base change of a reduced scheme over $\mathbb{Z}$ I'm reading through the stacks project and came across a lemma along these lines: Let X be a scheme over a perfect field k. Then, $X$ is reduced implies $X$ is geometrically reduced. 
here is my question: does this lemma extend to schemes over rings of char. 0? for instance, if $X$ is a reduced scheme over $\mathbb{Z}$ and $k$ is a field containing $\mathbb{Z}$, is it true that that $X_{k}=X\times_\mathbb{Z} k$ is a reduced scheme over $k$? A proof (or reference) would be great.

By the theorem you cite we are immediately reduced to the case $k = \mathbb{Q}$. (Note that if you don't insist that $k$ has characteristic zero the result is false, e.g. consider $\mathbb{Z}[x]/(x^p-1)$. )
Now, let $A$ be an abelian group. Since $\mathbb{Q}$ is flat over $\mathbb{Z}$, an element $f \in A$ maps to $0$ in $A \otimes \mathbb{Q}$ if and only if $f$ is a torsion element.
If $A$ is a $\mathbb{Z}$-algebra (i.e. a ring), $f \in A$, and $f^n = 0 \in A \otimes \mathbb{Q}$, then $f^n$ is torsion in $A$. Suppose $mf^n = 0$. Then $m^nf^n = 0$, so $(mf)^n = 0$. If $A$ is reduced, we conclude that $mf = 0$ in $A$, i.e. $f$ is torsion in $A$, i.e. $f = 0 \in A \otimes \mathbb{Q}$, and thus $A \otimes \mathbb{Q}$ is also reduced.
