 Solution

Compute each scheme-theoretic fibre by tensoring with the residue field of the chosen base point.

For the first map, put $L=\kappa (\mathfrak{p})$ for a point $\mathfrak{p}$ of $\mathrm{Spec}k[T]$ and let $t$ be the image of $T$ in $L$. ThenIf the characteristic of a field is not two and $t\ne 0$, the two factors $U-t$ and $U+t$ are coprime, so the Chinese remainder theorem gives $L[U]/(U^2-t^2)\simeq L\times L$: the scheme-theoretic fibre is two distinct $L$-points. If $t=0$, its ring is $L[U]/(U^2)$, a dual number ring, so it is a nonreduced double point. In characteristic two, $U^2-t^2=(U-t{)}^2$ at every point, giving a nonreduced double point in every scheme-theoretic fibre. This covers the generic point, where $L=k(T)$, as well as closed points defined by irreducible polynomials.

For the arithmetic map, the generic scheme-theoretic fibre isOver a closed point $(p)$ it is $\mathrm{Spec}{\mathbb{F}}_p[T]/(T^2+1)$. At $p=2$ this is $\mathrm{Spec}{\mathbb{F}}_2[ϵ]/({ϵ}^2)$, since $T^2+1=(T+1{)}^2$. For odd $p$, the finite field multiplicative group is cyclic, and $-1$ is a square exactly when $p\equiv 1(\mathrm{mod}4)$. ThusIn the second case there is one degree-two closed point over ${\mathbb{F}}_p$, which becomes two points after extending the residue field to an algebraic closure. The case $p=2$ remains nonreduced after such extension.

For $\mathrm{Spec}\mathbb{C}\to \mathrm{Spec}\mathbb{Z}$, the unique source point maps to the generic point $(0)$. Since all nonzero integers are invertible in $\mathbb{C}$,Indeed $\mathbb{C}{\otimes }_{\mathbb{Z}}\mathbb{Q}=\mathbb{C}$, whereas $\mathbb{C}{\otimes }_{\mathbb{Z}}{\mathbb{F}}_p=\mathbb{C}/p\mathbb{C}=0$. The distinction between a reduced split scheme-theoretic fibre and a nonreduced double point is essential in the first two examples.