= Solution
Use the <ring homomorphism> $k[t]\to k[x]$ given by $t\mapsto x^2$. The source <scheme> $X=\operatorname{Spec}k[x]$ is an <integral scheme>. At the rational point $y=(t)$, its <residue field> is $k$, and its <scheme-theoretic fibre> is
$$
\boxed{X_y=\operatorname{Spec}(k[x]\otimes_{k[t]}k)
=\operatorname{Spec}(k[x]/(x^2)).}
$$
The class of $x$ is nonzero but has square zero, so this is not a <reduced scheme>. This example works over every field. An integral total space therefore need not have reduced fibres.
Back to article page
