TITLE: Geometrically reduced variety
QUESTION [5 upvotes]: What is a geometrically reduced variety (or geometrically reduced algebraic set if you will, a variety has not been assumed to be irreducible in this definition)? I tried looking up on the internet as well as a few books like Atiyah-MacDonald, Eisenbud and Hartshorne but to no avail. The last one has an exercise for schemes which I have no idea about. I just need the definition (along with some examples and counterexamples, if possible) for my simple setting of varieties. Thank you :-)

REPLY [13 votes]: A scheme $X$ over a field $k$ is said to be geometrically reduced if after an arbitrary field extension $k\to K$ the base changed scheme $X_K=X\times_kK$ is reduced.
 Of course by taking $K=k$ we see that $X=X\times_k k$ itself must then be reduced!
Also, it suffices to check  for just one algebraically closed extension $k\to \Omega$ that $X_\Omega$ is reduced to conclude that $X$ is geometrically reduced : this  is a not completely  trivial result.
The notion "geometrically reduced"  remains valid if you restrict it to some subcategory of schemes, like varieties.
The definition of variety however is unfortunately not quite standardized: it is a variation on " a reduced and irreducible finite-type $k$-scheme".
One must of course realize, for this to be interesting, that the properties  reduced (and, since we are on the subject, irreducible)  are not necessarily preseved by base-change:   
$\bullet$  The affine (irreducible) variety $V(x^2+y^2)\subset \mathbb A^2_\mathbb R=\operatorname {Spec }(\mathbb R[x,y])$ becomes reducible (but remains reduced) after the base change $\mathbb R \to \mathbb C$ since $x^2+y^2=(x+iy)(x-iy)$ over $\mathbb C$.
$\bullet \bullet$ More relevantly for your question, the affine  variety  $X=V(x^2+t)\subset \mathbb A^1_k=\operatorname {Spec } k[x]$ over the rational function field $k=\mathbb F_2(t)$ is reduced and irreducible but not geometrically reduced:
 Indeed for the extension $k=\mathbb F_2( t)\to K=\mathbb F_2(\sqrt t) $ we have $$X_K=V(x^2+t)=V((x+\sqrt t)^2)\subset  \mathbb A^1_K$$ so that $X_K$ is a non-reduced (but still irreducible) double point embedded in the affine line over $K$. 
And to end on a reassuring note, let me mention that over a perfect field  $k$ (for example any field of characteristic zero, every algebraically closed field, any finite field, ...) a scheme is geometrically reduced as soon as it is just reduced.
 This is why in my  example above I had to take  the slightly unusual base field $k=\mathbb F_2(t)$ .