}}_{,k}$ is an
isomorphism. Let $\sigma\in\Delta_{0}$ be a cone and $\sigma_{0}$ denote
$(\mathrm{pr}_{1})_{\mathbb{R}}(\sigma)$. Then the above isomorphism induces
the following three Cartesian diagrams:

${X(\sigma)}$${X(\sigma_{0})\times{\mathbb{G}_{m}^{1}}_{,k}}$${X(\Delta_{0})}$${X(\Delta_{(0)})\times{\mathbb{G}_{m}^{1}}_{,k}}$$\scriptstyle{p}$
${\overline{O_{\sigma}}\cap
X(\Delta_{0})}$${\overline{O_{\sigma_{0}}}\times{\mathbb{G}_{m}^{1}}_{,k}}$${X(\Delta_{0})}$${X(\Delta_{(0)})\times{\mathbb{G}_{m}^{1}}_{,k}}$$\scriptstyle{p}$
${O_{\sigma}}$${O_{\sigma_{0}}\times{\mathbb{G}_{m}^{1}}_{,k}}$${X(\Delta_{0})}$${X(\Delta_{(0)})\times{\mathbb{G}_{m}^{1}}_{,k}}$$\scriptstyle{p}$

where $p$ denote $(\mathrm{pr}_{1})_{*}\times(\mathrm{pr}_{2})_{*}$.

  3. (c)

Let $\Psi$ denote the composition of the following maps:

$X_{k(t)}(\Delta_{(0)})\rightarrow
X(\Delta_{(0)})\times{\mathbb{G}_{m}^{1}}_{,k}\rightarrow
X(\Delta_{0})\hookrightarrow X(\Delta),$

where the first morphism is a base change of the generic fiber
$\operatorname{Spec}(k(t))\rightarrow\mathbb{G}_{m}^{1}$ along
$(\mathrm{pr}_{2})_{*}\colon
X(\Delta_{(0)})\times{\mathbb{G}_{m}^{1}}_{,k}\rightarrow\mathbb{G}_{m}^{1}$,
the second one is the inverse morphism of the morphism in (b), and the third
one is an open immersion. Then the large square of the following diagram is a
Cartesian square:

${X_{k(t)}(\Delta_{(0)})}$${X(\Delta_{(0)})\times{\mathbb{G}_{m}^{1}}_{,k}}$${X(\Delta_{0})}$${X(\Delta)}$${\operatorname{Spec}(k(t))}$
${\mathbb{G}_{m}^{1}}$${\mathbb{A}^{1}}$$\scriptstyle{\sim}$$\scriptstyle{\mathrm{pr}_{2}}$$\scriptstyle{(\mathrm{pr}_{2})_{*}}$

  4. (d)

A morphism $\Psi$ induces the following three Cartesian diagrams:

${X_{k(t)}(\sigma_{0})}$${X(\sigma)}$${X_{k(t)}(\Delta_{(0)})}$${X(\Delta)}$$\scriptstyle{\Psi}$
${\overline{O_{\sigma_{0},k(t)}}}$${\overline{O_{\sigma}}}$${X_{k(t)}(\Delta_{(0)})}$${X(\Delta)}$$\scriptstyle{\Psi}$
${O_{\sigma_{0},k(t)}}$${O_{\sigma}}$${X_{k(t)}(\Delta_{(0)})}$${X(\Delta)}$$\scriptstyle{\Psi}$

###### Proof.

We prove the statements from (a) to (d) in order.

  1. (a)

Because $(\mathrm{pr}_{1})_{\mathbb{R}}|_{\operatorname{Supp}(\Delta_{0})}$ is
injective, $\Delta_{(0)}$ is a polyhedral convex fan in $N_{\mathbb{R}}$ from
Lemma 10.1(c). Because $\Delta$ is a rational convex fan and
$\mathrm{pr}_{1}\colon(N\oplus\mathbb{Z})\rightarrow N$ is a morphism of
lattices, $\Delta_{(0)}$ is a rational convex fan. Moreover, $\Delta_{(0)}$ is
strongly convex from Lemma 10.1(d).

  2. (b)

We can regard $\Delta_{0}$ as a direct product of fans $\Delta_{(0)}$ and
$\\{\\{0\\}\\}$, so $(\mathrm{pr}_{1})_{*}\times(\mathrm{pr}_{2})_{*}$ is
isomorphic. The latter part of the statement follows from this identification.

  3. (c)

The right square is a Cartesian square, and the morphism
$X(\Delta_{(0)})\times{\mathbb{G}_{m}^{1}}_{,k}\rightarrow X(\Delta_{0})$ is
an isomorphism from (b). Thus, the statement holds.

  4. (d)

Let $\theta$ denote a morphism $X_{k(t)}(\Delta_{(0)})\rightarrow
X(\Delta_{(0)})\times{\mathbb{G}_{m}^{1}}_{,k}$ in the diagram in (c). Then
there exists the following Cartesian diagram:

${X_{k(t)}(\sigma_{0})}$${X(\sigma_{0})\times{\mathbb{G}_{m}^{1}}_{,k}}$${X_{k(t)}(\Delta_{(0)})}$${X(\Delta_{(0)})\times{\mathbb{G}_{m}^{1}}_{,k}}$$\scriptstyle{\theta}$

Thus, from (b), all small squares in the following diagram are Cartesian
squares:

${X_{k(t)}(\sigma_{0})}$${X(\sigma_{0})\times{\mathbb{G}_{m}^{1}}_{,k}}$${X(\sigma)}$${X(\sigma)}$${X_{k(t)}(\Delta_{(0)})}$${X(\Delta_{(0)})\times{\mathbb{G}_{m}^{1}}_{,k}}$${X(\Delta_{0})}$${X(\Delta)}$$\scriptstyle{\sim}$$\scriptstyle{\theta}$$\scriptstyle{p^{-1}}$

where $p$ denotes $(\mathrm{pr}_{1})_{*}\times(\mathrm{pr}_{2})_{*}$. Thus,
the first diagram in the statement is a Cartesian diagram. Similarly, the
second one and the third one in the statement are Cartesian diagrams.

∎

###  10.2. Toric varieties Part 2

Let $N$ be a lattice of finite rank, $M$ be the dual lattice of $N$, $\sigma$
be a strongly convex rational polyhedral cone in $N_{\mathbb{R}}$, $f$ be a
Laurent polynomial in $k[M]$, $H^{\circ}_{X(\sigma),f}$ denote a hypersurface
in $T_{N}$ defined by $f=0$, and $H_{X(\sigma),f}$ denote the scheme theoretic
closure of $H^{\circ}_{X(\sigma),f}$ in $X(\sigma)$. In this subsection, we
consider how to determine whether $O_{\sigma}\subset H_{X(\sigma),f}$ or not.
In fact, this condition can be determined by torus invariant valuations(cf.
Lemma10.18). The method of determination itself is likely to be a well-known
fact, as the proof is not particularly challenging. However, the author was
unable to find the literature on it.

A summary of this subsection follows: In Definition 10.11 and Lemma 10.12, we
construct a cone $C^{\sigma}_{f}$ and consider its property. The cone
$C^{\sigma}_{f}$ is important to define the function
$\operatorname{val}^{\sigma}_{f}$ at Definition10.14 and construct the
refinement of $\sigma$ along $f$ at Lemma10.16. Next, we show Lemma10.18. For
the proof, the refinement of $\sigma$ along $f$ plays an important role. After
showing Lemma10.18, we define new two properties of $f$ for a strongly convex
rational polyhedral fan $\Delta$ at Definition 10.19.

For $v\in\sigma\cap N$, we can define $v(f)$. At first, we consider the
configuation of $\\{(v,v(f))\\}_{v\in\sigma\cap N}$ in
$(N\oplus\mathbb{Z})_{\mathbb{R}}$ by constructing a new cone
$C^{\sigma}_{f}$.

######  Definition 10.11.

Let $N$ be a lattice of finite rank, $M$ be the dual lattice of $N$, and
$\sigma$ be a rational polyhedral convex cone in $N_{\mathbb{R}}$. Let $f\in
k[M]$ be a nonzero Laurant polynomial as follows:

$f=\sum_{\omega\in M}d_{\omega}\chi^{\omega},$

where $d_{\omega}\in k$ for $\omega\in M$. Let $D^{\sigma}_{f}$ be a rational
polyhedral convex cone in $(M\oplus\mathbb{Z})_{\mathbb{R}}$ generated by the
following set:

$\\{(\omega^{\prime},0)\in(M\oplus\mathbb{Z})_{\mathbb{R}}\mid\omega^{\prime}\in\sigma^{\vee}\\}\cup\\{(\omega,1)\in(M\oplus\mathbb{Z})_{\mathbb{R}}\mid
d_{\omega}\neq 0\\}$

Let $C^{\sigma}_{f}$ denote a dual cone of $D^{\sigma}_{f}$ in
$(N\oplus\mathbb{Z})_{\mathbb{R}}$. We remark that $C^{\sigma}_{f}$ is also a
rational polyhedral convex cone in $(N\oplus\mathbb{Z})_{\mathbb{R}}$.

From Lemma10.12(c), we can compute the configuation of
$\\{(v,v(f))\\}_{v\in\sigma\cap N}$ in $(N\oplus\mathbb{Z})_{\mathbb{R}}$. In
fact, they are on the boundary of a cone. Moreover, the cone $C^{\sigma}_{f}$
constructs the refinement of $\sigma$. This refinement plays an important role
in showing Lemma10.18.

######  Lemma 10.12.

We keep the notation in Definition 10.11, the following statements follow:

  1. (a)

For any $(v,a)\in C^{\sigma}_{f}$, we have that $v\in\sigma$.

  2. (b)

Let $v\in\sigma$ be an element. Then the following subset of $\mathbb{R}$ is
non-empty and has the minimum value:

$\\{a\in\mathbb{R}\mid(v,a)\in C^{\sigma}_{f}\\}$

From now on, let $s(v)$ denote the minimum value of the set above.

  3. (c)

For any $v\in\sigma\cap N$, we have $v(f)+s(v)=0$.

  4. (d)

Let $\tau\preceq C^{\sigma}_{f}$ be a face. We assume that $(0,1)\notin\tau$.
Then for any $(v,a)\in\tau$, we have $a=s(v)$.

  5. (e)

For any $v\in\sigma$, there exists a face $\tau\preceq C^{\sigma}_{f}$ such
that $(v,s(v))\in\tau$ and $(0,1)\notin\tau$.

  6. (f)

For any $r\in\mathbb{R}$ and $v\in\sigma$, we have $r\cdot s(v)=s(r\cdot v)$.

  7. (g)

Let $\Delta^{\sigma,f}$ denote the following set:

$\Delta^{\sigma,f}=\\{\tau\preceq C^{\sigma}_{f}\mid(0,1)\notin\tau\\}$

We remark that $\Delta^{\sigma,f}$ is a rational polyhedral fan in
$N_{\mathbb{R}}\oplus\mathbb{R}$. Let $\pi$ denote a first projection from
$N\oplus\mathbb{Z}$ to $N$. Then the following subset $\Delta^{\sigma}_{f}$ is
a rational polyhedral fan in $N_{\mathbb{R}}$ and a refinement of a fan which
consists of all faces of $\sigma$:

$\Delta^{\sigma}_{f}=\\{\pi_{\mathbb{R}}(\tau)\mid\tau\in\Delta^{\sigma,f}\\}$

  8. (h)

Let $E^{\sigma}_{f}$ denote a convex cone in
$(N\oplus\mathbb{Z})_{\mathbb{R}}$ generated by the following set:

$\\{(v,s(v))\mid v\in\sigma\\}\cup\\{(0,1)\\}$

Then $E^{\sigma}_{f}$ is $C^{\sigma}_{f}$.

  9. (i)

The cone $C^{\sigma}_{f}$ is strongly convex if and only if
$\Delta^{\sigma}_{f}$ is a strongly convex fan in $N_{\mathbb{R}}$. In
particular, if $\sigma$ is strongly convex, $\Delta^{\sigma}_{f}$ is a
strongly convex fan in $N_{\mathbb{R}}$.

###### Proof.

We prove the statements from (a) to (i) in order.

  1. (a)

Let $(v,a)\in C^{\sigma}_{f}$ be an element. For any
$\omega^{\prime}\in\sigma^{\vee}$, we have $(\omega^{\prime},0)\in
D^{\sigma}_{f}$. Thus, $\langle
v,\omega^{\prime}\rangle=\langle(v,a),(\omega^{\prime},0)\rangle\geq 0$. In
particular, $v\in\sigma$.

  2. (b)

Let $a_{v}\in\mathbb{R}$ denote the following number:

$a_{v}=\max_{d_{\omega}\neq 0}\\{-\langle v,\omega\rangle\\}$

Then we can check that $(v,a_{v})\in C^{\sigma}_{f}$ and $a\geq a_{v}$ for any
$(v,a)\in C^{\sigma}_{f}$ because of the definition of $C^{\sigma}_{f}$.

  3. (c)

Let $v\in\sigma\cap N$ be an element. Because $v$ is a torus invariant
valuation, $v(f)=\min_{d_{\omega}\neq 0}\\{\langle v,\omega\rangle\\}$. Thus,
$v(f)+s(v)=0$ from the proof of (b).

  4. (d)

Let $(v,a)\in\tau$ be an element. From (b), $a\geq s(v)$. Moreover,
$(v,s(v)),$ and $(0,a-s(v))\in C^{\sigma}_{f}$. Because $\tau$ is a face of
$C^{\sigma}_{f}$, we have that $(v,s(v))$ and $(0,a-s(v))\in\tau$. Thus, from
the assumption of $\tau$, we have that $a=s(v)$.

  5. (e)

Let $\tau\preceq C^{\sigma}_{f}$ be a face such that
$(v,s(v))\in\tau^{\circ}$. If $\tau$ contains $(0,1)$, then there exists
$\epsilon>0$ such that $(v,s(v)-\epsilon)\in\tau$ because
$(v,s(v))\in\tau^{\circ}$. However, it is a contradiction to the definition of
$s(v)$ from (b). Thus, we have that $(0,1)\notin\tau$.

  6. (f)

From the proof of (b), we can check it easily.

  7. (g)

We remark that $\pi_{\mathbb{R}}|_{\operatorname{Supp}(\Delta^{\sigma,f})}$ is
injective from (d). Then from Lemma 10.1(c), $\Delta^{\sigma,f}$ is a
polyhedral convex fan. A morphism $\pi$ is a lattice morphism, and
$C^{\sigma}_{f}$ is a rational polyhedral fan so that $\Delta^{\sigma}_{f}$ is
a rational polyhedral convex fan. From (a) and (e), we have that
$\operatorname{Supp}(\Delta^{\sigma}_{f})=\sigma$.

  8. (h)

We can check easily that $E^{\sigma}_{f}\subset C^{\sigma}_{f}$ from (b). Let
$(v,a)\in C^{\sigma}_{f}$ be an element. From (a), $v\in\sigma$. Moreover,
from (b), $a-s(v)\geq 0$. Thus, $(v,a)=(v,s(v))+(0,a-s(v))\in E^{\sigma}_{f}$.

  9. (i)

We remark that if $\sigma$ is strongly convex, then $D^{\sigma}_{f}$ is a full
cone from its construction, so that $C^{\sigma}_{f}$ is strongly convex.
Because $\pi_{\mathbb{R}}|_{\operatorname{Supp}(\Delta^{\sigma,f})}$ is
injective, the strong convexity of $\Delta^{\sigma,f}$ is equivalent to that
of $\Delta^{\sigma}_{f}$ from Lemma 10.1(d). Let $\tau$ denote the minimal
face of $C^{\sigma}_{f}$. This is a linear subspace of
$N_{\mathbb{R}}\oplus\mathbb{R}$. Because $C^{\sigma}_{f}$ does not contain
$(0,-1)$, $\tau$ does not contain $(0,1)$. Thus, $\tau\in\Delta^{\sigma,f}$
and the statement follows.

∎

The following lemma is cited many times in the main argument.

######  Lemma 10.13.

Let $N$ be a lattice of finite rank and $\sigma$ be a rational polyhedral
convex cone in $N_{\mathbb{R}}$. Let $s$ and $t$ be maps from $\sigma$ to
$\mathbb{R}$. We assume that $s$ and $t$ satisfy the following 3 conditions:

  1. (1)

These two maps are continuous.

  2. (2)

For any $r\in\mathbb{R}$ and $v\in\sigma$, $r\cdot s(v)=s(r\cdot v)$ and
$r\cdot t(v)=t(r\cdot v)$.

  3. (3)

For any $v\in\sigma\cap N$, we have that $s(v)=t(v)$.

Then we have $s=t$.

###### Proof.

From the conditions (2) and (3), $s(v)=t(v)$ for any $v\in\sigma\cap
N_{\mathbb{Q}}$. Because $\sigma$ is a rational polyhedral convex cone,
$\sigma\cap N_{\mathbb{Q}}$ is a dense subset of $\sigma$. Thus, $s=t$ because
$\mathbb{R}$ is a Hausdorff space and continuity of $s$ and $t$. ∎

For any $v\in\sigma$, we define the virtual valuation
$\operatorname{val}^{\sigma}_{f}(v)$.

######  Definition 10.14.

We keep the notation in Definition 10.11.

  1. (1)

Let $\operatorname{val}^{\sigma}_{f}$ denote a function from $\sigma$ to
$\mathbb{R}$ defined as $-s$.

  2. (2)

Let $\Omega^{\sigma}_{f}$ be the following subset of $M_{\mathbb{Q}}$:

$\Omega^{\sigma}_{f}=\\{\omega\in
M_{\mathbb{Q}}\mid\operatorname{val}^{\sigma}_{f}(v)\leq\langle
v,\omega\rangle\quad\forall v\in\sigma\\}$

  3. (3)

For $\omega\in\Omega^{\sigma}_{f}$, let $C^{\sigma}_{f}(\omega)$ denote the
following subset of $\sigma$:

$C^{\sigma}_{f}(\omega)=\\{v\in\sigma\mid\operatorname{val}^{\sigma}_{f}(v)=\langle
v,\omega\rangle\\}$

From the following lemma,
$\\{C^{\sigma}_{f}(\omega)\\}_{\omega\in\Omega^{\sigma}_{f}}$ is a rational
polyhedral convex fan in $N_{\mathbb{R}}$ and this is a refinement of the fan
which consists of all faces of $\sigma$.

######  Lemma 10.15.

We keep the notation in Definition 10.11 and Definition 10.14, the following
statements follow:

  1. (a)

For any $\omega\in\Omega^{\sigma}_{f}$, there exists
$\tau\in\Delta^{\sigma,f}$ such that
$C^{\sigma}_{f}(\omega)=\pi_{\mathbb{R}}(\tau)$. In particular,
$C^{\sigma}_{f}(\omega)$ is a rational polyhedral convex cone in
$N_{\mathbb{R}}$.

  2. (b)

Conversely, for any $\tau\in\Delta^{\sigma,f}$, there exists
$\omega\in\Omega^{\sigma}_{f}$ such that
$\pi_{\mathbb{R}}(\tau)=C^{\sigma}_{f}(\omega)$. In particular,
$\\{C^{\sigma}_{f}(\omega)\\}_{\omega\in\Omega^{\sigma}_{f}}$ is a rational
polyhedral convex fan in $N_{\mathbb{R}}$ and this is a refinement of the fan
which consists of all faces of $\sigma$.

  3. (c)

The function $\operatorname{val}^{\sigma}_{f}$ is a continuous map from
$\sigma$ to $\mathbb{R}$. In particular, $\operatorname{val}^{\sigma}_{f}$ is
a map that satisfies conditions (1) and (2) in Lemma 10.13. Moreover,
$\operatorname{val}^{\sigma}_{f}(v)=v(f)$ for any $v\in\sigma\cap N$.

  4. (d)

For any $\sigma\in\Delta$, $\operatorname{val}^{\sigma}_{f}$ is an upper
convex function.

  5. (e)

Let $\tau$ be a face of $\sigma$. Then
$val^{\sigma}_{f}|_{\tau}=val^{\tau}_{f}$.

  6. (f)

Let $\tau\preceq\sigma$ be a face and $\omega\in\Omega^{\sigma}_{f}$ be an
element. Then $\omega\in\Omega^{\tau}_{f}$ and
$C^{\sigma}_{f}(\omega)\cap\tau=C^{\tau}_{f}(\omega)$. In particular,
$C^{\tau}_{f}(\omega)\preceq C^{\sigma}_{f}(\omega)$.

  7. (g)

For any $\omega\in\Omega^{\sigma}_{f}$, there exists $\omega_{0}\in M$ such
that $\operatorname{val}^{\sigma}_{f}(v)=\langle v,\omega_{0}\rangle$ for any
$v\in C^{\sigma}_{f}(\omega)$.

###### Proof.

We prove the statements from (a) to (g) in order.

  1. (a)

Let $\omega\in\Omega^{\sigma}_{f}$ be an element. From the definition of
$\Omega^{\sigma}_{f}$ and Lemma 10.12(h), $(\omega,1)\in D^{\sigma}_{f}$. Let
$\tau$ denote $C^{\sigma}_{f}\cap(\omega,1)^{\perp}$. We remark that
$\tau\preceq C^{\sigma}_{f}$ and $(0,1)\notin\tau$. Thus
$C^{\sigma}_{f}(\omega)=\pi_{\mathbb{R}}(\tau)$. Indeed, for any $v\in\sigma$,
the equation $\langle(v,s(v)),(\omega,1)\rangle=0$ is equivalent to the
equation $\operatorname{val}^{\sigma}_{f}(v)=\langle v,\omega\rangle$ because
of the definition of $\operatorname{val}^{\sigma}_{f}$.

  2. (b)

Let $\tau\in\Delta^{\sigma,f}$ be a cone. Then there exists $(\omega,b)\in
D^{\sigma}_{f}$ such that $\omega\in M_{\mathbb{Q}}$, $b\in\mathbb{Q}_{\geq
0}$ and $\tau=C^{\sigma}_{f}\cap(\omega,b)^{\perp}$. If $b=0$, then
$(0,1)\in\tau$. However, it is a contradiction to the definition of
$\Delta^{\sigma,f}$. Thus, we may assume that $b=1$. From the definition of
$\Omega^{\sigma}_{f}$ and Lemma 10.12 (h), we have
$\omega\in\Omega^{\sigma}_{f}$. Thus, from the same argument in the proof of
(a), $\pi_{\mathbb{R}}(\tau)=C^{\sigma}_{f}(\omega)$. Therefore, from Lemma
10.12(g),
$\Delta^{\sigma}_{f}=\\{C^{\sigma}_{f}(\omega)\\}_{\omega\in\Omega^{\sigma}_{f}}$
and $\\{C^{\sigma}_{f}(\omega)\\}_{\omega\in\Omega^{\sigma}_{f}}$ is a
rational polyhedral convex fan which is a refinement of $\sigma$.

  3. (c)

From (b), the following equation holds:

$\sigma=\bigcup_{\omega\in\Omega^{\sigma}_{f}}C^{\sigma}_{f}(\omega)$

For each $\omega\in\Omega^{\sigma}_{f}$, $\operatorname{val}^{\sigma}_{f}$ is
a linear map on $C^{\sigma}_{f}(\omega)$. In addition to this,
$C^{\sigma}_{f}(\omega)$ is a closed subset of $\sigma$. Thus,
$\operatorname{val}^{\sigma}_{f}$ is a continuous map. From Lemma 10.12(f),
$\operatorname{val}^{\sigma}_{f}$ satisfies conditions (2) in Lemma 10.13.
From 10.12(c), the latter part of the statement follows.

  4. (d)

From the definition of $\operatorname{val}^{\sigma}_{f}$, it is enough to show
that $s$ is a lower convex function. From the convexity of $C^{\sigma}_{f}$
and 10.12(b), $s$ is a lower convex function.

  5. (e)

From (c), both functions satisfy conditions (1) and (2) in Lemma 10.13. Thus,
it is enough to show that $val^{\sigma}_{f}(v)=val^{\tau}_{f}(v)$ for any
$v\in\tau\cap N$. It follows from Lemma 10.12 (c).

  6. (f)

From (e), $\Omega^{\sigma}_{f}\subset\Omega^{\tau}_{f}$ and
$C^{\sigma}_{f}(\omega)\cap\tau=C^{\tau}_{f}(\omega)$ for any
$\omega\in\Omega^{\sigma}_{f}$. Let $\omega_{\tau}\in\sigma^{\vee}$ be an
element such that $\tau=\sigma\cap\omega_{\tau}^{\perp}$. Then
$\omega_{\tau}\in(C^{\sigma}_{f}(\omega))^{\vee}$ and
$C^{\sigma}_{f}(\omega)\cap\omega_{\tau}^{\perp}\preceq
C^{\sigma}_{f}(\omega)$. Thus,
$C^{\sigma}_{f}(\omega)\cap\omega_{\tau}^{\perp}=C^{\sigma}_{f}(\omega)\cap\tau=C^{\tau}_{f}(\omega)$
is a face of $C^{\sigma}_{f}(\omega)$.

  7. (g)

Let $v\in\langle C^{\sigma}_{f}(\omega)\rangle\cap N$ be an element. Then
there exists $v_{1}$ and $v_{2}\in C^{\sigma}_{f}(\omega)\cap N$ such that
$v=v_{1}-v_{2}$ because $C^{\sigma}_{f}(\omega)$ is a rational polyhedral
cone. From the definition of $C^{\sigma}_{f}(\omega)$ and (c), the following
equation holds:

$\displaystyle\langle v,\omega\rangle$ $\displaystyle=\langle
v_{1},\omega\rangle-\langle v_{2},\omega\rangle$
$\displaystyle=\operatorname{val}^{\sigma}_{f}(v_{1})-\operatorname{val}^{\sigma}_{f}(v_{2})$
$\displaystyle=v_{1}(f)-v_{2}(f)\in\mathbb{Z}$

Let $M^{\prime}$ be the dual lattice of $\langle
C^{\sigma}_{f}(\omega)\rangle\cap N$ and $\iota^{*}$ denote a group morphism
$M\rightarrow M^{\prime}$ induced by an inclusion $\iota\colon\langle
C^{\sigma}_{f}(\omega)\rangle\cap N\hookrightarrow N$. Then we can identify
$\langle-,\omega\rangle|_{\langle C^{\sigma}_{f}(\omega)\rangle\cap N}$ as an
element in $M^{\prime}$. We remark that $\iota^{*}$ is surjective because
$N/(\langle C^{\sigma}_{f}(\omega)\rangle\cap N)$ is torsion free. Thus, there
exists $\omega_{0}\in M$ such that
$\iota^{*}(\omega_{0})=\langle-,\omega\rangle|_{\langle
C^{\sigma}_{f}(\omega)\rangle\cap N}$. Hence, the following equation holds for
any $v^{\prime}\in C^{\sigma}_{f}(\omega)\cap N$:

$\displaystyle\operatorname{val}^{\sigma}_{f}(v^{\prime})$
$\displaystyle=\langle v^{\prime},\omega\rangle$ $\displaystyle=\langle
v^{\prime},\iota^{*}(\omega_{0})\rangle$ $\displaystyle=\langle
v^{\prime},\omega_{0}\rangle$

Thus, from Lemma 10.13, $\operatorname{val}^{\sigma}_{f}(v^{\prime})=\langle
v^{\prime},\omega_{0}\rangle$ for any $v^{\prime}\in C^{\sigma}_{f}(\omega)$.

∎

For a strongly convex rational polyhedral fan $\Delta$, we can construct a
refinement of $\Delta$ along $f$.

######  Lemma 10.16.

Let $N$ be a lattice of finite rank, $\Delta$ be a strongly convex rational
polyhedral fan in $N_{\mathbb{R}}$, and $f\in k[M]$ be a polynomial. Then the
following set $\Delta_{f}$ is a strongly convex rational polyhedral fan and a
refinement of $\Delta$:

$\Delta_{f}=\bigcup_{\sigma\in\Delta}\\{C^{\sigma}_{f}(\omega)\\}_{\omega\in\Omega^{\sigma}_{f}}$

###### Proof.

For any $\sigma\in\Delta$,
$\\{C^{\sigma}_{f}(\omega)\\}_{\omega\in\Omega^{\sigma}_{f}}$ is a strongly
convex polyhedral fan in $N_{\mathbb{R}}$ from Lemma 10.12 (i) and Lemma 10.15
(b). Thus, all faces of a cone in $\Delta_{f}$ are contained in $\Delta_{f}$.
Let $\sigma$ and $\tau\in\Delta$ be cones, $\omega_{1}\in\Omega^{\sigma}_{f}$
and $\omega_{2}\in\Omega^{\tau}_{f}$ be elements. We show that
$C^{\sigma}_{f}(\omega_{1})\cap C^{\tau}_{f}(\omega_{2})$ is a common face of
$C^{\sigma}_{f}(\omega_{1})$ and $C^{\tau}_{f}(\omega_{2})$. Indeed,
$C^{\sigma}_{f}(\omega_{1})\cap\tau=C^{\sigma\cap\tau}_{f}(\omega_{1})$,
$C^{\tau}_{f}(\omega_{2})\cap\sigma=C^{\sigma\cap\tau}_{f}(\omega_{2})$, and
$\\{\omega_{1},\omega_{2}\\}\subset\Omega^{\sigma\cap\tau}_{f}$ from Lemma
10.15(f). Thus, $C^{\sigma}_{f}(\omega_{1})\cap
C^{\tau}_{f}(\omega_{2})=C^{\sigma\cap\tau}_{f}(\omega_{1})\cap
C^{\sigma\cap\tau}_{f}(\omega_{2})$. Moreover, the cone
$C^{\sigma\cap\tau}_{f}(\omega_{1})\cap C^{\sigma\cap\tau}_{f}(\omega_{2})$ is
a common face of $C^{\sigma\cap\tau}_{f}(\omega_{1})$ and
$C^{\sigma\cap\tau}_{f}(\omega_{2})$ from Lemma 10.15(b). Thus, from Lemma
10.15(f), this cone is a common face of $C^{\sigma}_{f}(\omega_{1})$ and
$C^{\tau}_{f}(\omega_{2})$. ∎

From the definition of $\operatorname{val}^{\sigma}_{f}$, we can check that
$\operatorname{val}^{\sigma}_{f}$ is a piecewise linear function on $\sigma$.
From now on, let’s consider the case when $\operatorname{val}^{\sigma}_{f}$ is
a linear map on $\sigma$. The following lemma reveals that the linearity of
$\operatorname{val}^{\sigma}_{f}$ is also preserved for the factors of $f$.

######  Lemma 10.17.

We keep the notation in Definition 10.11 and Definition 10.14. Let
$f,f_{1},\ldots,f_{r}\in k[M]$ be nonzero Laurant polynomials. We assume that
$f=f_{1}\cdot f_{2}\cdot\cdots\cdot f_{r}$ and $\sigma$ is strongly convex.
Then the following statements are equivalent:

  1. (a)

There exists $\omega\in M$ such that $v(f\cdot\chi^{\omega})=0$ for any
$v\in\sigma\cap N$.

  2. (b)

For any $1\leq i\leq r$, there exists $\omega_{i}\in M$ such that
$v(f_{i}\cdot\chi^{\omega_{i}})=0$ for any $v\in\sigma\cap N$.

###### Proof.

One of the directions is obvious, so we show the statement of (b) holds when
the statement of (a) holds. If it is necessary, we replace $f$ with
$\chi^{\omega}f$ and $f_{1}$ with $\chi^{\omega}f_{1}$ and we may assume that
$\omega=0$. Then $val^{\sigma}_{f}\equiv 0$ because of Lemma 10.13 and Lemma
10.15(c). Let $v_{0}\in\sigma^{\circ}\cap N$ be an element. Then from Lemma
10.15(b), for any $1\leq i\leq r$, there exists
$\omega_{i}\in\Omega^{\sigma}_{f_{i}}$ such that $v_{0}\in
C^{\sigma}_{f_{i}}(\omega_{i})$. From Lemma 10.13 and Lemma 10.15(c),
$val^{\sigma}_{f}=\sum_{1\leq i\leq r}val^{\sigma}_{f_{i}}$. Then the
following equation and inequality holds:

$\displaystyle 0=\sum_{1\leq i\leq r}\operatorname{val}^{\sigma}_{f_{i}}(v)$
$\displaystyle\leq\langle v,\omega^{\prime}\rangle\quad(v\in\sigma)$
$\displaystyle 0=\sum_{1\leq i\leq
r}\operatorname{val}^{\sigma}_{f_{i}}(v_{0})$ $\displaystyle=\langle
v_{0},\omega^{\prime}\rangle$ ,

where $\omega^{\prime}$ denotes $\sum_{1\leq i\leq r}\omega_{i}\in
M_{\mathbb{Q}}$. These equation and inequality show that
$\omega^{\prime}\in\sigma^{\vee}$ and $\omega^{\prime}\in\sigma^{\perp}$
because $v_{0}\in\sigma^{\circ}$. Thus, $\langle v,\omega^{\prime}\rangle=0$
for any $v\in\sigma$.

We show that $\bigcap_{1\leq i\leq r}C^{\sigma}_{f_{i}}(\omega_{i})=\sigma$ by
contradiction. We assume that $w\in\sigma\setminus\bigcap_{1\leq i\leq
r}C^{\sigma}_{f_{i}}(\omega_{i})$ exists. Then from the definition of
$C^{\sigma}_{f_{i}}(\omega_{i})$, $\sum_{1\leq i\leq
r}\operatorname{val}^{\sigma}_{f_{i}}(w)<\langle w,\omega^{\prime}\rangle=0$.
However, $\sum_{1\leq i\leq
r}val^{\sigma}_{f_{i}}(w)=\operatorname{val}^{\sigma}_{f}(w)=0$, so it is a
contradiction.

Thus, for any $1\leq i\leq r$, $C^{\sigma}_{f_{i}}(\omega_{i})=\sigma$. Hence,
from Lemma 10.15(g), for any $1\leq i\leq r$, there exists
$\omega^{\prime}_{i}\in M$ such that
$\operatorname{val}^{\sigma}_{f_{i}}(v)=\langle v,\omega^{\prime}_{i}\rangle$
for any $v\in\sigma$. In particular,
$v(f_{i}\cdot\chi^{-\omega^{\prime}_{i}})=0$ for any $v\in\sigma\cap N$. ∎

The problem that was stated at the beginning of the subsection can be
characterized by the linearity of $\operatorname{val}^{\sigma}_{f}$ on
$\sigma$ from Lemma10.18.

######  Lemma 10.18.

We keep the notation in Definition 10.11. We assume that $\sigma$ is strongly
convex. Let $H_{0}$ be a closed subscheme of $T_{N}$ defined by $f=0$ and $H$
be a scheme theoretic closure of $H_{0}$ in $X(\sigma)$. Then the following
statements are equivalent:

  1. (a)

A closed subscheme $H$ does not contain $O_{\sigma}$

  2. (b)

There exists $\omega\in M$ such that $v(f\cdot\chi^{\omega})=0$ for any
$v\in\sigma\cap N$.

In this case, the ideal of $\Gamma(X(\sigma),\mathscr{O}_{X(\sigma)})$
associated with a closed subscheme $H$ of $X(\sigma)$ is generated by
$f\cdot\chi^{\omega}$.

###### Proof.

Let $k[\sigma^{\vee}\cap M]$ denote
$\Gamma(X(\sigma),\mathscr{O}_{X(\Delta)})$, $I_{f}\subset k[M]$ denote an
ideal generated by $f$, and $\mathfrak{p}_{\sigma}\subset k[\sigma^{\vee}\cap
M]$ denote the prime ideal associated to $O_{\sigma}\subset X(\sigma)$.

First, we assume that the condition of (a) holds. Then there exists $g\in
I_{f}\cap k[\sigma^{\vee}\cap M]$ such that $g\notin\mathfrak{p}_{\sigma}$. In
particular, for $v_{0}\in\sigma^{\circ}\cap N$, we have $v_{0}(g)=0$.
Moreover, for any $v\in\sigma\cap N$, $v(g)\geq 0$ because $g\in
k[\sigma^{\vee}\cap M]$. Therefore, $\operatorname{val}^{\sigma}_{g}\equiv 0$
from Lemma 10.15(d). On the other hand, there exists $h\in k[M]$ such that
$g=fh$ because $g\in I_{f}$. Thus, from Lemma 10.17, the condition of (b)
holds.

Next, we assume that the condition of (b) holds. From the assumption,
$f\cdot\chi^{\omega}\in I_{f}\cap k[\sigma^{\vee}\cap M]$. Thus,
$f\cdot\chi^{\omega}\notin\mathfrak{p}_{\sigma}$ and the condition of (a)
holds.

In this case, for any $g\in I_{f}\cap k[\sigma^{\vee}\cap M]$, there exists
$h\in k[M]$ such that $g=(f\cdot\chi^{\omega})h$. Because,
$v(f\cdot\chi^{\omega})=0$ and $v(g)\geq 0$ for any $v\in\sigma\cap N$. Hence,
for any $v\in\sigma\cap N$, $v(h)\geq 0$ from the additivity of valuations.
Thus, $h\in k[\sigma^{\vee}\cap M]$ and this shows that $I_{f}\cap
k[\sigma^{\vee}\cap M]$ is generated by $f\cdot\chi^{\omega}$. ∎

After showing Lemma10.18, we define two properties of $f$ for $\Delta$. These
two properties are important for mock toric varieties.

######  Definition 10.19.

Let $N$ be a lattice of a finite rank, $\Delta$ be a strongly convex rational
polyhedral fan in $N_{\mathbb{R}}$, $f\in k[M]$ be a Laurent polynomial,
$H^{\circ}_{X(\Delta),f}$ denote the closed subscheme of $T_{N}$ defined by
$f=0$ and $H_{X(\Delta),f}$ denote the scheme theoretic closure of
$H^{\circ}_{X(\Delta),f}$ in $X(\Delta)$. If for any $\sigma\in\Delta$,
$H_{X(\Delta),f}$ does not contain $O_{\sigma}$, then we call that $f$ is fine
for $\Delta$.

######  Definition 10.20.

We keep the notation in Definition 10.19, and we assume that $f$ is fine for
$\Delta$. If $H_{X(\Delta),f}\cap O_{\sigma}$ is smooth over $k$ for any
$\sigma\in\Delta$, we call that $f$ is non-degenerate for $\Delta$.

For the notation above, the fineness or non-degeneracy of $f$ is inherited for
the closure of the orbit.

######  Lemma 10.21.

We keep the notation in Definition 10.19, and we assume that $f$ is fine for
$\Delta$. Let $\sigma\in\Delta$ be a cone. Then there exists $\omega\in M$
such that $v(\chi^{\omega}f)=0$ for any $v\in\sigma\cap N$ from Lemma 10.18.
Let
$p^{\sigma}\colon\Gamma(X(\sigma),\mathscr{O}_{X(\Delta)})\rightarrow\Gamma(O_{\sigma},\mathscr{O}_{O_{\sigma}})$
be a quotient map associated with the closed immersion
$O_{\sigma}\hookrightarrow X(\sigma)$ and let $f^{\sigma}$ denote
$p^{\sigma}(\chi^{\omega}f)$. Then the following statements follow:

  1. (a)

As a closed subscheme of $O_{\sigma}$, $H_{X(\Delta),f}\cap
O_{\sigma}=H^{\circ}_{\overline{O_{\sigma}},f^{\sigma}}$.

  2. (b)

We use the notation in Proposition 10.8, $f^{\sigma}$ is fine for
$\Delta[\sigma]$.

  3. (c)

As a closed subscheme of $\overline{O_{\sigma}}$,
$H_{X(\Delta),f}\cap\overline{O_{\sigma}}=H_{\overline{O_{\sigma}},f^{\sigma}}$.

  4. (d)

If $f$ is non-degenerate for $\Delta$, then $f^{\sigma}$ is non-degenerate for
$\Delta[\sigma]$.

###### Proof.

We show the statements from (a) to (d) in order.

  1. (a)

From Lemma 10.18, $\chi^{\omega}f$ is a generator of an ideal of
$k[X(\sigma)]$ associated with a closed subscheme $H_{X(\sigma),f}$ of
$X(\sigma)$. Thus, $f^{\sigma}$ is a generator of an ideal of $k[O_{\sigma}]$
associated with a closed subscheme $H_{X(\sigma),f}\cap O_{\sigma}$ of
$O_{\sigma}$.

  2. (b)

From the definition of $H_{\overline{O_{\sigma}},f^{\sigma}}$ and (a),
$H_{\overline{O_{\sigma}},f^{\sigma}}\subset
H_{X(\Delta),f}\cap\overline{O_{\sigma}}$. Thus,
$H_{\overline{O_{\sigma}},f^{\sigma}}$ does not contain $O_{\tau}$ for any
$\sigma\subset\tau\in\Delta$ because $f$ is fine for $\Delta$.

  3. (c)

The closed subscheme $\overline{O_{\sigma}}\subset X(\Delta)$ is covered by
$\\{X(\tau)\\}_{\sigma\subset\tau\in\Delta}$. Thus, for each such $\tau$, it
is enough to show that the scheme theoretic closure of
$H^{\circ}_{\overline{O_{\sigma}},f^{\sigma}}$ in $\overline{O_{\sigma}}\cap
X(\tau)$ is equal to $H_{X(\Delta),f}\cap\overline{O_{\sigma}}\cap X(\tau)$
from Lemma 10.30. For such $\tau$, there exists the following commutative
diagram:

${\Gamma(X(\tau),\mathscr{O}_{X(\Delta)})}$${\Gamma(\overline{O_{\sigma}}\cap
X(\tau),\mathscr{O}_{\overline{O_{\sigma}}})}$${\Gamma(X(\sigma),\mathscr{O}_{X(\Delta)})}$${\Gamma(O_{\sigma},\mathscr{O}_{\overline{O_{\sigma}}})}$$\scriptstyle{p^{\sigma}}$$\scriptstyle{p^{\sigma}}$

, where horizontal maps are induced by two closed immersions
$O_{\sigma}\hookrightarrow X(\sigma)$ and $\overline{O_{\sigma}}\cap
X(\tau)\hookrightarrow X(\tau)$. From Lemma 10.18, there exists
$\omega^{\prime}\in M$ such that $w(\chi^{\omega^{\prime}}f)=0$ for any
$w\in\tau\cap N$ because $f$ is fine for $\Delta$. Because
$\sigma\subset\tau$, $v(\chi^{\omega-\omega^{\prime}})=0$ for any
$v\in\sigma\cap N$. Thus,
$\chi^{\omega-\omega^{\prime}}\in\Gamma(X(\sigma),\mathscr{O}_{X(\Delta)})^{*}$.
Let $\eta$ denote $p^{\sigma}(\chi^{\omega-\omega^{\prime}})$ and $f^{\tau}$
denote $p^{\sigma}(\chi^{\omega^{\prime}}f)$. Then
$f^{\tau}\in\Gamma(\overline{O_{\sigma}}\cap
X(\tau),\mathscr{O}_{\overline{O_{\sigma}}})$ and $f^{\sigma}=\eta f^{\tau}$.
In particular,
$\eta\in\Gamma(O_{\sigma},\mathscr{O}_{\overline{O_{\sigma}}})^{*}$. Thus,
$H^{\circ}_{\overline{O_{\sigma}},f^{\sigma}}=H^{\circ}_{\overline{O_{\sigma}},f^{\tau}}$.

On the other hand, the ideal of $\Gamma(\overline{O_{\sigma}}\cap
X(\tau),\mathscr{O}_{\overline{O_{\sigma}}})$ associated with a closed
subscheme $H_{X(\Delta),f}\cap\overline{O_{\sigma}}\cap
X(\tau)\subset\overline{O_{\sigma}}\cap X(\tau)$ is generated by $f^{\tau}$
because the ideal of $\Gamma(X(\tau),\mathscr{O}_{X(\Delta)})$ associated with
a closed subscheme $H_{X(\Delta),f}\cap X(\tau)\subset X(\tau)$ is generated
by $\chi^{\omega^{\prime}}f$. Let $\tau^{\prime}$ denote
$\pi^{\sigma}_{\mathbb{R}}(\tau)$ in the notation of Proposition 10.8. We
remark that $X(\tau^{\prime})=\overline{O_{\sigma}}\cap X(\tau)$. If
$\operatorname{val}^{\tau^{\prime}}_{f^{\tau}}(w^{\prime})>0$ for some
$w^{\prime}\in{\tau^{\prime}}^{\circ}\cap N[\sigma]$, then
$H_{X(\Delta),f}\cap\overline{O_{\sigma}}\cap X(\tau)$ should contain
$O_{\tau}$. However, it is a contradiction to the assumption that $f$ is fine
for $\Delta$. Thus, $\operatorname{val}^{\tau^{\prime}}_{f^{\tau}}\equiv 0$,
so that the ideal of $\Gamma(\overline{O_{\sigma}}\cap
X(\tau),\mathscr{O}_{\overline{O_{\sigma}}})$ associated with the scheme
theoretic closure of $H^{\circ}_{\overline{O_{\sigma}},f^{\tau}}$ in
$\overline{O_{\sigma}}\cap X(\tau)$ is generated by $f^{\tau}$ too from Lemma
10.18. Therefore, the scheme theoretic closure of
$H^{\circ}_{\overline{O_{\sigma}},f^{\sigma}}$ in $\overline{O_{\sigma}}\cap
X(\tau)$ is equal to $H_{X(\Delta),f}\cap\overline{O_{\sigma}}\cap X(\tau)$ as
closed subschemes of $\overline{O_{\sigma}}\cap X(\tau)$.

  4. (d)

We use the notation in the proof of (c), and we have that
$H_{\overline{O_{\sigma}},f^{\sigma}}\cap
O_{\tau^{\prime}}=H_{X(\Delta),f}\cap\overline{O_{\sigma}}\cap
O_{\tau^{\prime}}=H_{X(\Delta),f}\cap O_{\tau}$. Thus, from the assumption,
$H_{\overline{O_{\sigma}},f^{\sigma}}\cap O_{\tau^{\prime}}$ is smooth over
$k$.

∎

From the following lemma, we can check that the fineness or non-degeneracy of
$f$ is inherited for the toric resolution.

######  Lemma 10.22.

Let $N$ and $N^{\prime}$ be lattices of finite rank, $\alpha\colon
N^{\prime}\rightarrow N$ be a surjective morphism, $\Delta$ and
$\Delta^{\prime}$ be strongly convex rational polyhedral fans in
$N_{\mathbb{R}}$ and $N^{\prime}_{\mathbb{R}}$ respectively. We assume that
$\alpha$ is compatible with the fans $\Delta^{\prime}$ and $\Delta$. Let $f\in
k[M]$ be a polynomial and let $\alpha^{*}$ denote a ring morphism
$k[M]\rightarrow k[M^{\prime}]$ induced by $\alpha$. We assume that $f$ is
fine for $\Delta$. Then the following statements follow:

  1. (a)

The polynomial $\alpha^{*}(f)$ is fine for $\Delta^{\prime}$.

  2. (b)

As a closed subscheme of $X(\Delta^{\prime})$,
$H_{X(\Delta),f}\times_{X(\Delta)}X(\Delta^{\prime})=H_{X(\Delta^{\prime}),\alpha^{*}(f)}$.

  3. (c)

If $f$ is non-degenerate for $\Delta$, $\alpha^{*}(f)$ is non-degenerate for
$\Delta^{\prime}$.

###### Proof.

We show the statements from (a) to (c) in order.

  1. (a)

Let $\tau\in\Delta^{\prime}$ and $\sigma\in\Delta$ be cones such that
$\alpha_{\mathbb{R}}(\tau)\subset\sigma$. Then there exists $\omega\in M$ such
that $v(\chi^{\omega}f)=0$ for any $v\in\sigma\cap N$ from Lemma 10.18. We
remark that for any $w\in\tau\cap N^{\prime}$,
$w(\alpha^{*}(\chi^{\omega}f))=\alpha(w)(\chi^{\omega}f)$. Thus,
$w(\chi^{\alpha^{*}(\omega)}\alpha^{*}(f))=0$ for any $w\in\tau\cap
N^{\prime}$ because $\alpha(w)\in\sigma\cap N$. Therefore, from Lemma 10.18,
$\alpha^{*}(f)$ is fine for $\Delta^{\prime}$.

  2. (b)

We use the notation in the proof of (a), and the ideal of $k[X(\tau)]$
associated with the closed subscheme $H_{X(\Delta^{\prime}),\alpha^{*}(f)}\cap
X(\tau)$ of $X(\tau)$ is generated by $\chi^{\alpha^{*}(\omega)}\alpha^{*}(f)$
from Lemma 10.18. Thus, the following diagram is a Cartesian diagram:

${H_{X(\Delta^{\prime}),\alpha^{*}(f)}\cap
X(\tau)}$${X(\tau)}$${H_{X(\Delta),f}\cap
X(\sigma)}$${X(\sigma)}$$\scriptstyle{\alpha_{*}}$$\scriptstyle{\alpha_{*}}$

Then the statement follows from Lemma 10.30.

  3. (c)

We use the notation in the proof of (a). There exists $\sigma\in\Delta$ such
that $\alpha_{\mathbb{R}}(\tau)\cap\sigma^{\circ}\neq\emptyset$. We remark
that $\alpha_{*}(O_{\tau})=O_{\sigma}$ because $\alpha$ is surjective. Thus,
there exists the following Cartesian diagram from the diagram in the proof of
(b):

${H_{X(\Delta^{\prime}),\alpha^{*}(f)}\cap
O_{\tau}}$${O_{\tau}}$${H_{X(\Delta),f}\cap
O_{\sigma}}$${O_{\sigma}}$$\scriptstyle{\alpha_{*}}$$\scriptstyle{\alpha_{*}}$

Because $\alpha_{*}\colon O_{\tau}\rightarrow O_{\sigma}$ is a trivial
algebraic torus fibration, the smoothness of
$H_{X(\Delta^{\prime}),\alpha^{*}(f)}\cap O_{\tau}$ and the smoothness of
$H_{X(\Delta),f}\cap O_{\sigma}$ are equivalent.

∎

In general, $H_{X(\Delta),f}$ contains some torus orbits of $X(\Delta)$. The
following lemma indicates that a toric resolution $X(\Delta_{f})\rightarrow
X(\Delta)$ remove this obstacle.

######  Lemma 10.23.

We use the notation in Lemma 10.16. Let $\Delta^{\prime}$ be a refinement of
$\Delta_{f}$. Then $f$ is fine for $\Delta^{\prime}$. Moreover, if $f$ is non-
degenerate for $\Delta_{f}$, $f$ is non-degenerate for $\Delta^{\prime}$ too.

###### Proof.

From Lemma 10.22(a) and (c), it is enough to show that $f$ is fine for
$\Delta_{f}$. Let $\tau\in\Delta_{f}$ be a cone. Then there exists
$\sigma\in\Delta$ and $\omega^{\prime}\in\Omega^{\sigma}_{f}$ such that
$\tau=C^{\sigma}_{f}(\omega^{\prime})$. From Lemma 10.15(g), there exists
$\omega\in M$ such that $\operatorname{val}^{\sigma}_{f}(v)=\langle
v,\omega\rangle$ for any $v\in\tau$. In particular, $v(\chi^{-\omega}f)=0$ for
any $v\in\tau\cap N$. Thus, from Lemma 10.18, $H_{X(\tau),f}$ does not contain
$O_{\tau}$. Therefore, $f$ is fine for $\Delta_{f}$. ∎

###  10.3. Toric varieties Part. 3

In Definition 10.20, the non-degeneracy of $f$ is determined by orbit-wise. In
this subsection, we connect the ”smoothness” of $H_{X(\Delta),f}$ and this
non-degeneracy of $f$. Note this ”smoothness” does not indicate true
smoothness.

######  Definition 10.24.

Let $S$ be a monoid. If there exists a lattice $N$ of finite rank and a
strongly convex rational polyhedral cone $\sigma$ in $N_{\mathbb{R}}$ such
that $S$ is isomorphic to $\sigma^{\vee}\cap M$, where $M$ is the dual
lattice, then we call that $S$ is a toric monoid. In this situation, let
$k[S]$ denote a $k$-algebra generated by $S$. For $\omega\in S$, let
$\chi^{\omega}$ denote an element of $k[S]$ associated with $\omega$.

The ”smoothness” mentioned at the beginning of this subsection is explained in
the following lemma.

######  Lemma 10.25.

Let $N$ denote a lattice of finite rank, $\sigma$ be a strongly convex
rational polyhedral cone in $N_{\mathbb{R}}$, $f\in k[M]$ be a Laurent
polynomial, and $V$ be a non-empty open subset of $X(\sigma)$. We assume that
$H_{X(\sigma),f}$ does not contain $O_{\sigma}$ and $H_{X(\sigma),f}\cap
O_{\sigma}\cap V$ is smooth over $k$. Then there exist

  * •

A toric monoid $S$

  * •

A morphism $p\colon X(\sigma)\rightarrow\operatorname{Spec}(k[S])$ over
$\operatorname{Spec}(k)$

  * •

A open subset $U$ of $V$

such that

  1. (1)

$H_{X(\sigma),f}\cap O_{\sigma}\cap V\subset U$.

  2. (2)

The restriction $p|_{H_{X(\sigma),f}\cap U}\colon H_{X(\sigma),f}\cap
U\rightarrow\operatorname{Spec}(k[S])$ is smooth.

###### Proof.

From Lemma10.18, there exists $\omega\in M$ such that $\chi^{\omega}f\in
k[\sigma^{\vee}\cap M]$ and $H_{X(\sigma),f}=H_{X(\sigma),\chi^{\omega}f}$.
Thus, we may replace $f$ with $\chi^{\omega}f$ and assume that $f\in
k[\sigma^{\vee}\cap M]$. There exists a sublattice $N_{0}$ of $N$ such that
$(\langle\sigma\rangle\cap N)\oplus N_{0}=N$. Let $\pi$ denote the quotient
morphism $N\rightarrow N/N_{0}$ and $s$ denote a section of $\pi$ such that
$s\circ\pi(v)=v$ for any $v\in\langle\sigma\rangle\cap N$. Let
$\sigma^{\prime}$ denote $\pi_{\mathbb{R}}(\sigma)$. Then from Lemma 10.3,
$\sigma^{\prime}$ is a full and strongly convex rational polyhedral cone in
$(N/N_{0})_{\mathbb{R}}$ and the toric morphism $\pi_{*}\colon
X(\sigma)\rightarrow X(\sigma^{\prime})$ is a trivial algebraic torus
fibration. Because $\sigma^{\prime}$ is a full cone in
$(N/N_{0})_{\mathbb{R}}$, a closed orbit $O_{\sigma^{\prime}}\subset
X(\sigma^{\prime})$ is a closed point of $X(\sigma^{\prime})$. Let $x$ denote
the closed point $O_{\sigma^{\prime}}\in X(\sigma^{\prime})$. Then
$(\pi_{*})^{-1}(x)=O_{\sigma}$. In particular,
$(H_{X(\sigma),f})_{\kappa(x)}\cap V_{\kappa(x)}=H_{X(\sigma),f}\cap
O_{\sigma}\cap V$. Moreover, $H_{X(\sigma),f}\cap O_{\sigma}\cap V\neq
O_{\sigma}\cap V$ because $H_{X(\sigma),f}$ does not contain $O_{\sigma}$. We
remark that $X(\sigma^{\prime})$ is an affine scheme of finite type over $k$.
Thus, from Corollary 10.37, there exists an open subset $U$ of $V$ such that

  * •

$H_{X(\sigma),f}\cap O_{\sigma}\cap V\subset U$.

  * •

The restriction $\pi_{*}|_{H_{X(\sigma),f}\cap U}\colon H_{X(\sigma),f}\cap
U\rightarrow X(\sigma^{\prime})$ is smooth.

Therefore, the statement holds. ∎

The following elementary lemma is needed in Lemma 10.27.

######  Lemma 10.26.

Let $N$ be a lattice of finite rank and $N^{\prime}$ be a sublattice of
$N\oplus\mathbb{Z}$. We assume that $(N\oplus\mathbb{Z})/N^{\prime}$ is tosion
free and there exists $v\in N$ such that $(v,1)\in N^{\prime}$. Then there
exists a sub lattice $N^{\prime\prime}$ of $N$ such that
$N^{\prime}\oplus(N^{\prime\prime}\times\\{0\\})=N\oplus\mathbb{Z}$.

###### Proof.

Let $\iota$ denote a canonical inclusion $N\rightarrow N\oplus\mathbb{Z}$ such
that $\iota(w)=(w,0)$ for any $w\in N$. Let $N_{1}$ be an inverse image of
$N^{\prime}$ along $\iota$. We claim that
$(N_{1}\times\\{0\\})\oplus\mathbb{Z}(v,1)=N^{\prime}$. Indeed, for any
$(w,a)\in N^{\prime}$, we have $(av,a)\in\mathbb{Z}(v,1)$ and
$(w-av,0)\in(N_{1}\times\\{0\\})$. Moreover, $N/N_{1}$ is torsion free because
$(N\oplus\mathbb{Z})/N^{\prime}$ is torsion free. Thus, there exists a sub
lattice $N_{2}$ of $N$ such that $N_{1}\oplus N_{2}=N$. Therefore,
$(N_{1}\times\\{0\\})\oplus(N_{2}\times\\{0\\})\oplus\mathbb{Z}(v,1)=N\oplus\mathbb{Z}$,
so let $N^{\prime\prime}$ denote $N_{2}$. ∎

Let $X(\Delta)\rightarrow\mathbb{A}^{1}$ be a toric morphism. The following
lemma indicates how to check the reduced-ness of the fiber of the closed orbit
of $\mathbb{A}^{1}_{k}$ combinatorially.

######  Lemma 10.27.

Let $\sigma$ be a strongly convex rational polyhedral cone in
$(N\oplus\mathbb{Z})_{\mathbb{R}}$. We assume that $\sigma\subset
N_{\mathbb{R}}\times\mathbb{R}_{\geq 0}$ and $\sigma\not\subset
N_{\mathbb{R}}\times\\{0\\}$, and for every 1-dimensional ray
$\gamma\preceq\sigma$ with
$\gamma\cap(N_{\mathbb{R}}\times\\{1\\})\neq\emptyset$, we have
$\gamma\cap(N\times\\{1\\})\neq\emptyset$. Then the following statements
follow:

  1. (a)

There exists a sublattice $N^{\prime}$ of $N$ such that $N/N^{\prime}$ is
torsion free and the restriction
$(\pi,\mathrm{id})|_{\langle\sigma\rangle\cap(N\oplus\mathbb{Z})}\colon\langle\sigma\rangle\cap(N\oplus\mathbb{Z})\rightarrow
N/N^{\prime}\oplus\mathbb{Z}$ is isomorphic, where $\pi\colon N\rightarrow
N/N^{\prime}$ be the quotient map and $(\pi,\mathrm{id}_{\mathbb{Z}})\colon
N\oplus\mathbb{Z}\rightarrow N/N^{\prime}\oplus\mathbb{Z}$ is a direct product
morphism $\pi$ and ${\mathrm{id}}_{\mathbb{Z}}$.

  2. (b)

For such $N^{\prime}$ in (a), let $\sigma^{\prime}$ denote
$(\pi,\mathrm{id}_{\mathbb{Z}})_{\mathbb{R}}(\sigma)$. Let $F$ denote
$(\mathrm{pr}_{2})_{*}^{-1}(0)$ which is the fiber of the toric morphism
$(\mathrm{pr}_{2})_{*}\colon X(\sigma^{\prime})\rightarrow\mathbb{A}^{1}$ at a
closed orbit of $\mathbb{A}^{1}_{k}$. Then $\sigma^{\prime}$ is a strongly
convex rational polyhedral cone in
$(N/N^{\prime}\oplus\mathbb{Z})_{\mathbb{R}}$ and a second projection
$\mathrm{pr}_{2}\colon(N/N^{\prime}\oplus\mathbb{Z})\rightarrow\mathbb{Z}$ is
compatible with the cones $\sigma^{\prime}$ and $[0,\infty)$. Moreover, $F$ is
reduced.

###### Proof.

We prove the statements from (a) to (b).

  1. (a)

From the assumption of $\sigma$, there exists a 1-dimensional ray
$\gamma\preceq\sigma$ such that $\gamma\not\subset
N_{\mathbb{R}}\times\\{0\\}$. Because
$\gamma\cap(N\times\\{1\\})\neq\emptyset$, there exists $v\in N$ such that
$(v,1)\in\sigma$. Thus, from Lemma 10.26, there exists a sublattice $N_{0}$ of
$N$ such that
$(\langle\sigma\rangle\cap(N\oplus\mathbb{Z}))\oplus(N_{0}\times\\{0\\})=N\oplus\mathbb{Z}$.
Therefore, the statement holds.

  2. (b)

From Lemma 10.3(a), $\sigma^{\prime}$ is a full and strongly convex rational
polyhedral cone in $(N/N^{\prime}\oplus\mathbb{Z})_{\mathbb{R}}$. Because
$\sigma\subset N_{\mathbb{R}}\times\mathbb{R}_{\geq 0}$, we have
$\sigma^{\prime}\subset(N/N^{\prime})_{\mathbb{R}}\times\mathbb{R}_{\geq 0}$.
Thus, a second projection
$\mathrm{pr}_{2}\colon(N/N^{\prime}\oplus\mathbb{Z})\rightarrow\mathbb{Z}$ is
compatible with the cones $\sigma^{\prime}$ and $[0,\infty)$. From Lemma
10.3(b) and (c), there is a one-to-one correspondence with 1-dimensional rays
of $\sigma$ and those of $\sigma^{\prime}$. Thus, from the assumption of
$\sigma$ and $N^{\prime}$, for every 1-dimensional ray
$\gamma^{\prime}\preceq\sigma^{\prime}$ with
$\gamma^{\prime}\cap(N/N^{\prime})_{\mathbb{R}}\times\\{1\\}\neq\emptyset$, we
have $\gamma^{\prime}\cap(N/N^{\prime})\times\\{1\\}\neq\emptyset$.

We show that $F$ is reduced. Let $t\in k[\mathbb{Z}]$ be a torus invariant
monomial associated with $1\in\mathbb{Z}$. The irreducible component of the
fiber of the toric morphism $X(\sigma^{\prime})\rightarrow\mathbb{A}^{1}$ at a
closed orbit of $\mathbb{A}^{1}_{k}$ is an orbit closure of
$X(\sigma^{\prime})$ associated with a 1-dimensional ray of $\sigma^{\prime}$
such that it intersects $(N/N^{\prime})_{\mathbb{R}}\times\\{1\\}$. Let
$\gamma^{\prime}\preceq\sigma^{\prime}$ be a 1-dimensional ray with
$\gamma^{\prime}\cap(N/N^{\prime})_{\mathbb{R}}\times\\{1\\}\neq\emptyset$ and
$\eta^{\prime}\in X(\sigma)$ denote the generic point of
$O_{\gamma^{\prime}}$. Let $v^{\prime}\in(N/N^{\prime})\oplus\mathbb{Z}$
denote a minimal generator of $\gamma^{\prime}$. Then $v^{\prime}$ is a unique
torus invariant valuation of $X(\sigma^{\prime})$ such that the valuation ring
associated with $v^{\prime}$ is $\mathscr{O}_{X(\sigma^{\prime}),\eta}$ and
the image of $v^{\prime}$ is $\mathbb{Z}$. Now, we show that
$v^{\prime}(t)=1$. Indeed, there exists $v^{\prime\prime}\in N/N^{\prime}$
such that $v^{\prime}=(v^{\prime\prime},1)$ from the above argument. Because
the second component of $v^{\prime}$ is $1$, we have $v^{\prime}(t)=1$. In
particular, $\mathscr{O}_{F,\eta}$ is reduced. Thus, $F$ has a property
$(R_{0})$. Moreover, $X(\sigma^{\prime})$ is a Cohen-Macauray and an integral
scheme. Hence, $F$ is a Cohen-Macauray scheme. In particular, $F$ has a
property $(S_{1})$. Therefore, $F$ is reduced.

∎

When we compute the stable birational volume, it is important to construct a
strictly toroidal scheme (cf. Definition7.6). The following lemma indicates
that the sufficient conditions for the reduced-ness of the fiber of the closed
orbit can be given by the combinatorial setting.

######  Lemma 10.28.

We keep the notation and the assumption in Lemma 10.27. We identify with
$k[M\oplus\mathbb{Z}]$ and $k[M]\otimes_{k}k[t,t^{-1}]$, where $t$ is a torus
invariant monomial associated with $1\in\mathbb{Z}$. Let $f\in
k[M\oplus\mathbb{Z}]$ be a polynomial and $V$ be an open subscheme of
$X(\sigma)$. We assume that $H_{X(\sigma),f}$ does not contain $O_{\sigma}$
and $H_{X(\sigma),f}\cap O_{\sigma}\cap V$ is smooth over $k$. We regard
$X(\sigma)$ is a scheme over $\operatorname{Spec}(k[t])=\mathbb{A}^{1}$ by the
toric morphism $(\mathrm{pr}_{2})_{*}\colon
X(\sigma)\rightarrow\mathbb{A}^{1}$.

Then there exist

  * •

A toric monoid $S$

  * •

An alement $\omega\in S$

  * •

A morphism $p\colon
X(\sigma)\rightarrow\operatorname{Spec}(k[t][S]/(t-\chi^{\omega}))$ over
$\operatorname{Spec}(k[t])$

  * •

An open subset $U$ of $V$

such that

  1. (1)

$H_{X(\sigma),f}\cap O_{\sigma}\cap V\subset U$

  2. (2)

The restriction $p|_{H_{X(\sigma),f}\cap U}\colon H_{X(\sigma),f}\cap
U\rightarrow\operatorname{Spec}(k[t][S]/(t-\chi^{\omega}))$ is smooth.

  3. (3)

A ring $k[S]/(\chi^{\omega})$ is reduced.

###### Proof.

From Lemma10.18, there exists $\eta\in M\oplus\mathbb{Z}$ such that
$\chi^{\eta}f\in k[\sigma^{\vee}\cap(M\oplus\mathbb{Z})]$ and
$H_{X(\sigma),f}=H_{X(\sigma),\chi^{\eta}f}$. Thus, we may replace $f$ with
$\chi^{\eta}f$ and assume that $f\in k[\sigma^{\vee}\cap(M\oplus\mathbb{Z})]$.
We use the notation in Lemma 10.27. Let $S$ be a toric monoid associated with
$X(\sigma^{\prime})$, $g\colon
X(\sigma^{\prime})\rightarrow\operatorname{Spec}(k[S])$ be a natural
isomorphism, and $\omega\in S$ be an element associated with $t$. We remark
that $(\pi,\mathrm{id})_{*}\colon X(\sigma)\rightarrow X(\sigma^{\prime})$ is
a trivial algebraic torus fibration. We regard $k[S]$ as a $k[t]$-algebra by a
$k$-morphism $q\colon k[t]\rightarrow k[S]$ such that $q(t)=\chi^{\omega}$.
Then there exists the following commutative diagram:

${X(\sigma^{\prime})}$${\operatorname{Spec}(k[S])}$${\operatorname{Spec}(k[t])}$$\scriptstyle{g}$$\scriptstyle{(\mathrm{pr}_{2})_{*}}$$\scriptstyle{q^{*}}$

where $q^{*}$ is a morphism induced by $q$.

Let $r$ denote $q\otimes\mathrm{id}_{k[S]}\colon k[t][S]\rightarrow k[S]$. We
regard $k[t][S]$ as a $k[t]$-algebra by a $k$-morphism $s\colon
k[t]\rightarrow k[t][S]$ such that $s(t)=t$. We remark that $r$ is a
$k[t]$-morphism. Let $r_{0}$ denote the quotient map
$k[t][S]/(t-\chi^{\omega})\rightarrow k[S]$ induced by $r$ and $s_{0}$ denote
a morphism $k[t]\rightarrow k[t][S]/(t-\chi^{\omega})$ induced by $s$. Because
of the definition of $r$, $r_{0}$ is an isomorphism. Thus, there exists the
following commutative diagram:

${X(\sigma^{\prime})}$${\operatorname{Spec}(k[S])}$${\operatorname{Spec}(k[t])}$${\operatorname{Spec}(k[t][S]/(t-\chi^{\omega}))}$$\scriptstyle{g}$$\scriptstyle{(\mathrm{pr}_{2})_{*}}$$\scriptstyle{q^{*}}$$\scriptstyle{r^{*}_{0}}$$\scriptstyle{s^{*}_{0}}$

where $r^{*}_{0}$ and $s^{*}_{0}$ are morphisms induced by $r_{0}$ and
$s_{0}$. Let $p$ be a composition $r^{*}_{0}\circ
g\circ(\pi,\mathrm{id}_{\mathbb{Z}})_{*}\colon
X(\sigma)\rightarrow\operatorname{Spec}(k[t][S]/(t-\chi^{\omega}))$. Then from
the proof of Lemma 10.25, there exists an open subset of $U$ of $V$ such that
$H_{X(\sigma),f}\cap O_{\sigma}\cap V\subset U$ and the restriction
$p|_{H_{X(\sigma),f}\cap U}\colon H_{X(\sigma),f}\cap
U\rightarrow\operatorname{Spec}(k[t][S]/(t-\chi^{\omega}))$ is smooth.
Moreover, $k[S]/(\chi^{\omega})$ is isomorphic to a global section ring of the
fiber of $(\mathrm{pr}_{2})_{*}\colon
X(\sigma^{\prime})\rightarrow\mathbb{A}^{1}$ at a closed orbit of
$\mathbb{A}^{1}_{k}$. Thus, from Lemma 10.27, $k[S]/(\chi^{\omega})$ is
reduced. ∎

###  10.4. Scheme theoretic image of quasi-compact morphism

In this subsection, we prove some lemmas about the scheme theoretic image
which we use in this article.

######  Lemma 10.29.

Let $f\colon X\rightarrow Y$ be a quasi-compact morphism of schemes, and $Z$
be a scheme theoretic image of $f$, and $V$ be an open subscheme of $Y$. Then
$Z\cap V$ is a scheme theoretic image of $f|_{f^{-1}(V)}\colon
f^{-1}(V)\rightarrow V$.

###### Proof.

Because $f$ is quasi-compact, $\mathscr{I}=\ker(\mathscr{O}_{Y}\rightarrow
f_{*}\mathscr{O}_{X})$ is a quasi-coherent ideal sheaf associated with the
closed subscheme $Z$ of $Y$. Because
$\mathscr{I}|_{V}=\ker(\mathscr{O}_{V}\rightarrow(f|_{f^{-1}(V)})_{*}\mathscr{O}_{f^{-1}(V)})$
and $\mathscr{I}|_{V}$ is a quasi-coherent ideal sheaf associated with the
closed subscheme $Z\cap V$ of $V$, $Z\cap V$ is a scheme theoretic image of
$f|_{f^{-1}(V)}$. ∎

######  Lemma 10.30.

Let $f\colon X\rightarrow Y$ be a quasi-compact morphism of schemes, $Z$ be a
closed subscheme of $Y$, and $\\{V_{i}\\}_{i\in I}$ be an open covering of
$Y$. Then the following statements are equivalent.

  * (a)

A closed subscheme $Z$ of $Y$ is a scheme theoretic image of $f$.

  * (b)

For any $i\in I$, a closed subscheme $Z\cap V_{i}$ of $V_{i}$ is a scheme
theoretic image of $f|_{f^{-1}(V_{i})}$.

###### Proof.

When the statement (a) holds, the statement (b) holds from Lemma 10.29. Then
we assume the statement (b). Let $\mathscr{I}_{Z}\subset\mathscr{O}_{Y}$ be a
quasi-coherent ideal sheaf associated with $Z$ of $Y$. From the assumption,
for $i\in I$,
$\mathscr{I}_{Z}|_{V_{i}}=\ker(\mathscr{O}_{V_{i}}\rightarrow(f|_{f^{-1}(V_{i})})_{*}\mathscr{O}_{f^{-1}(V_{i})})$.
Thus, $\mathscr{I}_{Z}=\ker(\mathscr{O}_{Y}\rightarrow f_{*}\mathscr{O}_{X})$
because $\\{V_{i}\\}_{i\in I}$ is an open covering of $Y$. Therefore, $Z$ is a
scheme theoretic image of $f$. ∎

###  10.5. Some lemmas about valuation

######  Lemma 10.31.

Let $X,X^{\prime},Y,Y^{\prime},Z$, and $Z^{\prime}$ be integral schemes,
$\iota\colon Z\rightarrow X$, $q\colon X\rightarrow Y$, $\iota^{\prime}\colon
Z^{\prime}\rightarrow X^{\prime}$, $q^{\prime}\colon X^{\prime}\rightarrow
Y^{\prime}$, $g\colon X^{\prime}\rightarrow X$, $h\colon Y^{\prime}\rightarrow
Y$, and $f\colon Z^{\prime}\rightarrow Z$ are morphisms such that the
following diagram is commutative:

${Z^{\prime}}$${X^{\prime}}$${Y^{\prime}}$${Z}$${X}$${Y}$$\scriptstyle{\iota^{\prime}}$$\scriptstyle{f}$$\scriptstyle{q^{\prime}}$$\scriptstyle{g}$$\scriptstyle{h}$$\scriptstyle{\iota}$$\scriptstyle{q}$

Let $r$ denote $q\circ\iota$ and $r^{\prime}$ denote
$q^{\prime}\circ\iota^{\prime}$. We assume that $f,g,h,q,$ and $q^{\prime}$
are dominant and $r$ and $r^{\prime}$ are open immersions. Then the following
equation follows as a map from $\operatorname{Val}(X^{\prime})$ to
$\operatorname{Val}(Z)$:

$f_{\natural}\circ({r^{\prime}}^{-1})_{\natural}\circ(q^{\prime})_{\natural}=(r^{-1})_{\natural}\circ
q_{\natural}\circ g_{\natural}$

###### Proof.

From the commutativity of the above diagram, $h\circ q^{\prime}=q\circ g$ and
$f\circ(r^{\prime})^{-1}=r^{-1}\circ h$. Thus, we can check that the statement
follows. ∎

###  10.6. Some lemmas about valuation ring

We use the notation as follows:

  * •

Let $k$ be an algebraically closed field with $\operatorname{char}(k)=0$.

  * •

Let $t$ be an indeterminate invariant.

  * •

Let $l$ be a positive integer and $R_{l}$ be a valuation ring defined as
follows:

$R_{l}=k[[t^{\frac{1}{l}}]]$

  * •

Let $l$ be a positive integer and $K_{l}$ be a fraction field of $R_{l}$. We
remark that $K$ is written explicitly as follows:

$K_{l}=k((t^{\frac{1}{l}}))$

  * •

Let $\mathscr{R}$ be a valuation ring defined as follows:

$\mathscr{R}=\bigcup_{n\in\mathbb{Z}_{>0}}R_{n}$

  * •

Let $\mathscr{K}$ be a fraction field of $\mathscr{R}$. A field $\mathscr{K}$
is a field of Puiseux series over $k$. We remark that $\mathscr{K}$ is written
explicitly as follows:

$\mathscr{K}=\bigcup_{n\in\mathbb{Z}_{>0}}K_{n}$

In this subsection, we prove some lemmas about schemes over
$\operatorname{Spec}(\mathscr{R})$.

The following lemma is referenced in [4, Remark. 4.6]; however, for the sake
of thoroughness, a detailed proof is provided here.

######  Lemma 10.32.

[4, Remark. 4.6] Let $\mathfrak{X}$ be a flat scheme over
$\operatorname{Spec}(\mathscr{R})$, $X$ denote a base change of $\mathfrak{X}$
along
$\operatorname{Spec}(\mathscr{K})\rightarrow\operatorname{Spec}(\mathscr{R})$,
and $Y$ be a closed subscheme of $X$. Then there exists a unique closed
subscheme $\mathfrak{Y}$ of $\mathfrak{X}$ such that the following conditions
hold:

  1. (1)

A scheme $\mathfrak{Y}$ is flat over $\operatorname{Spec}(\mathscr{R})$.

  2. (2)

A base change of $\mathfrak{Y}$ along
$\operatorname{Spec}(\mathscr{K})\rightarrow\operatorname{Spec}(\mathscr{R})$
is equal to $Y$ as a closed subscheme of $X$.

In fact, such $\mathfrak{Y}$ is a scheme theoretic closure of $Y$ in
$\mathfrak{X}$.

###### Proof.

We remark that the open immersion $X\hookrightarrow\mathfrak{X}$ is quasi-
compact because
$\operatorname{Spec}(\mathscr{K})\rightarrow\operatorname{Spec}(\mathscr{R})$
is quasi-compact. Thus, the composition $Y\hookrightarrow
X\rightarrow\mathfrak{X}$ is quasi-compact too. We already know that the
statement holds in the case when $\mathfrak{X}$ is an affine scheme from [4,
Proposition. 4.4].

First, we show the existence of $\mathfrak{Y}$. Let $\mathfrak{Y}_{0}$ be a
scheme theoretic closure of $Y$ in $\mathfrak{X}$. Thus, for any affine open
subset $U\subset\mathfrak{X}$, the scheme theoretic closure of $U\cap Y$ in
$U$ is $U\cap\mathfrak{Y}_{0}$ from Lemma 10.29. For such $U$, $U$ is flat
over $\operatorname{Spec}(\mathscr{R})$, so that
$(U\cap\mathfrak{Y}_{0})_{\mathscr{K}}=U\cap Y$ and $U\cap\mathfrak{Y}_{0}$ is
flat over $\operatorname{Spec}(\mathscr{R})$ from [4, Proposition. 4.4]. By
considering one affine open covering of $\mathfrak{X}$, $\mathfrak{Y}_{0}$ is
flat over $\operatorname{Spec}(\mathscr{R})$ and the generic fiber of
$\mathfrak{Y}_{0}$ is equal to $Y$ as a closed subscheme of $X$.

Second, we show the uniqueness of $\mathfrak{Y}$. Let $\mathfrak{Y}_{1}$ be a
closed subscheme of $\mathfrak{X}$ such that $\mathfrak{Y}_{1}$ is flat over
$\operatorname{Spec}(\mathscr{R})$ and the generic fiber of $\mathfrak{Y}_{1}$
is equal to $Y$ as a closed subscheme of $X$. Let $U$ be an affine open subset
of $\mathfrak{X}$. Then $U\cap\mathfrak{Y}_{1}$ is flat over
$\operatorname{Spec}(\mathscr{R})$ and the generic fiber of
$U\cap\mathfrak{Y}_{1}$ is equal to $U\cap Y$ as a closed subscheme of
$U\times_{\mathscr{R}}\mathscr{K}$. Because $U$ is flat over
$\operatorname{Spec}(\mathscr{R})$, $U\cap\mathfrak{Y}_{1}$ is the scheme
theoretic closure of $U\cap Y$ in $U$ from [4, Proposition. 4.4]. Thus, by
considering one affine open covering of $\mathfrak{X}$, $\mathfrak{Y}_{1}$ is
the scheme theoretic closure of $Y$ in $\mathfrak{X}$ from Lemma 10.30. ∎

The following lemma is referenced in [13, Tag 055C]. We remark that if $S$ in
the following lemma is replaced with $\mathscr{R}$, then the same proof does
not hold.

######  Lemma 10.33.

[13, Tag 055C] Let $S$ be a DVR, $u$ be its uniformizer, $Q$ be the fraction
field of $S$, and $\kappa$ be the residue field of $S$. Let $B$ be a flat
$S$-algebra and $x\in B\otimes_{S}Q$ be an idempotent element. We remark that
extension morphism $B\rightarrow B\otimes_{S}Q$ is injective, and $u$ is a
non-zero divisor of $B$ because $B$ is flat over $S$. We assume that
$B\otimes_{S}\kappa$ is reduced. Then we have $x\in B$.

###### Proof.

Let $a\in B$ be an element and $n\in\mathbb{Z}_{\geq 0}$ be a non-negative
integer such that $x=\frac{a}{u^{n}}$ in $B\otimes_{S}Q$. We can take $n$
minimal. Because $x$ is an idempotent element,
$\frac{a}{u^{n}}=\frac{a^{2}}{u^{2n}}$ in $B\otimes_{S}Q$. Then $a^{2}=au^{n}$
in $B$ because $u$ is a non-zero divisor of $B$. If $n\geq 1$, this shows that
$a+(u)B$ is a nilpotent element in $B\otimes_{S}\kappa$. However, from the
assumption, $a\in(u)B$ and this is a contradiction to the minimality of $n$.
Thus, we can take $n$ as $0$, so that $x\in B$. ∎

It is not trivial whether Lemma 10.33 holds when $S=\mathscr{R}$. In fact, if
$B$ is flat and of finite presentation over $\mathscr{R}$, then Lemma 10.33
holds. This claim are proven in Lemma 10.35. The following lemma serves as a
preparation for the proof of Lemma 10.35.

######  Lemma 10.34.

Let $A$ be a flat and of finite presentation ring over $\mathscr{R}$ and $x\in
A\otimes_{\mathscr{R}}\mathscr{K}$ be an element. Then there exists a positive
integer $l$ and a sub $R_{l}$-algebra $C$ of $A$ such that

  * •

A $R_{l}$-algebra $C$ is flat over $R_{l}$.

  * •

All morphisms in the following diagram are injective:

${C}$${C\otimes_{R_{l}}K_{l}}$${A}$${A\otimes_{\mathscr{R}}\mathscr{K}}$

where the right one is induced by the base of $R_{l}\rightarrow K_{l}$ along
$C\hookrightarrow A$

  * •

A subring $C\otimes_{R_{l}}K_{l}$ of $A\otimes_{\mathscr{R}}\mathscr{K}$
contains $x$.

  * •

The inclusion $C\hookrightarrow A$ induce the isomorphism
$C\otimes_{R_{l}}k\cong A\otimes_{\mathscr{R}}{k}$

###### Proof.

From the assumption, there exist $N\in\mathbb{Z}_{>0}$ and
$f_{1},\ldots,f_{r}\in\mathscr{R}[X_{1},\ldots,X_{N}]$ such that
$\mathscr{R}[X_{1},\ldots,X_{N}]/(f_{1},\ldots,f_{r})\cong A$ as an
$\mathscr{R}$-algebra. Moreover, there exists
$g\in\mathscr{K}[X_{1},\ldots,X_{N}]$ such that $g+(f_{1},\ldots,f_{r})=x$ by
the identification with $\mathscr{K}[X_{1},\ldots,X_{N}]/(f_{1},\ldots,f_{r})$
and $A\otimes_{\mathscr{R}}\mathscr{K}$. Thus, from the definition of
$\mathscr{R}$ and $\mathscr{K}$, there exists $l\in\mathbb{Z}_{>0}$ such that
$f_{1},\ldots,f_{r}\in R_{l}[X_{1},\ldots,X_{N}]$ and $g\in
K_{l}[X_{1},\ldots,X_{N}]$. Let $C$ denote a $R_{l}$-algebra
$R_{l}[X_{1},\ldots,X_{N}]/(f_{1},\ldots,f_{r})$. We remark that
$C\otimes_{R_{l}}\mathscr{R}\cong\mathscr{R}[X_{1},\ldots,X_{N}]/(f_{1},\ldots,f_{r})$
as an $\mathscr{R}$-algebra. We remark that
$\mathscr{R}\otimes_{R_{l}}K_{l}=\mathscr{R}[t^{-\frac{1}{l}}]=\mathscr{K}$.

First, we show that $C$ is flat over $R_{l}$. Because $R_{l}$ is a DVR, it is
enough to show that $C$ is a torsion-free $R_{l}$-module. Let $f\colon
C\rightarrow C$ denote an endomorphism of $C$ defined by the scalar product of
$t^{\frac{1}{l}}$. Then it is enough to show that $f$ is injective. Because
$\mathscr{R}$ is a faithfully flat $R_{l}$-module, it is enough to show that
$f\otimes\mathrm{id}_{\mathscr{R}}$, which is induced by $f$, is injective.
From the assumption, $\mathscr{R}[X_{1},\ldots,X_{N}]/(f_{1},\ldots,f_{r})$ is
flat over $\mathscr{R}$. Thus, $f\otimes\mathrm{id}_{\mathscr{R}}$ is
injective.

Second, the extension morphism
$C\rightarrow\mathscr{R}[X_{1},\ldots,X_{N}]/(f_{1},\ldots,f_{r})$ is
injective because $C$ is flat over $R_{l}$. Thus, we can regard $C$ as an
$R_{l}$-sub algebra of $A$ on the identification with
$\mathscr{R}[X_{1},\ldots,X_{N}]/(f_{1},\ldots,f_{r})$ and $A$. We can easily
check that this inclusion $C\hookrightarrow A$ induces an isomorphism
$C\otimes_{R_{l}}k\cong A\otimes_{\mathscr{R}}k$ because both rings are
isomorphic to
$k[X_{1},\ldots,X_{N}]/(\overline{f_{1}},\ldots,\overline{f_{r}})$, where
$\overline{f_{i}}\in k[X_{1},\ldots,X_{N}]$ is a polynomial whose each
coefficient is replaced by constant terms of coefficients of $f_{i}$.

Third, in the above diagram, we show that all morphisms are injective. Indeed,
the left one is injective from the above argument. The upper one is injective
because $C$ is flat over $R_{l}$. The right one is injective because $K_{l}$
is flat over $R_{l}$. At this point, we remark that
$A\otimes_{\mathscr{R}}\mathscr{K}=A\otimes_{\mathscr{R}}\mathscr{R}\otimes_{R_{l}}K_{l}=A\otimes_{R_{l}}K_{l}$.
The lower one is injective because $A$ is flat over $\mathscr{R}$.

Finally, we show that $x\in C\otimes_{R_{l}}K_{l}$. Indeed,
$g+(f_{1},\ldots,f_{r})\in C\otimes_{R_{l}}K_{l}$ and this element go to $x$
along the right morphism in the above diagram. ∎

From now on, we prove the following lemma, which is derived from the
Lemma10.33 by replacing the condition that $S$ is a DVR with the condition
that $S=\mathscr{R}$ and adding one condition.

######  Lemma 10.35.

Let $\eta$ denote a generic point of $\operatorname{Spec}(\mathscr{R})$. Let
$X$ be a flat and locally of finite presentation scheme over
$\operatorname{Spec}(\mathscr{R})$. For an open subscheme $U\subset X$, let
$U_{\eta}$ denote a generic fiber of $U$. We assume that $X$ is a connected
scheme and the closed fiber $X_{k}$ is reduced. Then the following statements
follow:

  1. (a)

Let $U$ be an open subscheme of $X$. Then the following ring morphism induced
by an open immersion $U_{\eta}\rightarrow U$ is injective.

$\Gamma(U,\mathscr{O}_{X})\rightarrow\Gamma(U_{\eta},\mathscr{O}_{X})$

  2. (b)

A scheme $X_{\eta}$ is connected.

###### Proof.

We prove the statements from (a) to (b).

  1. (a)

When $U$ is an affine open subscheme of $X$, the statement holds. Indeed, let
$A$ be a ring such that $\operatorname{Spec}(A)=X$. Then $A$ is flat over
$\mathscr{R}$. Thus, an extension map $A\rightarrow
A\otimes_{\mathscr{R}}\mathscr{K}$ is injective. In the general case, $U$ can
be covered by open affine subschemes of $X$. Because $\mathscr{O}_{X}$ is a
sheaf over $X$, the statement holds.

  2. (b)

We show that the statement holds by a contradiction. We assume that $X_{\eta}$
isn’t connected. Then there exists an idempotent element
$e\in\Gamma(X_{\eta},\mathscr{O}_{X})$ such that $e\neq 0$ and $e\neq 1$. Let
$U$ be an open affine subscheme of $X$ and $A$ be a ring such that
$U=\operatorname{Spec}(A)$ and $A$ is of finite presentation over
$\mathscr{R}$. We remark that $U_{\eta}=\operatorname{Spec}(A[\frac{1}{t}])$,
so that $e|_{U_{\eta}}\in A\otimes_{\mathscr{R}}\mathscr{K}$. Then for $A$ and
$e|_{U_{\eta}}\in A\otimes_{\mathscr{R}}\mathscr{K}$, there exists
$l\in\mathbb{Z}_{>0}$ and a sub $R_{l}$-algebra $C$ of $A$ such that $l$ and
$C$ satisfy the conditions in Lemma 10.34. Thus, $e|_{U_{\eta}}\in
C\otimes_{R_{l}}K_{l}$ and $C\otimes_{R_{l}}K_{l}\rightarrow
A\otimes_{\mathscr{R}}\mathscr{K}$ is injective. Hence, $e|_{U_{\eta}}$ is an
idempotent elements of $C\otimes_{R_{l}}K_{l}$. Moreover, $C$ is a flat over a
DVR $R_{l}$ and $C\otimes_{R_{l}}k\cong A\otimes_{\mathscr{R}}k$ is reduced
from the assumption. Thus, from Lemma 10.33, we have $e|_{U_{\eta}}\in C$ and
$e|_{U_{\eta}}\in A$. This shows that
$e|_{U_{\eta}}\in\Gamma(U,\mathscr{O}_{X})$ for any affine open subscheme $U$
of $X$. Hence, from (a), we can check that $e\in\Gamma(X,\mathscr{O}_{X})$. An
element $e\in\Gamma(X,\mathscr{O}_{X})$ is a non-trivial idempotent element of
$\Gamma(X,\mathscr{O}_{X})$, but it is a contradiction to the assumption that
$X$ is connected.

∎

###  10.7. ”Smoothness”

In this subsection, we prove the lemmas are needed in Lemma10.25 and 10.28.

######  Lemma 10.36.

We itemize the notation as follows:

  * •

Let $k$ be a field.

  * •

Let $A$ be a ring of finite type over $k$.

  * •

Let $X$ denote $\operatorname{Spec}(A)$.

  * •

For $x\in X$, let $\kappa(x)$ denote a residue field of $x$.

  * •

Let $A[X_{1},\ldots,X_{n}]$ denote a global section ring of
$X\times_{k}\mathbb{A}^{n}_{k}$ and $0\neq f\in A[X_{1},\ldots,X_{n}]$ be a
polynomial.

  * •

Let $Y$ be a closed subscheme of $X\times_{k}\mathbb{A}^{n}_{k}$ defined by
$f=0$.

  * •

Let $V$ be a non-empty open subset of $X\times_{k}\mathbb{A}^{n}_{k}$.

  * •

Let $\pi$ denote the first projection
$X\times_{k}\mathbb{A}^{n}_{k}\rightarrow X$.

We assume that $V_{\kappa(x)}\cap Y_{\kappa(x)}\neq V_{\kappa(x)}$ and
$V_{\kappa(x)}\cap Y_{\kappa(x)}$ is smooth over $\kappa(x)$.

Then there exists a non-empty open subset of $U\subset V$ such that
$V_{\kappa(x)}\cap Y_{\kappa(x)}\subset U$ and the restriction $\pi|_{U\cap
Y}\colon U\cap Y\rightarrow X$ is a smooth morphism.

###### Proof.

Let $s$ denote a natural morphism
$A[X_{1},\ldots,X_{n}]\rightarrow\kappa(x)[X_{1},\ldots,X_{n}]$ induced by
$A\rightarrow\kappa(x)$. Then from the assumption, $s(f)\neq 0$ in
$\kappa(x)[X_{1},\ldots,X_{n}]$. We remark that for any $1\leq i\leq n$ and
$g\in A[X_{1},\ldots,X_{n}]$, the following equation holds in
$\kappa(x)[X_{1},\ldots,X_{n}]$:

$\frac{\partial}{\partial X_{i}}s(g)=s\bigl{(}\frac{\partial g}{\partial
X_{i}}\bigr{)}$

Let $Z$ denote a closed subscheme of $X\times_{k}\mathbb{A}^{n}_{k}$ defined
by an ideal generated by $\\{\frac{\partial f}{\partial X_{i}}\\}_{1\leq i\leq
n}$. Now, we show that $V_{\kappa(x)}\cap Y_{\kappa(x)}\cap
Z_{\kappa(x)}=\emptyset$ by a contradiction. We assume that $V_{\kappa(x)}\cap
Y_{\kappa(x)}\cap Z_{\kappa(x)}\neq\emptyset$. Then
$V_{\overline{\kappa(x)}}\cap Y_{\overline{\kappa(x)}}\cap
Z_{\overline{\kappa(x)}}\neq\emptyset$. From above remark, this shows that a
hypersurface $Y_{\overline{\kappa(x)}}$ of
$\mathbb{A}^{n}_{\overline{\kappa(x)}}$has a singular point in
$V_{\overline{\kappa(x)}}\cap Y_{\overline{\kappa(x)}}$, but this is a
contradiction to the assumption that $V_{\kappa(x)}\cap Y_{\kappa(x)}$ is
smooth over $\kappa(x)$.

Let $U$ denote $V\cap(X\times_{k}\mathbb{A}^{n}_{k}\setminus Z)$ and $W_{i}$
denote a fundamental affine open subset of $X\times_{k}\mathbb{A}^{n}_{k}$
defined by $\frac{\partial f}{\partial X_{i}}$ for $1\leq i\leq n$. Thus, $U$
contains $V_{\kappa(x)}\cap Y_{\kappa(x)}$. Moreover, for any $1\leq i\leq n$,
the restriction $\pi\colon Y\cap W_{i}\rightarrow X$ is a smooth morphism.
Thus, the restriction $\pi|_{U\cap Y}\colon U\cap Y\rightarrow X$ is a smooth
morphism too. ∎

######  Corollary 10.37.

We itemize the notation as follows:

  * •

Let $k$ be a field.

  * •

Let $A$ be a ring of finite type over $k$.

  * •

Let $X$ denote $\operatorname{Spec}(A)$.

  * •

For $x\in X$, let $\kappa(x)$ denote a residue field of $x$.

  * •

Let $A[X_{1}^{\pm},\ldots,X_{n}^{\pm}]$ denote a global section ring of
$X\times_{k}{\mathbb{G}_{m}^{n}}_{,k}$ and $0\neq f\in
A[X_{1}^{\pm},\ldots,X_{n}^{\pm}]$ be a Laurant polynomial.

  * •

Let $Y$ be a closed subscheme of $X\times_{k}{\mathbb{G}_{m}^{n}}_{,k}$
defined by $f=0$.

  * •

Let $V$ be a non-empty open subset of $X\times_{k}{\mathbb{G}_{m}^{n}}_{,k}$.

  * •

Let $\pi$ denote the first projection
$X\times_{k}{\mathbb{G}_{m}^{n}}_{,k}\rightarrow X$.

We assume that $V_{\kappa(x)}\cap Y_{\kappa(x)}\neq V_{\kappa(x)}$ and
$V_{\kappa(x)}\cap Y_{\kappa(x)}$ is smooth over $\kappa(x)$.

Then there exists a non-empty open subset of $U\subset V$ such that
$V_{\kappa(x)}\cap Y_{\kappa(x)}\subset U$ and the restriction $\pi|_{U\cap
Y}\colon U\cap Y\rightarrow X$ is a smooth morphism.

###### Proof.

The natural inclusion $A[X_{1},\ldots,X_{n}]\hookrightarrow
A[X_{1}^{\pm},\ldots,X_{n}^{\pm}]$ induces an open immersion
$X\times_{k}{\mathbb{G}_{m}^{n}}_{,k}\hookrightarrow
X\times_{k}\mathbb{A}^{n}_{k}$. Thus, on the above immersion, there exists
$g\in A[X_{1},\ldots,X_{n}]$ such that
$Z\cap(X\times_{k}{\mathbb{G}_{m}^{n}}_{,k})=Y$ as a closed subscheme of
$X\times_{k}{\mathbb{G}_{m}^{n}}_{,k}$, where $Z$ is a closed subscheme of
$X\times_{k}\mathbb{A}^{n}_{k}$ defined by $g=0$. Moreover, we can regard $V$
as an open subscheme of $X\times_{k}\mathbb{A}^{n}_{k}$. Thus, from Lemma
10.36, the statement holds. ∎

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