= 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 $\operatorname{Spec}k[T]$ and let $t$ be the image of $T$ in $L$. Then
$$
\boxed{X_{\mathfrak p}=\operatorname{Spec}L[U]/(U^2-t^2).}
$$
If the <characteristic of a field> is not two and $t\ne0$, 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> is
$$
\boxed{\operatorname{Spec}\mathbb Q[T]/(T^2+1)=\operatorname{Spec}\mathbb Q(i).}
$$
Over a <closed point> $(p)$ it is $\operatorname{Spec}\mathbb F_p[T]/(T^2+1)$. At $p=2$ this is $\operatorname{Spec}\mathbb F_2[\epsilon]/(\epsilon^2)$, since $T^2+1=(T+1)^2$. For odd $p$, the <finite field> multiplicative group is cyclic, and $-1$ is a square exactly when $p\equiv1\pmod4$. Thus
$$
\boxed{X_{(p)}\simeq
\begin{cases}
\operatorname{Spec}(\mathbb F_p\times\mathbb F_p),&p\equiv1\pmod4,\\
\operatorname{Spec}\mathbb F_{p^2},&p\equiv3\pmod4,\\
\operatorname{Spec}\mathbb F_2[\epsilon]/(\epsilon^2),&p=2.
\end{cases}}
$$
In 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 $\operatorname{Spec}\mathbb C\to\operatorname{Spec}\mathbb Z$, the unique source point maps to the <generic point> $(0)$. Since all nonzero integers are invertible in $\mathbb C$,
$$
\boxed{X_{(0)}=\operatorname{Spec}\mathbb C,\qquad X_{(p)}=\varnothing\text{ for every prime }p.}
$$
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.
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