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An exercise in Hartshorne claims that a scheme $X$ of finite type over a field $k$ is geometrically irreducible (respectively geometrically reduced) if and only if $X \times_k K$ is irreducible (respectively reduced) for any field extension $K/k$. Now I'm perfectly fine with this if $K/k$ is algebraic, but what about transcendental extensions?

We already encounter some weird phenomena in "simple" examples such as $\text{Spec}(k(x) \otimes_k k(y))$, which is a pretty bizarre scheme. I'm not claiming this is a counterexample, since it ends up being an integral scheme (also $k(x)$ is not finitely generated over $k$), only that I don't have a good feel for what happens with transcendental extensions of the base field.

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Let me half-answer your question for geometrically reduced schemes first. This is essentially:

Lemma: Let $A$ be a reduced $k$-algebra, where $k$ is an algebraically closed field. Then if $K/k$ is any extension, $A \otimes_k K$ is reduced too.

The whole point is that we don't actually have to handle anything besides the case of $A$ a field! The reason is that any reduced noetherian ring imbeds into a (finite) product of fields (and in general, one can argue that $A$ is a filtered colimit of its noetherian subrings). As a result, one simply has to show that if $L, K$ are any two $k$-extensions, then $L \otimes_k K$ is a reduced ring.

The situation is more subtle in the case of irreducibility, and one needs a bit of general topology. Here the key lemma is:

Lemma: Let $X$ be a irreducible $k$-scheme, where $k$ is algebraically closed. Then $X \times_k K$ is irreducible for any extension $K/k$.

Here one may argue that $X \times_k K \to X $ is an open surjection (it is open because any map of discrete schemes is universally open). Moreover, one can show that the fibers are irreducible.

Now, if one has a continuous surjection $p: X \to Y$ such that the fibers $p^{-1}(y)$ are irreducible, and if $Y$ is irreducible, then $X$ is irreducible: to see this, consider a decomposition $X = X_1 \cup X_2$ for two proper closed subsets. We can consider the set of $y \in Y$ such that $X_y \subset X_1$ and the set such that $X_y \subset X_2$; these fill $Y$ as the fibers are irreducible. Moreover they are closed (as they are the complements of the images of $X - X_i$ and $p$ is open). It follows that one fills $Y$ and consequently one of $X_1, X_2$ fills $X$.

So there is again a bit of commutative algebra one needs, which is:

Lemma: Let $k$ be an algebraically closed field, and let $K, L$ be extensions of $k$. Then $\mathrm{Spec} K \otimes_k L$ is irreducible.

Both of these lemmas amount to saying that the tensor product of extensions of an algebraically closed field is an integral domain. I actually can't remember the proof of this theorem right now; I can look it up a bit later. But hopefully this clarifies the somewhat opaque Hartshorne exercise to a much more believable statement (at least, to me).

The finite type hypothesis isn't necessary, by the way.

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Very helpful! By the way, I tracked down the statement that a morphism with discrete target is universally open in EGA, and it seems to be somewhat nontrivial. I guess the key fact is that a flat morphism locally of finite type is open. –  Justin Campbell Jun 19 '11 at 21:42
As for the last lemma, how about this? Since $K,L$ are purely transcendental over $k$ they are the fraction fields of some (possibly infinite-dimensional) polynomial rings over $k$. Thus $K \otimes_k L$ is a localization of a polynomial ring over $k$, hence a domain. –  Justin Campbell Jun 19 '11 at 22:41
@Justin: Dear Justin, the problem is that they're not necessarily purely transcendental (think for instance of $k(x,y, \sqrt{y^2 - x^3})$). I realize I didn't get back to you about this; let me do that shortly. –  Akhil Mathew Jun 19 '11 at 23:18
@Justin: Also, yes, you're right: the point is that a flat morphism locally of finite presentation (I'm not sure that finite type is enough here, but maybe it is by limiting arguments in IV-11 that I don't fully understand yet) is universally open and from there it is some technical checking, but purely elementary. –  Akhil Mathew Jun 19 '11 at 23:19
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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 ( in the errata !

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Interesting. The reason I called $\text{Spec}(k(x) \otimes_k k(y))$ "bizarre" is because it looks pretty strange geometrically: set-theoretically it consists of those points of $\mathbb{A}^2$ which do not come from the setwise product $\mathbb{A}^1 \times \mathbb{A}^1$, plus the generic point. In particular, it has "diagonal" lines and conics but not horizontal or vertical lines. –  Justin Campbell Jun 19 '11 at 3:47
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