Geometrically reduced variety 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 :-)
 A: 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 $k=\mathbb F_2(t)$ .
A: A word of caution: it is very atypical to allow the definition of a variety to be nonreduced. The following is nonstandard.
The condition that a variety is reduced is local, i.e. can be investigated on an open cover. For a projective variety $V \subset \mathbb{P}_\mathbb{Q}^n$, this means we can dehomogenize the defining equations giving an affine variety $X \subset \mathbb{A}_\mathbb{Q}^n$. This is given by some finite collection $f_1, \dots, f_r$ of polynomials in $n$ variables, not necessarily homogeneous. The coordinate ring is then
 $$R = \mathbb{Q}[x_1, \dots, x_n]/(f_1, \dots, f_r).$$
We say that the variety $X$ is reduced if for any $r \in R$, $r^m = 0$ for some $m \geq 1$ implies that $r = 0$. In other words, $R$ has no nonzero nilpotents. We say that $V$ is reduced if, for all the standard affine charts, each affine variety is reduced.
An example of a nonreduced variety would be taking $r = 1$ and $f_1 = (x-1)^2$.
Any time you see "geometric" in algebraic geometry, it usually means that the condition must hold over an algebraically closed field. In this case, we consider $R \otimes_\mathbb{Q} \mathbb{C}$, which is
 $$\mathbb{C}[x_1, \dots, x_n]/(f_1, \dots, f_r).$$
To say that $X$ is geometrically reduced means that $R \otimes_\mathbb{Q} \mathbb{C}$ has no nonzero nilpotents.
In characteristic zero (or more generally, over a perfect field), reduced implies geometrically reduced. However, if the base field is not perfect, then this can certainly fail. Since the source you reference is over $\mathbb{Q}$, I do not believe this will concern you.
