If $X$ is an integral scheme over a field, then it can happen that $X$ becomes either nonreduced or reducible after extending the ground field. This does not happen for geometrically integral schemes.
Definition 33.9.1. Let $X$ be a scheme over the field $k$.
Let $x \in X$. We say $X$ is geometrically pointwise integral at $x$ if for every field extension $k'/k$ and every $x' \in X_{k'}$ lying over $x$ the local ring $\mathcal{O}_{X_{k'}, x'}$ is integral.
We say $X$ is geometrically pointwise integral if $X$ is geometrically pointwise integral at every point.
We say $X$ is geometrically integral over $k$ if the scheme $X_{k'}$ is integral for every field extension $k'$ of $k$.
The distinction between notions (2) and (3) is necessary. For example if $k = \mathbf{R}$ and $X = \mathop{\mathrm{Spec}}(\mathbf{C}[x])$, then $X$ is geometrically pointwise integral over $\mathbf{R}$ but of course not geometrically integral.
Lemma 33.9.2. Let $k$ be a field. Let $X$ be a scheme over $k$. Then $X$ is geometrically integral over $k$ if and only if $X$ is both geometrically reduced and geometrically irreducible over $k$.
Proof. See Properties, Lemma 28.3.4. $\square$
Lemma 33.9.3. Let $k$ be a field. Let $X$ be a proper scheme over $k$.
$A = H^0(X, \mathcal{O}_ X)$ is a finite dimensional $k$-algebra,
$A = \prod _{i = 1, \ldots , n} A_ i$ is a product of Artinian local $k$-algebras, one factor for each connected component of $X$,
if $X$ is reduced, then $A = \prod _{i = 1, \ldots , n} k_ i$ is a product of fields, each a finite extension of $k$,
if $X$ is geometrically reduced, then $k_ i$ is finite separable over $k$,
if $X$ is geometrically connected, then $A$ is geometrically irreducible over $k$,
if $X$ is geometrically irreducible, then $A$ is geometrically irreducible over $k$,
if $X$ is geometrically reduced and geometrically connected, then $A = k$, and
if $X$ is geometrically integral, then $A = k$.
Proof. By Cohomology of Schemes, Lemma 30.19.2 we see that $A = H^0(X, \mathcal{O}_ X)$ is a finite dimensional $k$-algebra. This proves (1).
Then $A$ is a product of local Artinian $k$-algebras by Algebra, Lemma 10.53.2 and Proposition 10.60.7. If $X = Y \amalg Z$ with $Y$ and $Z$ open in $X$, then we obtain an idempotent $e \in A$ by taking the section of $\mathcal{O}_ X$ which is $1$ on $Y$ and $0$ on $Z$. Conversely, if $e \in A$ is an idempotent, then we get a corresponding decomposition of $X$. Finally, as $X$ has a Noetherian underlying topological space its connected components are open. Hence the connected components of $X$ correspond $1$-to-$1$ with primitive idempotents of $A$. This proves (2).
If $X$ is reduced, then $A$ is reduced. Hence the local rings $A_ i = k_ i$ are reduced and therefore fields (for example by Algebra, Lemma 10.25.1). This proves (3).
If $X$ is geometrically reduced, then $A \otimes _ k \overline{k} = H^0(X_{\overline{k}}, \mathcal{O}_{X_{\overline{k}}})$ (equality by Cohomology of Schemes, Lemma 30.5.2) is reduced. This implies that $k_ i \otimes _ k \overline{k}$ is a product of fields and hence $k_ i/k$ is separable for example by Algebra, Lemmas 10.44.2 and 10.44.4. This proves (4).
If $X$ is geometrically connected, then $A \otimes _ k \overline{k} = H^0(X_{\overline{k}}, \mathcal{O}_{X_{\overline{k}}})$ is a zero dimensional local ring by part (2) and hence its spectrum has one point, in particular it is irreducible. Thus $A$ is geometrically irreducible. This proves (5). Of course (5) implies (6).
If $X$ is geometrically reduced and geometrically connected, then $A = k_1$ is a field and the extension $k_1/k$ is finite separable and geometrically irreducible. However, then $k_1 \otimes _ k \overline{k}$ is a product of $[k_1 : k]$ copies of $\overline{k}$ and we conclude that $k_1 = k$. This proves (7). Of course (7) implies (8). $\square$
Here is a baby version of Stein factorization; actual Stein factorization will be discussed in More on Morphisms, Section 37.53.
Lemma 33.9.4. Let $X$ be a proper scheme over a field $k$. Set $A = H^0(X, \mathcal{O}_ X)$. The fibres of the canonical morphism $X \to \mathop{\mathrm{Spec}}(A)$ are geometrically connected.
Proof. Set $S = \mathop{\mathrm{Spec}}(A)$. The canonical morphism $X \to S$ is the morphism corresponding to $\Gamma (S, \mathcal{O}_ S) = A = \Gamma (X, \mathcal{O}_ X)$ via Schemes, Lemma 26.6.4. The $k$-algebra $A$ is a finite product $A = \prod A_ i$ of local Artinian $k$-algebras finite over $k$, see Lemma 33.9.3. Denote $s_ i \in S$ the point corresponding to the maximal ideal of $A_ i$. Choose an algebraic closure $\overline{k}$ of $k$ and set $\overline{A} = A \otimes _ k \overline{k}$. Choose an embedding $\kappa (s_ i) \to \overline{k}$ over $k$; this determines a $\overline{k}$-algebra map
Consider the base change
of $X$ to $\overline{S} = \mathop{\mathrm{Spec}}(\overline{A})$. By Cohomology of Schemes, Lemma 30.5.2 we have $\Gamma (\overline{X}, \mathcal{O}_{\overline{X}}) = \overline{A}$. If $\overline{s}_ i \in \mathop{\mathrm{Spec}}(\overline{A})$ denotes the $\overline{k}$-rational point corresponding to $\sigma _ i$, then we see that $\overline{s}_ i$ maps to $s_ i \in S$ and $\overline{X}_{\overline{s}_ i}$ is the base change of $X_{s_ i}$ by $\mathop{\mathrm{Spec}}(\sigma _ i)$. Thus we see that it suffices to prove the lemma in case $k$ is algebraically closed.
Assume $k$ is algebraically closed. In this case $\kappa (s_ i)$ is algebraically closed and we have to show that $X_{s_ i}$ is connected. The product decomposition $A = \prod A_ i$ corresponds to a disjoint union decomposition $\mathop{\mathrm{Spec}}(A) = \coprod \mathop{\mathrm{Spec}}(A_ i)$, see Algebra, Lemma 10.21.2. Denote $X_ i$ the inverse image of $\mathop{\mathrm{Spec}}(A_ i)$. It follows from Lemma 33.9.3 part (2) that $A_ i = \Gamma (X_ i, \mathcal{O}_{X_ i})$. Observe that $X_{s_ i} \to X_ i$ is a closed immersion inducing an isomorphism on underlying topological spaces (because $\mathop{\mathrm{Spec}}(A_ i)$ is a singleton). Hence if $X_{s_ i}$ isn't connected, then neither is $X_ i$. So either $X_ i$ is empty and $A_ i = 0$ or $X_ i$ can be written as $U \amalg V$ with $U$ and $V$ open and nonempty which would imply that $A_ i$ has a nontrivial idempotent. Since $A_ i$ is local this is a contradiction and the proof is complete. $\square$
Lemma 33.9.5. Let $k$ be a field. Let $X$ be a proper geometrically reduced scheme over $k$. The following are equivalent
$H^0(X, \mathcal{O}_ X) = k$, and
$X$ is geometrically connected.
Proof. By Lemma 33.9.4 we have (1) $\Rightarrow $ (2). By Lemma 33.9.3 we have (2) $\Rightarrow $ (1). $\square$
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