** Geometric irreductibility is Zariski local modulo connectedness. **

Lemma 33.8.5. Let $k$ be a field. Let $X$ be a scheme over $k$. The following are equivalent

$X$ is geometrically irreducible over $k$,

for every nonempty affine open $U$ the $k$-algebra $\mathcal{O}_ X(U)$ is geometrically irreducible over $k$ (see Algebra, Definition 10.47.4),

$X$ is irreducible and there exists an affine open covering $X = \bigcup U_ i$ such that each $k$-algebra $\mathcal{O}_ X(U_ i)$ is geometrically irreducible, and

there exists an open covering $X = \bigcup _{i \in I} X_ i$ with $I \not= \emptyset $ such that $X_ i$ is geometrically irreducible for each $i$ and such that $X_ i \cap X_ j \not= \emptyset $ for all $i, j \in I$.

Moreover, if $X$ is geometrically irreducible so is every nonempty open subscheme of $X$.

**Proof.**
An affine scheme $\mathop{\mathrm{Spec}}(A)$ 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 28.3.3. 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).
$\square$

## Comments (3)

Comment #775 by Keenan Kidwell on

Comment #793 by Johan on

Comment #1115 by Simon Pepin Lehalleur on

There are also: