hcju/stacksqa huggingface dataset:
Assume (1). This in particular implies that $\mathcal{O}_{X, x}$ is reduced. Let $k \subset k'$ be a finite purely inseparable field extension. Consider the ring $\mathcal{O}_{X, x} \otimes_k k'$. By Algebra, Lemma \ref{algebra-lemma-p-ring-map} its spectrum is the same as the spectrum of $\mathcal{O}_{X, x}$. Hence it is a local ring also (Algebra, Lemma \ref{algebra-lemma-characterize-local-ring}). Therefore there is a unique point $x' \in X_{k'}$ lying over $x$ and $\mathcal{O}_{X_{k'}, x'} \cong \mathcal{O}_{X, x} \otimes_k k'$. By assumption this is a reduced ring. Hence we deduce (2) by Algebra, Lemma \ref{algebra-lemma-geometrically-reduced-finite-purely-inseparable-extension}. \medskip\noindent Assume (2). Let $k \subset k'$ be a field extension. Since $\Spec(k') \to \Spec(k)$ is surjective, also $X_{k'} \to X$ is surjective (Morphisms, Lemma \ref{morphisms-lemma-base-change-surjective}). Let $x' \in X_{k'}$ be any point lying over $x$. The local ring $\mathcal{O}_{X_{k'}, x'}$ is a localization of the ring $\mathcal{O}_{X, x} \otimes_k k'$. Hence it is reduced by assumption and (1) is proved.
Assume (1). Then for every field extension $k \subset k'$ and every point $x' \in X_{k'}$ the local ring of $X_{k'}$ at $x'$ is reduced. In other words $X_{k'}$ is reduced. Hence (2). \medskip\noindent Assume (2). Let $U \subset X$ be an affine open. Then for every field extension $k \subset k'$ the scheme $X_{k'}$ is reduced, hence $U_{k'} = \Spec(\mathcal{O}(U)\otimes_k k')$ is reduced, hence $\mathcal{O}(U)\otimes_k k'$ is reduced (see Properties, Section \ref{properties-section-integral}). In other words $\mathcal{O}(U)$ is geometrically reduced, so (7) holds. \medskip\noindent Assume (7). For any field extension $k \subset k'$ the base change $X_{k'}$ is gotten by gluing the spectra of the rings $\mathcal{O}_X(U) \otimes_k k'$ where $U$ is affine open in $X$ (see Schemes, Section \ref{schemes-section-fibre-products}). Hence $X_{k'}$ is reduced. So (1) holds. \medskip\noindent This proves that (1), (2), and (7) are equivalent. These are equivalent to (3), (4), (5), and (6) because we can apply Algebra, Lemma \ref{algebra-lemma-geometrically-reduced-finite-purely-inseparable-extension} to $\mathcal{O}_X(U)$ for $U \subset X$ affine open.
Note that if $k \subset k'$ is purely inseparable, then $X_{k'} \to X$ induces a homeomorphism on underlying topological spaces, see Algebra, Lemma \ref{algebra-lemma-p-ring-map}. Whence the uniqueness of $x'$ lying over $x$ mentioned in the statement. Moreover, in this case $\mathcal{O}_{X_{k'}, x'} = \mathcal{O}_{X, x} \otimes_k k'$. Hence the lemma follows from Lemma \ref{lemma-geometrically-reduced-at-point} above and Algebra, Lemma \ref{algebra-lemma-geometrically-reduced-finite-purely-inseparable-extension}.
It is clear that (1) implies (2). Assume (2). Let $k \subset k''$ be a finite purely inseparable field extension and let $x'' \in X_{k''}$ be a point lying over $x$ (actually it is unique). We can find a common field extension $k \subset k'''$ (i.e.\ with both $k' \subset k'''$ and $k'' \subset k'''$) and a point $x''' \in X_{k'''}$ lying over both $x'$ and $x''$. Consider the map of local rings $$ \mathcal{O}_{X_{k''}, x''} \longrightarrow \mathcal{O}_{X_{k'''}, x''''}. $$ This is a flat local ring homomorphism and hence faithfully flat. By (2) we see that the local ring on the right is reduced. Thus by Algebra, Lemma \ref{algebra-lemma-descent-reduced} we conclude that $\mathcal{O}_{X_{k''}, x''}$ is reduced. Thus by Lemma \ref{lemma-check-only-finite-inseparable-extensions} we conclude that $X$ is geometrically reduced at $x$.
Part (1) follows from Lemma \ref{lemma-geometrically-reduced-at-point} and the fact that if $A$ is a geometrically reduced $k$-algebra, then $S^{-1}A$ is a geometrically reduced $k$-algebra for any multiplicative subset $S$ of $A$, see Algebra, Lemma \ref{algebra-lemma-geometrically-reduced-permanence}. \medskip\noindent Let $A = \mathcal{O}_{X, x}$. The assumptions (a) and (b) of (2) imply that $A$ is reduced, and that $A_{\mathfrak q}$ is geometrically reduced over $k$ for every minimal prime $\mathfrak q$ of $A$. Hence $A$ is geometrically reduced over $k$, see Algebra, Lemma \ref{algebra-lemma-generic-points-geometrically-reduced}. Thus $X$ is geometrically reduced at $x$, see Lemma \ref{lemma-geometrically-reduced-at-point}. \medskip\noindent Part (3) follows trivially from part (2).
Assume $X$ locally Noetherian and geometrically reduced at $x$. By Properties, Lemma \ref{properties-lemma-ring-affine-open-injective-local-ring} we can find an affine open neighbourhood $U \subset X$ of $x$ such that $R = \mathcal{O}_X(U) \to \mathcal{O}_{X, x}$ is injective. By Lemma \ref{lemma-geometrically-reduced-at-point} the assumption means that $\mathcal{O}_{X, x}$ is geometrically reduced over $k$. By Algebra, Lemma \ref{algebra-lemma-subalgebra-separable} this implies that $R$ is geometrically reduced over $k$, which in turn implies that $U$ is geometrically reduced.
To prove this lemma we may replace $X$ by its reduction $X_{red}$. Hence we may assume that $X$ is reduced and locally of finite type over $k$. Let $x_1, \ldots, x_n \in X$ be the generic points of the irreducible components of $X$. Note that for every purely inseparable algebraic extension $k \subset k'$ the morphism $(X_{k'})_{red} \to X$ is a homeomorphism, see Algebra, Lemma \ref{algebra-lemma-p-ring-map}. Hence the points $x'_1, \ldots, x'_n$ lying over $x_1, \ldots, x_n$ are the generic points of the irreducible components of $(X_{k'})_{red}$. As $X$ is reduced the local rings $K_i = \mathcal{O}_{X, x_i}$ are fields, see Algebra, Lemma \ref{algebra-lemma-minimal-prime-reduced-ring}. As $X$ is locally of finite type over $k$ the field extensions $k \subset K_i$ are finitely generated field extensions. Finally, the local rings $\mathcal{O}_{(X_{k'})_{red}, x'_i}$ are the fields $(K_i \otimes_k k')_{red}$. By Algebra, Lemma \ref{algebra-lemma-make-separable} we can find a finite purely inseparable extension $k \subset k'$ such that $(K_i \otimes_k k')_{red}$ are separable field extensions of $k'$. In particular each $(K_i \otimes_k k')_{red}$ is geometrically reduced over $k'$ by Algebra, Lemma \ref{algebra-lemma-characterize-separable-field-extensions}. At this point Lemma \ref{lemma-generic-points-geometrically-reduced} part (3) implies that $(X_{k'})_{red}$ is geometrically reduced.
The scheme theoretic fibres of $p$ are connected, since they are base changes of the geometrically connected scheme $X$ by field extensions. Moreover the scheme theoretic fibres are homeomorphic to the set theoretic fibres, see Schemes, Lemma \ref{schemes-lemma-fibre-topological}. By Morphisms, Lemma \ref{morphisms-lemma-scheme-over-field-universally-open} the map $p$ is open. Thus we may apply Topology, Lemma \ref{topology-lemma-connected-fibres-connected-components} to conclude.
Since $k$ is separably algebraically closed we see that $k'$ is geometrically connected over $k$, see Algebra, Lemma \ref{algebra-lemma-separably-closed-connected-implies-geometric}. Hence $Z = \Spec(k')$ is geometrically connected over $k$ by Lemma \ref{lemma-affine-geometrically-connected} above. Since $X_{k'} = Z \times_k X$ the result is a special case of Lemma \ref{lemma-bijection-connected-components}.
Assume $X_{\overline{k}}$ is connected. Let $k \subset k'$ be a field extension. There exists a field extension $\overline{k} \subset \overline{k}'$ such that $k'$ embeds into $\overline{k}'$ as an extension of $k$. By Lemma \ref{lemma-separably-closed-field-connected-components} we see that $X_{\overline{k}'}$ is connected. Since $X_{\overline{k}'} \to X_{k'}$ is surjective we conclude that $X_{k'}$ is connected as desired.
We remark that if $X$ is also quasi-separated this follows from Limits, Lemma \ref{limits-lemma-descend-opens}. Let $U_1, \ldots, U_n$ be finitely many affine opens of $X$ such that $V \subset \bigcup U_{i, A}$. Say $U_i = \Spec(R_i)$. Since $V$ is quasi-compact we can find finitely many $f_{ij} \in R_i \otimes_k A$, $j = 1, \ldots, n_i$ such that $V = \bigcup_i \bigcup_{j = 1, \ldots, n_i} D(f_{ij})$ where $D(f_{ij}) \subset U_{i, A}$ is the corresponding standard open. (We do not claim that $V \cap U_{i, A}$ is the union of the $D(f_{ij})$, $j = 1, \ldots, n_i$.) It is clear that we can find a finitely generated $k$-subalgebra $A' \subset A$ such that $f_{ij}$ is the image of some $f_{ij}' \in R_i \otimes_k A'$. Set $V' = \bigcup D(f_{ij}')$ which is a quasi-compact open of $X_{A'}$. Denote $\pi : X_A \to X_{A'}$ the canonical morphism. We have $\pi(V) \subset V'$ as $\pi(D(f_{ij})) \subset D(f_{ij}')$. If $x \in X_A$ with $\pi(x) \in V'$, then $\pi(x) \in D(f_{ij}')$ for some $i, j$ and we see that $x \in D(f_{ij})$ as $f_{ij}'$ maps to $f_{ij}$. Thus we see that $V = \pi^{-1}(V')$ as desired.
By Lemma \ref{lemma-descend-open} there exists a finite subextension $k \subset k' \subset \overline{k}$ and an open $V' \subset X_{k'}$ which pulls back to $V$. This proves (1). Since $\text{Gal}(\overline{k}/k')$ is open in $\text{Gal}(\overline{k}/k)$ part (2) is clear as well.
This lemma immediately reduces to the case where $X = \Spec(A)$ is affine. In this case, let $\overline{I} \subset A \otimes_k \overline{k}$ be the radical ideal corresponding to $\overline{T}$. Assumption (2) implies that $\sigma(\overline{I}) = \overline{I}$ for all $\sigma \in \text{Gal}(\overline{k}/k)$. Pick $x \in \overline{I}$. There exists a finite Galois extension $k \subset k'$ contained in $\overline{k}$ such that $x \in A \otimes_k k'$. Set $G = \text{Gal}(k'/k)$. Set $$ P(T) = \prod\nolimits_{\sigma \in G} (T - \sigma(x)) \in (A \otimes_k k')[T] $$ It is clear that $P(T)$ is monic and is actually an element of $(A \otimes_k k')^G[T] = A[T]$ (by basic Galois theory). Moreover, if we write $P(T) = T^d + a_1T^{d - 1} + \ldots + a_0$ the we see that $a_i \in I := A \cap \overline{I}$. By Algebra, Lemma \ref{algebra-lemma-polynomials-divide} we see that $x$ is contained in the radical of $I(A \otimes_k \overline{k})$. Hence $\overline{I}$ is the radical of $I(A \otimes_k \overline{k})$ and setting $T = V(I)$ is a solution.
It follows immediately from the definition that (1) implies (2). Assume that $X$ is not geometrically connected. Let $k \subset \overline{k}$ be a separable algebraic closure of $k$. By Lemma \ref{lemma-characterize-geometrically-connected} it follows that $X_{\overline{k}}$ is disconnected. Say $X_{\overline{k}} = \overline{U} \amalg \overline{V}$ with $\overline{U}$ and $\overline{V}$ open, closed, and nonempty. \medskip\noindent Suppose that $W \subset X$ is any quasi-compact open. Then $W_{\overline{k}} \cap \overline{U}$ and $W_{\overline{k}} \cap \overline{V}$ are open and closed in $W_{\overline{k}}$. In particular $W_{\overline{k}} \cap \overline{U}$ and $W_{\overline{k}} \cap \overline{V}$ are quasi-compact, and by Lemma \ref{lemma-Galois-action-quasi-compact-open} both $W_{\overline{k}} \cap \overline{U}$ and $W_{\overline{k}} \cap \overline{V}$ are defined over a finite subextension and invariant under an open subgroup of $\text{Gal}(\overline{k}/k)$. We will use this without further mention in the following. \medskip\noindent Pick $W_0 \subset X$ quasi-compact open such that both $W_{0, \overline{k}} \cap \overline{U}$ and $W_{0, \overline{k}} \cap \overline{V}$ are nonempty. Choose a finite subextension $k \subset k' \subset \overline{k}$ and a decomposition $W_{0, k'} = U_0' \amalg V_0'$ into open and closed subsets such that $W_{0, \overline{k}} \cap \overline{U} = (U'_0)_{\overline{k}}$ and $W_{0, \overline{k}} \cap \overline{V} = (V'_0)_{\overline{k}}$. Let $H = \text{Gal}(\overline{k}/k') \subset \text{Gal}(\overline{k}/k)$. In particular $\sigma(W_{0, \overline{k}} \cap \overline{U}) = W_{0, \overline{k}} \cap \overline{U}$ and similarly for $\overline{V}$. \medskip\noindent Having chosen $W_0$, $k'$ as above, for every quasi-compact open $W \subset X$ we set $$ U_W = \bigcap\nolimits_{\sigma \in H} \sigma(W_{\overline{k}} \cap \overline{U}), \quad V_W = \bigcup\nolimits_{\sigma \in H} \sigma(W_{\overline{k}} \cap \overline{V}). $$ Now, since $W_{\overline{k}} \cap \overline{U}$ and $W_{\overline{k}} \cap \overline{V}$ are fixed by an open subgroup of $\text{Gal}(\overline{k}/k)$ we see that the union and intersection above are finite. Hence $U_W$ and $V_W$ are both open and closed. Also, by construction $W_{\bar k} = U_W \amalg V_W$. \medskip\noindent We claim that if $W \subset W' \subset X$ are quasi-compact open, then $W_{\overline{k}} \cap U_{W'} = U_W$ and $W_{\overline{k}} \cap V_{W'} = V_W$. Verification omitted. Hence we see that upon defining $U = \bigcup_{W \subset X} U_W$ and $V = \bigcup_{W \subset X} V_W$ we obtain $X_{\overline{k}} = U \amalg V$ is a disjoint union of open and closed subsets. It is clear that $V$ is nonempty as it is constructed by taking unions (locally). On the other hand, $U$ is nonempty since it contains $W_0 \cap \overline{U}$ by construction. Finally, $U, V \subset X_{\bar k}$ are closed and $H$-invariant by construction. Hence by Lemma \ref{lemma-closed-fixed-by-Galois} we have $U = (U')_{\bar k}$, and $V = (V')_{\bar k}$ for some closed $U', V' \subset X_{k'}$. Clearly $X_{k'} = U' \amalg V'$ and we see that $X_{k'}$ is disconnected as desired.
Write $X_{\overline{k}} = \overline{U} \amalg \overline{V}$ with $\overline{U}$ and $\overline{V}$ open and closed. Denote $\overline{f} : T_{\overline{k}} \to X_{\overline{k}}$ the base change of $f$. Since $T_{\overline{k}}$ is connected we see that $T_{\overline{k}}$ is contained in either $\overline{f}^{-1}(\overline{U})$ or $\overline{f}^{-1}(\overline{V})$. Say $T_{\overline{k}} \subset \overline{f}^{-1}(\overline{U})$. \medskip\noindent Fix a quasi-compact open $W \subset X$. There exists a finite Galois subextension $k \subset k' \subset \overline{k}$ such that $\overline{U} \cap W_{\overline{k}}$ and $\overline{V} \cap W_{\overline{k}}$ come from quasi-compact opens $U', V' \subset W_{k'}$. Then also $W_{k'} = U' \amalg V'$. Consider $$ U'' = \bigcap\nolimits_{\sigma \in \text{Gal}(k'/k)} \sigma(U'), \quad V'' = \bigcup\nolimits_{\sigma \in \text{Gal}(k'/k)} \sigma(V'). $$ These are Galois invariant, open and closed, and $W_{k'} = U'' \amalg V''$. By Lemma \ref{lemma-closed-fixed-by-Galois} we get open and closed subsets $U_W, V_W \subset W$ such that $U'' = (U_W)_{k'}$, $V'' = (V_W)_{k'}$ and $W = U_W \amalg V_W$. \medskip\noindent We claim that if $W \subset W' \subset X$ are quasi-compact open, then $W \cap U_{W'} = U_W$ and $W \cap V_{W'} = V_W$. Verification omitted. Hence we see that upon defining $U = \bigcup_{W \subset X} U_W$ and $V = \bigcup_{W \subset X} V_W$ we obtain $X = U \amalg V$. It is clear that $V$ is nonempty as it is constructed by taking unions (locally). On the other hand, $U$ is nonempty since it contains $f(T)$ by construction.
Set $T = \Spec(\kappa(x))$. Let $k \subset \overline{k}$ be a separable algebraic closure of $k$. The assumption on $k \subset \kappa(x)$ implies that $T_{\overline{k}}$ is irreducible, see Algebra, Lemma \ref{algebra-lemma-field-extension-geometrically-irreducible}. Hence by Lemma \ref{lemma-geometrically-connected-criterion} we see that $X_{\overline{k}}$ is connected. By Lemma \ref{lemma-characterize-geometrically-connected} we conclude that $X$ is geometrically connected.
The image $p(T)$ is contained in some connected component $X'$ of $X$. Consider $X'$ as a closed subscheme of $X$ in any way. Then $T$ is also a connected component of $X'_K = p^{-1}(X')$ and we may therefore assume that $X$ is connected. The morphism $p$ is open (Morphisms, Lemma \ref{morphisms-lemma-scheme-over-field-universally-open}),  closed (Morphisms, Lemma \ref{morphisms-lemma-integral-universally-closed}) and the fibers of $p$ are finite sets (Morphisms, Lemma \ref{morphisms-lemma-finite-quasi-finite}). Thus we may apply Topology, Lemma \ref{topology-lemma-finite-fibre-connected-components} to conclude.
When $k \subset K$ is finite this is Lemma \ref{lemma-image-connected-component-finite-extension}. In general the proof is more difficult. \medskip\noindent Let $T \subset X$ be the connected component of $X$ containing the image of $\overline{T}$. We may replace $X$ by $T$ (with the induced reduced subscheme structure). Thus we may assume $X$ is connected. Let $A = H^0(X, \mathcal{O}_X)$. Let $L \subset A$ be the maximal weakly \'etale $k$-subalgebra, see More on Algebra, Lemma \ref{more-algebra-lemma-max-weakly-etale-subalgebra}. Since $A$ does not have any nontrivial idempotents we see that $L$ is a field and a separable algebraic extension of $k$ by More on Algebra, Lemma \ref{more-algebra-lemma-class-weakly-etale-over-field}. Observe that $L$ is also the maximal weakly \'etale $L$-subalgebra of $A$ (because any weakly \'etale $L$-algebra is weakly \'etale over $k$ by More on Algebra, Lemma \ref{more-algebra-lemma-composition-weakly-etale}). By Schemes, Lemma \ref{schemes-lemma-morphism-into-affine} we obtain a factorization $X \to \Spec(L) \to \Spec(k)$ of the structure morphism. \medskip\noindent Let $L'/L$ be a finite separable extension. By Cohomology of Schemes, Lemma \ref{coherent-lemma-finite-locally-free-base-change-cohomology} we have $$ A \otimes_L L' = H^0(X \times_{\Spec(L)} \Spec(L'), \mathcal{O}_{X \times_{\Spec(L)} \Spec(L')}) $$ The maximal weakly \'etale $L'$-subalgebra of $A \otimes_L L'$ is $L \otimes_L L' = L'$ by More on Algebra, Lemma \ref{more-algebra-lemma-change-fields-max-weakly-etale-subalgebra}. In particular $A \otimes_L L'$ does not have nontrivial idempotents (such an idempotent would generate a weakly \'etale subalgebra) and we conclude that $X \times_{\Spec(L)} \Spec(L')$ is connected. By Lemma \ref{lemma-characterize-geometrically-disconnected} we conclude that $X$ is geometrically connected over $L$. \medskip\noindent Let's give $\overline{T}$ the reduced induced scheme structure and consider the composition $$ \overline{T} \xrightarrow{i} X_K = X \times_{\Spec(k)} \Spec(K) \xrightarrow{\pi} \Spec(L \otimes_k K) $$ The image is contained in a connected component of $\Spec(L \otimes_k K)$. Since $K \to L \otimes_k K$ is integral we see that the connected components of $\Spec(L \otimes_k K)$ are points and all points are closed, see Algebra, Lemma \ref{algebra-lemma-integral-over-field}. Thus we get a quotient field $L \otimes_k K \to E$ such that $\overline{T}$ maps into $\Spec(E) \subset \Spec(L \otimes_k K)$. Hence $i(\overline{T}) \subset \pi^{-1}(\Spec(E))$. But $$ \pi^{-1}(\Spec(E)) = (X \times_{\Spec(k)} \Spec(K)) \times_{\Spec(L \otimes_k K)} \Spec(E) = X \times_{\Spec(L)} \Spec(E) $$ which is connected because $X$ is geometrically connected over $L$. Then we get the equality $\overline{T} = X \times_{\Spec(L)} \Spec(E)$ (set theoretically) and we conclude that $\overline{T} \to X$ is surjective as desired.
The action (\ref{equation-galois-action-base-change-kbar}) of $\text{Gal}(\overline{k}/k)$ on $X_{\overline{k}}$ induces an action on its connected components. Connected components are always closed (Topology, Lemma \ref{topology-lemma-connected-components}). Hence if $\overline{T}$ is as in (1), then by Lemma \ref{lemma-closed-fixed-by-Galois} there exists a closed subset $T \subset X$ such that $\overline{T} = T_{\overline{k}}$. Note that $T$ is geometrically connected over $k$, see Lemma \ref{lemma-characterize-geometrically-connected}. To see that $T$ is a connected component of $X$, suppose that $T \subset T'$, $T \not = T'$ where $T'$ is a connected component of $X$. In this case $T'_{k'}$ strictly contains $\overline{T}$ and hence is disconnected. By Lemma \ref{lemma-tricky} this means that $T'$ is disconnected! Contradiction. \medskip\noindent We omit the proof of the functoriality in (2).
Since the connected components are open, cover $X_{\overline{k}}$ (Topology, Lemma \ref{topology-lemma-connected-components}) and $X_{\overline{k}}$ is quasi-compact, we conclude that there are only finitely many of them. Thus (a) holds. By Lemma \ref{lemma-descend-open} these connected components are each defined over a finite subextension of $k \subset \overline{k}$ and we get (b). If $X$ is of finite type over $k$, then $X_{\overline{k}}$ is of finite type over $\overline{k}$ (Morphisms, Lemma \ref{morphisms-lemma-base-change-finite-type}). Hence $X_{\overline{k}}$ is a Noetherian scheme (Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian}). Thus $X_{\overline{k}}$ has finitely many irreducible components (Properties, Lemma \ref{properties-lemma-Noetherian-irreducible-components}) and a fortiori finitely many connected components (which are therefore open).
First, note that the scheme theoretic fibres of $p$ are irreducible, since they are base changes of the geometrically irreducible scheme $X$ by field extensions. Moreover the scheme theoretic fibres are homeomorphic to the set theoretic fibres, see Schemes, Lemma \ref{schemes-lemma-fibre-topological}. By Morphisms, Lemma \ref{morphisms-lemma-scheme-over-field-universally-open} the map $p$ is open. Thus we may apply Topology, Lemma \ref{topology-lemma-irreducible-fibres-irreducible-components} to conclude.
An affine scheme $\Spec(A)$ over $k$ is geometrically irreducible if and only if $A$ is geometrically irreducible over $k$; this is immediate from the definitions. Recall that if a scheme is irreducible so is every nonempty open subscheme of $X$, any two nonempty open subsets have a nonempty intersection. Also, if every affine open is irreducible then the scheme is irreducible, see Properties, Lemma \ref{properties-lemma-characterize-irreducible}. Hence the final statement of the lemma is clear, as well as the implications (1) $\Rightarrow$ (2), (2) $\Rightarrow$ (3), and (3) $\Rightarrow$ (4). If (4) holds, then for any field extension $k'/k$ the scheme $X_{k'}$ has a covering by irreducible opens which pairwise intersect. Hence $X_{k'}$ is irreducible. Hence (4) implies (1).
Assume (1). Recall that $\mathcal{O}_{X, \xi}$ is the filtered colimit of $\mathcal{O}_X(U)$ where $U$ runs over the nonempty open affine subschemes of $X$. Combining Lemma \ref{lemma-geometrically-irreducible-local} and Algebra, Lemma \ref{algebra-lemma-subalgebra-geometrically-irreducible} we see that $\mathcal{O}_{X, \xi}$ is geometrically irreducible over $k$. Since $\mathcal{O}_{X, \xi} \to \kappa(\xi)$ is a surjection with locally nilpotent kernel (see Algebra, Lemma \ref{algebra-lemma-minimal-prime-reduced-ring}) it follows that $\kappa(\xi)$ is geometrically irreducible, see Algebra, Lemma \ref{algebra-lemma-p-ring-map}. \medskip\noindent Assume (2). We may assume that $X$ is reduced. Let $U \subset X$ be a nonempty affine open. Then $U = \Spec(A)$ where $A$ is a domain with fraction field $\kappa(\xi)$. Thus $A$ is a $k$-subalgebra of a geometrically irreducible $k$-algebra. Hence by Algebra, Lemma \ref{algebra-lemma-subalgebra-geometrically-irreducible} we see that $A$ is geometrically irreducible over $k$. By Lemma \ref{lemma-geometrically-irreducible-local} we conclude that $X$ is geometrically irreducible over $k$.
Since $k$ is separably algebraically closed we see that $k'$ is geometrically irreducible over $k$, see Algebra, Lemma \ref{algebra-lemma-separably-closed-irreducible-implies-geometric}. Hence $Z = \Spec(k')$ is geometrically irreducible over $k$. by Lemma \ref{lemma-geometrically-irreducible-local} above. Since $X' = Z \times_k X$ the result is a special case of Lemma \ref{lemma-bijection-irreducible-components}.