algebraic-geometry
Sorry for my bad English.
Let $k$ be field, and $X$ be finite type scheme over $k$.
Now if $X_{\bar{k}}=X\times_k \bar{k}$ is irreducible,
we say $X$ is geometrically irreducible.
(as so reduced, connected, integral)
But I confuse what geometrically?
Please tell me origin or example ,thanks.
Here are a few observations that might help you get a feel for transcendental extensions of the base field.
A) Krull dimension Given two field extensions $k\to K , k\to K'$ the Krull dimesion of their tensor product is the minimum of their transcendence degrees over $k$ :
$$ Krulldim(K\otimes _k K')= min (trdeg_k K,trdeg_k K')$$
B) Tensor product with purely transcendental extension Given a field $k$, an arbitrary extension $k\to K$ and a purely transcendental extension $k\to k(\mathcal X)=k(...,X_i,...)$, the tensor product $K \otimes _k k(\mathcal X)$ is the sub-$k$-algebra of $K(\mathcal X)$ consisting of fractions of the form $\frac {F(\ldots,X_i,\ldots)}{f(\ldots,X_i,\ldots)}$ with $F(\ldots,X_i,\ldots)\in K[\mathcal X] \;$ and $f(\ldots,X_i,\ldots)\in k[\mathcal X]\setminus \{0\}$.
In particular $K \otimes _k k(\mathcal X)$ is an integral domain and the corresponding scheme $Spec(K \otimes _k k(\mathcal X))$ is reduced and irreducible. As a particular case you get your one-dimensional [cf. A)] scheme $Spec(k(x) \otimes _k k(y)$, which I hope now looks a little less "bizarre"
C) Examples Let me show how bad behaviour is detected by finite algebraic extensions.
i) Let $k\subset K$ be an algebraic extension of char. $p$ fields such that there is an $a\in K$ with $a \notin k$ but $a^p \in k$. Then $K\otimes_k k(a)$ is not reduced because $(a\otimes1-1\otimes a)^p=a^p\otimes1-1 \otimes a^p=0$ and so $a\otimes1-1\otimes a$ is a non zero nilpotent of $K\otimes_k k(a)$. Hence $Spec(K)$ is a reduced but not geometrically reduced $k$-scheme.
ii) Consider the extension $\mathbb Q\subset \mathbb Q(\sqrt 2) $. We have
$\mathbb Q(\sqrt 2) \otimes _{\mathbb Q} \mathbb Q(\sqrt 2)= \mathbb Q(\sqrt 2) \times \mathbb Q(\sqrt 2)$, a split (also called diagonal) $Q(\sqrt 2)$-algebra. Hence $Spec(\mathbb Q(\sqrt 2) \otimes _{\mathbb Q} \mathbb Q(\sqrt 2))$ is a reducible -even disconnected- scheme.Thus the $\mathbb Q$- scheme $Spec (\mathbb Q(\sqrt 2))$ is irreducible but not geometrically irreducible .
Interestingly for all non-trivial finite Galois extensions $k\subset K$ the $k$-scheme $Spec(K)$ is irreducible but not geometically irreducible . Indeed, a finite extension of fields $k\subset K$ of dimension $n$ is Galois if and only if it splits itself: $K\otimes_k K=K^n $
D) It is all just field theory ! Recall two possible attributes of a field extension $k\subset K$ :
separable if for all extensions (equivalently, all algebraic extensions) $k\to L$ the ring $K\otimes_k L$ is reduced i.e. has zero as sole nilpotent element.
primary a void condition in characteristic zero and in characteristic $p$ the condition that any $a\in K$ algebraic over $k$ is purely inseparable: $a^{p^r}\in k$ for some positive integer $r$.
With this terminology we reduce scheme properties to field properties:
Theorem 1 The $k$-scheme $X$ is geometrically irreducible if and only if it is irreducible and the residue field of its generic point $\eta$ is a primary extension $Rat(X)=\kappa(\eta)$ of $k$.
Theorem 2 The $k$-scheme $X $ is geometrically reduced if and only if it is reduced and the residue field of each of its its generic points $\eta_i$ is a separable extension $\kappa (\eta_i)$ of $k$.
Final remark A purely transcendental extension $k\subset k(\ldots,T_i,\ldots)$ is both primary and separable. This, in conjunction with Theorems 1 and 2, might help develop an intuition for non-algebraic base extensions .
Bibliography The formula for the Krull dimension of the tensor product of two fields is very difficult to locate in the literature. The only reference I know is EGA IV, Quatrième partie, where it is found on page 349 as Remarque (4.2.1.4) in the errata !
For the first part, it's an important fact in itself that $X_{\overline{k}} \rightarrow X_{k^{sep}}$ is a homeomorphism.
To do that, it's not an issue to reduce to the case of affine $X$, and thus you want to show that for any $k$-algebra $A$, we have a natural correspondance between the spectra of $A \otimes \overline{k}$ and $A \otimes k^{sep}$. To do that, the main ingredient is to note that for each $x \in A \otimes \overline{k}$, in prime characteristic $p > 0$, there is an integer $n \geq 0$ such that $x^{p^n} \in A \otimes k^{sep}$. (It's not entirely obvious how to go on after that, but it's not very difficult).
Now, we want to explain why, if $X_{\overline{k}}$ is irreducible, then for any extension $k'/k$, $X_{k'}$ is irreducible.
First, we can assume $k$ is algebraically closed. Indeed, let $K$ be an algebraic closure of $k'$, so that we have a map $\overline{k} \rightarrow K$ (respecting $k' \cap \overline{k}$). Now $K/\overline{k}$ is a field extension, and $X_{\overline{k}}$ is irreducible. If $X_K$ is indeed irreducible, then, as $X_K \rightarrow X_{k'}$ is surjective, it follows that $X_{k'}$ will be irreducible.
So assume that $k$ is an algebraically closed field and $X$ is an irreducible scheme over $k$. Let $K/k$ be a field extension, we want to show that $X_{K}$ is irreducible. It's enough to show it when $X$ is affine (because the affine subsets of a general irreducible $X$ are dense and irreducible, and their base changes will be irreducible, and will pairwise intersect, which forces $X_K$ to be irreducible by topology), and we can furthermore assume $X$ is reduced (so $X$ is the spectrum of an integral $k$-algebra $A$), and we will show that if $K/k$ is an extension, $A\otimes K$ is an integral domain.
Let $K/k$ be an extension and let $a,b \in A \otimes K$ have a null product. There is a finitely generated (and integral) $k$-subalgebra $B \subset K$ such that $a,b$ are in the image of the monomorphism (flatness) $A \otimes_k B \rightarrow A \otimes K$, so we want to show that, in $A \otimes B$, $a$ or $b$ must be zero when their product is zero.
Now, take a $k$-basis $e_i$ of $A$, and let $a_i,b_i$ be the coordinates of $a,b$ in the $B$-basis $e_i \otimes 1$. Then, $(a_i)_i,(b_i)_i$ are almost null sequences of elements of $B$. Assume that there are some nonzero $a_i,b_j$. Then $a_ib_j \in B$ is nonzero (and $B$ is a finitely generated $k$-algebra), hence is not contained in some maximal ideal $\mu$ by the Nullstellensatz. Consider the reduction map mod $\mu$: $A \otimes B \rightarrow A$ (as $k$ is algebraically closed): then the images of neither $a$ nor $b$ are zero, but their product is, so $A$ isn't an integral domain, so we have a contradiction, which shows that $A \otimes K$ is a domain and we are done.
Best Answer
"Geometric" typically means happening over an algebraically closed field (or something that happens once one passes to an algebraically closed field). I don't have a precise reference for the way this terminology is used - it's common though not totally standardized in algebraic geometry, and it is frequently accompanied by a statement explaining exactly what the author means. As far as history goes, this dates back at least to EGA/SGA1 and the term "geometric point", so it's been around quite a while.