The family xy=t

The total ring. Let $\phi :k[x,y,t]\to k[x,y]$ be $t\mapsto xy$, $x\mapsto x$, $y\mapsto y$. The defining polynomial $F=xy-t$ is monic of degree one in $t$ over $k[x,y]$, so division by $F$ writes any $p$ as $qF+r$ with $r\in k[x,y]$ and $\phi (p)=r$; hence $\ker \phi =(F)$ and $\phi$ induces an isomorphism $C:=k[x,y,t]\mathrm{/}(xy-t)\cong k[x,y]$, sending the class of $t$ to $xy$. In particular $C$ is a domain, the map $k[t]\to C$ is injective (as $xy\mathrm{\ne }0$ is not algebraic over $k$), and $xy\in (x,y{)}^2$.

Flatness and finite presentation. If $0\mathrm{\ne }p\in k[t]$ and $c\in C$ satisfy $p\cdot c=0$, then, under the identification $C=k[x,y]$ of step 1.1, $p(xy)c=0$ in the domain $k[x,y]$, so $c=0$ because $p(xy)\mathrm{\ne }0$. Thus $C$ is torsion-free over the principal ideal domain $k[t]$, hence flat over $k[t]$ by [F2]; the single affine chart $\mathrm{Spec}C\to \mathrm{Spec}k[t]$ then gives flatness of $f$ by [F1]. Moreover $k[t,x,y]$ is a finitely presented $k[t]$-algebra and $C$ is its quotient by the principal ideal $(xy-t)$, so $C$ is finitely presented over $k[t]$ and $f$ is locally of finite presentation by [F8].

The special fibre. By [F7] the fibre of $f$ over the prime $(t)$ is $\mathrm{Spec}(C{\otimes }_{k[t]}k)=\mathrm{Spec}k[x,y]\mathrm{/}(xy)$. The prime ${\mathfrak{q}}_0=(t,x,y)\subseteq C$ lies over $(t)$ and corresponds to the maximal ideal $\mathfrak{m}=(x,y)k[x,y]\mathrm{/}(xy)$ of the fibre; write $A=k[x,y]\mathrm{/}(xy)$ and $R=A_{\mathfrak{m}}$.

The fibre local ring at the origin is not regular. The ring $R$ is a Noetherian local ring by [F9]. In $A$ the elements $x,y$ are nonzero (their classes are not in the ideal $(xy)$) and satisfy $x\mathrm{\ne }0$, $y\mathrm{\ne }0$, $xy=0$; the same holds in the localisation $R$, so $R$ has zero divisors and is not a domain. By [F6] a regular local ring is a domain, so $R$ is not regular.

Smoothness away from the origin. Let $\mathfrak{q}\mathrm{\ne }{\mathfrak{q}}_0$ be a point of $\mathrm{Spec}C$. If both $x$ and $y$ belonged to $\mathfrak{q}$, then $t=xy\in \mathfrak{q}$, so $\mathfrak{q}\supseteq (t,x,y)$; since $(t,x,y)$ is a maximal ideal of $C$, this forces $\mathfrak{q}={\mathfrak{q}}_0$. Hence $x\mathrm{\notin }\mathfrak{q}$ or $y\mathrm{\notin }\mathfrak{q}$; put $h:=x$ in the first case and $h:=y$ in the second, so that $h\in C\setminus \mathfrak{q}$ and the image of $h$ in $C_h$ is a unit. In the affine chart $C=k[t][x,y]\mathrm{/}(xy-t)$ over $A=k[t]$ the Jacobian of the single equation $xy-t$ with respect to $(x,y)$ is the row $(y,x)$, whose $1\times 1$ minors are $y$ and $x$; the minor $h$ (namely $x$ or $y$) is a unit of $C_h$, and $C_h=(A[x,y]\mathrm{/}(xy-t){)}_h$ is a localisation of the displayed presentation. Since $f$ is locally of finite presentation by step 2.1, the criterion [F3] applies and yields that $f$ is smooth at $\mathfrak{q}$.

Failure of smoothness at the origin. If the fibre were geometrically regular at the image of ${\mathfrak{q}}_0$, then by the case $K=k=\kappa ((t))$ of [F5] the localisation $R$ would be regular; step 3.1 shows it is not. Since $f$ is locally of finite presentation and flat (step 2.1), [F4] implies that $f$ is not smooth at ${\mathfrak{q}}_0$.

Conclusion. Steps 4.1 and 3.2 show that $f$ is smooth at every point of $\mathrm{Spec}C$ except the origin prime ${\mathfrak{q}}_0$, where it fails to be smooth although it is flat. The Axiom of Choice [F10] is assumed in the Statement and is used exactly through the Jacobian criterion [F3] in step 3.2 and the domain theorem [F6] in step 3.1. [F3, F6, F10, step 4.1, step 3.2] $\mathrm{\square }$