## Tag `03IG`

## 59.11. Points on spaces

In this section we prove some properties of points on decent algebraic spaces.

Lemma 59.11.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Consider the map $$ \{\mathop{\mathrm{Spec}}(k) \to X \text{ monomorphism}\} \longrightarrow |X| $$ This map is always injective. If $X$ is decent then this map is a bijection.

Proof.We have seen in Properties of Spaces, Lemma 57.4.11 that the map is an injection in general. By Lemma 59.5.1 it is surjective when $X$ is decent (actually one can say this is part of the definition of being decent). $\square$The following lemma tells us that the henselian local ring of a point on a decent algebraic space is defined.

Lemma 59.11.2. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. For every point $x \in |X|$ there exists an étale morphism $$ (U, u) \longrightarrow (X, x) $$ where $U$ is an affine scheme, $u$ is the only point of $U$ lying over $x$, and the induced morphism $\mathop{\mathrm{Spec}}(\kappa(u)) \to X$ is a monomorphism.

Proof.We may assume that $X$ is quasi-compact by replacing $X$ with a quasi-compact open containing $x$. Recall that $x$ can be represented by a quasi-compact (mono)morphism from the spectrum a field (by definition of decent spaces). Thus the lemma follows from Lemma 59.8.3. $\square$Definition 59.11.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in X$ be a point. An

elementary étale neighbourhoodis an étale morphism $(U, u) \to (X, x)$ where $U$ is a scheme, $u \in U$ is a point mapping to $x$, and $\mathop{\mathrm{Spec}}(\kappa(u)) \to X$ is a monomorphism. Amorphism of elementary étale neighbourhoods$(U, u) \to (U', u')$ is defined as a morphism $U \to U'$ over $X$ mapping $u$ to $u'$.If $X$ is not decent then the category of elementary étale neighbourhoods may be empty.

Lemma 59.11.4. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Let $x$ be a point of $X$. The category of elementary étale neighborhoods of $(X, x)$ is cofiltered (see Categories, Definition 4.20.1).

Proof.The category is nonempty by Lemma 59.11.2. Suppose that we have two elementary étale neighbourhoods $(U_i, u_i) \to (X, x)$. Then consider $U = U_1 \times_X U_2$. Since $\mathop{\mathrm{Spec}}(\kappa(u_i)) \to X$, $i = 1, 2$ are both monomorphisms in the class of $x$, we see that $$ u = \mathop{\mathrm{Spec}}(\kappa(u_1)) \times_X \mathop{\mathrm{Spec}}(\kappa(u_2)) $$ is the spectrum of a field $\kappa(u)$ such that the induced maps $\kappa(u_i) \to \kappa(u)$ are isomorphisms. Then $u \to U$ is a point of $U$ and we see that $(U, u) \to (X, x)$ is an elementary étale neighbourhood dominating $(U_i, u_i)$. If $a, b : (U_1, u_1) \to (U_2, u_2)$ are two morphisms between our elementary étale neighbourhoods, then we consider the scheme $$ U = U_1 \times_{(a, b), (U_2 \times_X U_2), \Delta} U_2 $$ Using Properties of Spaces, Lemma 57.16.6 we see that $U \to X$ is étale. Moreover, in exactly the same manner as before we see that $U$ has a point $u$ such that $(U, u) \to (X, x)$ is an elementary étale neighbourhood. Finally, $U \to U_1$ equalizes $a$ and $b$ and the proof is finished. $\square$Definition 59.11.5. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Let $x \in |X|$. The

henselian local ring of $X$ at $x$, is $$ \mathcal{O}_{X, x}^h = \mathop{\mathrm{colim}}\nolimits \Gamma(U, \mathcal{O}_U) $$ where the colimit is over the elementary étale neighbourhoods $(U, u) \to (X, x)$.To be sure, the henselian local ring of $X$ at $x$ is equal to the henselization $\mathcal{O}_{U, u}^h$ of the local ring $\mathcal{O}_{U, u}$ of any elementary étale neighbourhood. This follows from the definition, Lemma 59.11.4, and More on Morphisms, Lemma 36.31.5.

The following lemma shows that specialization of points behaves well on decent algebraic spaces. Spaces, Example 56.14.9 shows that this is

nottrue in general.Lemma 59.11.6. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Let $U \to X$ be an étale morphism from a scheme to $X$. If $u, u' \in |U|$ map to the same point of $|X|$, and $u' \leadsto u$, then $u = u'$.

Proof.Combine Lemmas 59.5.1 and 59.7.1. $\square$Lemma 59.11.7. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Let $x, x' \in |X|$ and assume $x' \leadsto x$, i.e., $x$ is a specialization of $x'$. Then for every étale morphism $\varphi : U \to X$ from a scheme $U$ and any $u \in U$ with $\varphi(u) = x$, exists a point $u'\in U$, $u' \leadsto u$ with $\varphi(u') = x'$.

Proof.Combine Lemmas 59.5.1 and 59.7.2. $\square$Lemma 59.11.8. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Then $|X|$ is Kolmogorov (see Topology, Definition 5.8.4).

Proof.Let $x_1, x_2 \in |X|$ with $x_1 \leadsto x_2$ and $x_2 \leadsto x_1$. We have to show that $x_1 = x_2$. Pick a scheme $U$ and an étale morphism $U \to X$ such that $x_1, x_2$ are both in the image of $|U| \to |X|$. By Lemma 59.11.7 we can find a specialization $u_1 \leadsto u_2$ in $U$ mapping to $x_1 \leadsto x_2$. By Lemma 59.11.7 we can find $u_2' \leadsto u_1$ mapping to $x_2 \leadsto x_1$. This means that $u_2' \leadsto u_2$ is a specialization between points of $U$ mapping to the same point of $X$, namely $x_2$. This is not possible, unless $u_2' = u_2$, see Lemma 59.11.6. Hence also $u_1 = u_2$ as desired. $\square$Proposition 59.11.9. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Then the topological space $|X|$ is sober (see Topology, Definition 5.8.4).

Proof.We have seen in Lemma 59.11.8 that $|X|$ is Kolmogorov. Hence it remains to show that every irreducible closed subset $T \subset |X|$ has a generic point. By Properties of Spaces, Lemma 57.12.4 there exists a closed subspace $Z \subset X$ with $|Z| = |T|$. By definition this means that $Z \to X$ is a representable morphism of algebraic spaces. Hence $Z$ is a decent algebraic space by Lemma 59.5.3. By Theorem 59.10.2 we see that there exists an open dense subspace $Z' \subset Z$ which is a scheme. This means that $|Z'| \subset T$ is open dense. Hence the topological space $|Z'|$ is irreducible, which means that $Z'$ is an irreducible scheme. By Schemes, Lemma 25.11.1 we conclude that $|Z'|$ is the closure of a single point $\eta \in T$ and hence also $T = \overline{\{\eta\}}$, and we win. $\square$For decent algebraic spaces dimension works as expected.

Lemma 59.11.10. Let $S$ be a scheme. Dimension as defined in Properties of Spaces, Section 57.9 behaves well on decent algebraic spaces $X$ over $S$.

- If $x \in |X|$, then $\dim_x(|X|) = \dim_x(X)$, and
- $\dim(|X|) = \dim(X)$.

Proof.Proof of (1). Choose a scheme $U$ with a point $u \in U$ and an étale morphism $h : U \to X$ mapping $u$ to $x$. By definition the dimension of $X$ at $x$ is $\dim_u(|U|)$. Thus we may pick $U$ such that $\dim_x(X) = \dim(|U|)$. Let $d$ be an integer. If $\dim(U) \geq d$, then there exists a sequence of nontrivial specializations $u_d \leadsto \ldots \leadsto u_0$ in $U$. Taking the image we find a corresponding sequence $h(u_d) \leadsto \ldots \leadsto h(u_0)$ each of which is nontrivial by Lemma 59.11.6. Hence we see that the image of $|U|$ in $|X|$ has dimension at least $d$. Conversely, suppose that $x_d \leadsto \ldots \leadsto x_0$ is a sequence of specializations in $|X|$ with $x_0$ in the image of $|U| \to |X|$. Then we can lift this to a sequence of specializations in $U$ by Lemma 59.11.7.Part (2) is an immediate consequence of part (1), Topology, Lemma 5.10.2, and Properties of Spaces, Section 57.9. $\square$

Lemma 59.11.11. Let $S$ be a scheme. Let $X \to Y$ be a locally quasi-finite morphism of algebraic spaces over $S$. Let $x \in |X|$ with image $y \in |Y|$. Then the dimension of the local ring of $Y$ at $y$ is $\geq$ to the dimension of the local ring of $X$ at $x$.

Proof.The definition of the dimension of the local ring of a point on an algebraic space is given in Properties of Spaces, Definition 57.10.2. Choose an étale morphism $(V, v) \to (Y, y)$ where $V$ is a scheme. Choose an étale morphism $U \to V \times_Y X$ and a point $u \in U$ mapping to $x \in |X|$ and $v \in V$. Then $U \to V$ is locally quasi-finite and we have to prove that $$ \dim(\mathcal{O}_{V, v}) \geq \dim(\mathcal{O}_{U, u}) $$ This is Algebra, Lemma 10.124.4. $\square$Lemma 59.11.12. Let $S$ be a scheme. Let $X \to Y$ be a locally quasi-finite morphism of algebraic spaces over $S$. Then $\dim(X) \leq \dim(Y)$.

Proof.This follows from Lemma 59.11.11 and Properties of Spaces, Lemma 57.10.3. $\square$The following lemma is a tiny bit stronger than Properties of Spaces, Lemma 57.15.3. We will improve this lemma in Lemma 59.13.2.

Lemma 59.11.13. Let $S$ be a scheme. Let $k$ be a field. Let $X$ be an algebraic space over $S$ and assume that there exists a surjective étale morphism $\mathop{\mathrm{Spec}}(k) \to X$. If $X$ is decent, then $X \cong \mathop{\mathrm{Spec}}(k')$ where $k' \subset k$ is a finite separable extension.

Proof.The assumption implies that $|X| = \{x\}$ is a singleton. Since $X$ is decent we can find a quasi-compact monomorphism $\mathop{\mathrm{Spec}}(k') \to X$ whose image is $x$. Then the projection $U = \mathop{\mathrm{Spec}}(k') \times_X \mathop{\mathrm{Spec}}(k) \to \mathop{\mathrm{Spec}}(k)$ is a monomorphism, whence $U = \mathop{\mathrm{Spec}}(k)$, see Schemes, Lemma 25.23.11. Hence the projection $\mathop{\mathrm{Spec}}(k) = U \to \mathop{\mathrm{Spec}}(k')$ is étale and we win. $\square$

The code snippet corresponding to this tag is a part of the file `decent-spaces.tex` and is located in lines 2118–2426 (see updates for more information).

```
\section{Points on spaces}
\label{section-points}
\noindent
In this section we prove some properties of points on decent algebraic spaces.
\begin{lemma}
\label{lemma-decent-points-monomorphism}
Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.
Consider the map
$$
\{\Spec(k) \to X \text{ monomorphism}\}
\longrightarrow
|X|
$$
This map is always injective. If $X$ is decent then this map
is a bijection.
\end{lemma}
\begin{proof}
We have seen in
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-points-monomorphism}
that the map is an injection in general.
By Lemma \ref{lemma-bounded-fibres} it is surjective when $X$ is
decent (actually one can say this is part of the definition
of being decent).
\end{proof}
\noindent
The following lemma tells us that the henselian local ring of a point
on a decent algebraic space is defined.
\begin{lemma}
\label{lemma-decent-space-elementary-etale-neighbourhood}
Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$.
For every point $x \in |X|$ there exists an \'etale morphism
$$
(U, u) \longrightarrow (X, x)
$$
where $U$ is an affine scheme, $u$ is the only point of $U$ lying
over $x$, and the induced morphism $\Spec(\kappa(u)) \to X$
is a monomorphism.
\end{lemma}
\begin{proof}
We may assume that $X$ is quasi-compact by replacing $X$ with a
quasi-compact open containing $x$. Recall that $x$ can be
represented by a quasi-compact (mono)morphism
from the spectrum a field (by definition of decent spaces). Thus the
lemma follows from Lemma \ref{lemma-filter-quasi-compact}.
\end{proof}
\begin{definition}
\label{definition-elemenary-etale-neighbourhood}
Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.
Let $x \in X$ be a point. An {\it elementary \'etale neighbourhood}
is an \'etale morphism $(U, u) \to (X, x)$ where $U$ is a scheme,
$u \in U$ is a point mapping to $x$, and $\Spec(\kappa(u)) \to X$
is a monomorphism. A {\it morphism of elementary \'etale neighbourhoods}
$(U, u) \to (U', u')$ is defined as a morphism $U \to U'$
over $X$ mapping $u$ to $u'$.
\end{definition}
\noindent
If $X$ is not decent then the category of elementary \'etale neighbourhoods
may be empty.
\begin{lemma}
\label{lemma-elementary-etale-neighbourhoods}
Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$.
Let $x$ be a point of $X$.
The category of elementary \'etale neighborhoods of $(X, x)$
is cofiltered (see
Categories, Definition \ref{categories-definition-codirected}).
\end{lemma}
\begin{proof}
The category is nonempty by
Lemma \ref{lemma-decent-space-elementary-etale-neighbourhood}.
Suppose that we have two elementary \'etale neighbourhoods
$(U_i, u_i) \to (X, x)$.
Then consider $U = U_1 \times_X U_2$. Since
$\Spec(\kappa(u_i)) \to X$, $i = 1, 2$ are both monomorphisms
in the class of $x$, we see that
$$
u = \Spec(\kappa(u_1)) \times_X \Spec(\kappa(u_2))
$$
is the spectrum of a field $\kappa(u)$ such that the induced maps
$\kappa(u_i) \to \kappa(u)$ are isomorphisms. Then $u \to U$ is a point
of $U$ and we see that $(U, u) \to (X, x)$ is an elementary
\'etale neighbourhood dominating $(U_i, u_i)$.
If $a, b : (U_1, u_1) \to (U_2, u_2)$ are two morphisms between
our elementary \'etale neighbourhoods, then we consider the scheme
$$
U = U_1 \times_{(a, b), (U_2 \times_X U_2), \Delta} U_2
$$
Using Properties of Spaces, Lemma
\ref{spaces-properties-lemma-etale-permanence}
we see that $U \to X$ is \'etale. Moreover, in exactly the same manner
as before we see that $U$ has a point $u$
such that $(U, u) \to (X, x)$ is an elementary
\'etale neighbourhood. Finally, $U \to U_1$ equalizes $a$ and $b$
and the proof is finished.
\end{proof}
\begin{definition}
\label{definition-henselian-local-ring}
Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$.
Let $x \in |X|$. The {\it henselian local ring of $X$ at $x$}, is
$$
\mathcal{O}_{X, x}^h = \colim \Gamma(U, \mathcal{O}_U)
$$
where the colimit is over the elementary \'etale neighbourhoods
$(U, u) \to (X, x)$.
\end{definition}
\noindent
To be sure, the henselian local ring of $X$ at $x$ is equal to
the henselization $\mathcal{O}_{U, u}^h$
of the local ring $\mathcal{O}_{U, u}$ of any elementary \'etale
neighbourhood. This follows from the definition,
Lemma \ref{lemma-elementary-etale-neighbourhoods}, and
More on Morphisms, Lemma \ref{more-morphisms-lemma-describe-henselization}.
\medskip\noindent
The following lemma shows that specialization of points behaves well
on decent algebraic spaces.
Spaces, Example \ref{spaces-example-infinite-product}
shows that this is {\bf not} true in general.
\begin{lemma}
\label{lemma-decent-no-specializations-map-to-same-point}
Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$.
Let $U \to X$ be an \'etale morphism from a scheme to $X$.
If $u, u' \in |U|$ map to the same point of $|X|$, and
$u' \leadsto u$, then $u = u'$.
\end{lemma}
\begin{proof}
Combine Lemmas \ref{lemma-bounded-fibres} and
\ref{lemma-no-specializations-map-to-same-point}.
\end{proof}
\begin{lemma}
\label{lemma-decent-specialization}
Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$.
Let $x, x' \in |X|$ and assume $x' \leadsto x$, i.e., $x$ is a
specialization of $x'$. Then for every \'etale morphism
$\varphi : U \to X$ from a scheme $U$ and any $u \in U$ with
$\varphi(u) = x$, exists a point $u'\in U$, $u' \leadsto u$ with
$\varphi(u') = x'$.
\end{lemma}
\begin{proof}
Combine Lemmas \ref{lemma-bounded-fibres} and
\ref{lemma-specialization}.
\end{proof}
\begin{lemma}
\label{lemma-kolmogorov}
Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$.
Then $|X|$ is Kolmogorov (see
Topology, Definition \ref{topology-definition-generic-point}).
\end{lemma}
\begin{proof}
Let $x_1, x_2 \in |X|$ with $x_1 \leadsto x_2$ and $x_2 \leadsto x_1$.
We have to show that $x_1 = x_2$. Pick a scheme $U$ and an \'etale morphism
$U \to X$ such that $x_1, x_2$ are both in the image of $|U| \to |X|$.
By Lemma \ref{lemma-decent-specialization} we can find a specialization
$u_1 \leadsto u_2$ in $U$ mapping to $x_1 \leadsto x_2$.
By Lemma \ref{lemma-decent-specialization} we can find
$u_2' \leadsto u_1$ mapping to $x_2 \leadsto x_1$. This means that
$u_2' \leadsto u_2$ is a specialization between points of $U$ mapping to
the same point of $X$, namely $x_2$. This is not possible, unless
$u_2' = u_2$, see
Lemma \ref{lemma-decent-no-specializations-map-to-same-point}. Hence
also $u_1 = u_2$ as desired.
\end{proof}
\begin{proposition}
\label{proposition-reasonable-sober}
Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$.
Then the topological space $|X|$ is sober (see
Topology, Definition \ref{topology-definition-generic-point}).
\end{proposition}
\begin{proof}
We have seen in Lemma \ref{lemma-kolmogorov} that $|X|$ is Kolmogorov.
Hence it remains to show that every irreducible closed subset
$T \subset |X|$ has a generic point. By
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-reduced-closed-subspace}
there exists a closed subspace $Z \subset X$ with $|Z| = |T|$.
By definition this means that $Z \to X$ is a representable morphism
of algebraic spaces. Hence $Z$ is a decent algebraic space
by Lemma \ref{lemma-representable-properties}. By
Theorem \ref{theorem-decent-open-dense-scheme}
we see that there exists an open dense subspace $Z' \subset Z$
which is a scheme. This means that $|Z'| \subset T$ is open dense.
Hence the topological space $|Z'|$ is irreducible, which means that
$Z'$ is an irreducible scheme. By
Schemes, Lemma \ref{schemes-lemma-scheme-sober}
we conclude that $|Z'|$ is the closure of a single point $\eta \in T$
and hence also $T = \overline{\{\eta\}}$, and we win.
\end{proof}
\noindent
For decent algebraic spaces dimension works as expected.
\begin{lemma}
\label{lemma-dimension-decent-space}
Let $S$ be a scheme. Dimension as defined in
Properties of Spaces, Section \ref{spaces-properties-section-dimension}
behaves well on decent algebraic spaces $X$ over $S$.
\begin{enumerate}
\item If $x \in |X|$, then $\dim_x(|X|) = \dim_x(X)$, and
\item $\dim(|X|) = \dim(X)$.
\end{enumerate}
\end{lemma}
\begin{proof}
Proof of (1). Choose a scheme $U$ with a point $u \in U$
and an \'etale morphism $h : U \to X$ mapping $u$ to $x$.
By definition the dimension of $X$ at $x$ is $\dim_u(|U|)$.
Thus we may pick $U$ such that $\dim_x(X) = \dim(|U|)$.
Let $d$ be an integer. If $\dim(U) \geq d$, then
there exists a sequence of nontrivial specializations
$u_d \leadsto \ldots \leadsto u_0$ in $U$. Taking the image
we find a corresponding sequence
$h(u_d) \leadsto \ldots \leadsto h(u_0)$
each of which is nontrivial by
Lemma \ref{lemma-decent-no-specializations-map-to-same-point}.
Hence we see that the image of $|U|$ in $|X|$ has dimension at least $d$.
Conversely, suppose that $x_d \leadsto \ldots \leadsto x_0$ is a
sequence of specializations in $|X|$ with $x_0$ in the image of
$|U| \to |X|$. Then we can lift this to a sequence of specializations
in $U$ by Lemma \ref{lemma-decent-specialization}.
\medskip\noindent
Part (2) is an immediate consequence of part (1),
Topology, Lemma \ref{topology-lemma-dimension-supremum-local-dimensions},
and Properties of Spaces, Section \ref{spaces-properties-section-dimension}.
\end{proof}
\begin{lemma}
\label{lemma-dimension-local-ring-quasi-finite}
Let $S$ be a scheme. Let $X \to Y$ be a locally quasi-finite morphism
of algebraic spaces over $S$. Let $x \in |X|$ with image $y \in |Y|$.
Then the dimension of the local ring of $Y$ at $y$ is $\geq$ to the
dimension of the local ring of $X$ at $x$.
\end{lemma}
\begin{proof}
The definition of the dimension of the local ring of a point on an
algebraic space is given in Properties of Spaces, Definition
\ref{spaces-properties-definition-dimension-local-ring}.
Choose an \'etale morphism $(V, v) \to (Y, y)$ where $V$ is a scheme.
Choose an \'etale morphism $U \to V \times_Y X$ and a point $u \in U$
mapping to $x \in |X|$ and $v \in V$. Then $U \to V$ is locally
quasi-finite and we have to prove that
$$
\dim(\mathcal{O}_{V, v}) \geq \dim(\mathcal{O}_{U, u})
$$
This is Algebra, Lemma \ref{algebra-lemma-dimension-inequality-quasi-finite}.
\end{proof}
\begin{lemma}
\label{lemma-dimension-quasi-finite}
Let $S$ be a scheme. Let $X \to Y$ be a locally quasi-finite morphism
of algebraic spaces over $S$. Then $\dim(X) \leq \dim(Y)$.
\end{lemma}
\begin{proof}
This follows from Lemma \ref{lemma-dimension-local-ring-quasi-finite}
and Properties of Spaces, Lemma \ref{spaces-properties-lemma-dimension}.
\end{proof}
\noindent
The following lemma is a tiny bit stronger than
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-point-like-spaces}.
We will improve this lemma in Lemma \ref{lemma-when-field}.
\begin{lemma}
\label{lemma-decent-point-like-spaces}
Let $S$ be a scheme. Let $k$ be a field. Let $X$ be an algebraic space
over $S$ and assume that there exists a surjective \'etale morphism
$\Spec(k) \to X$. If $X$ is decent, then $X \cong \Spec(k')$
where $k' \subset k$ is a finite separable extension.
\end{lemma}
\begin{proof}
The assumption implies that $|X| = \{x\}$ is a singleton. Since
$X$ is decent we can find a quasi-compact monomorphism $\Spec(k') \to X$
whose image is $x$. Then the projection
$U = \Spec(k') \times_X \Spec(k) \to \Spec(k)$
is a monomorphism, whence $U = \Spec(k)$, see
Schemes, Lemma \ref{schemes-lemma-mono-towards-spec-field}.
Hence the projection $\Spec(k) = U \to \Spec(k')$ is \'etale and
we win.
\end{proof}
```

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