A flat family with a nodal special fibre is not smooth at the node

We first identify the total ring. Let $\phi :k[t,x,y]\to k[x,y]$ be the $k$-algebra map with $\phi (t)=h:=y^2-x^2(x+1)$, $\phi (x)=x$, $\phi (y)=y$. Since $F=y^2-x^2(x+1)-t=h-t$ has unit leading coefficient $-1$ and degree one as a polynomial in $t$ over $k[x,y]$, division of any $p\in k[t,x,y]$ by $F$ gives $p=q\cdot F+r$ with $r\in k[x,y]$ and $\phi (p)=r$; hence $\ker \phi =(F)$ and $\phi$ induces an isomorphism $B\cong k[x,y]$ under which the class of $t$ is $h$. For any nonzero $p(t)\in k[t]$ of degree $d$ with leading coefficient $c_d$, the polynomial $p(h)$ has degree $2d$ in $y$ and leading coefficient $c_d$, so $p(h)\mathrm{\ne }0$. Thus $B$ is a domain and $k[t]\to B$ is injective; also $h\in (x,y{)}^2$.

The special fibre. Under the isomorphism of step 1.1 the ideal $tB$ corresponds to $hk[x,y]$, so by [F9] the fibre over the prime $(t)$ of $k[t]$ is $B{\otimes }_{k[t]}k\cong B\mathrm{/}tB\cong k[x,y]\mathrm{/}(h)$. The prime $\mathfrak{q}=(t,x,y)\subseteq B$ corresponds to the maximal ideal $(x,y)\subseteq k[x,y]$ (as $h\in (x,y)$), and its image in the fibre is $\mathfrak{m}=(x,y)A$, where $A:=k[x,y]\mathrm{/}(h)$.

Flatness. The ring $k[t]$ is a principal ideal domain by [F3]. If $p\in k[t]$ is nonzero and $b\in B$ satisfies $p\cdot b=0$, then under the identification $B=k[x,y]$ of step 1.1 this reads $p(h)b=0$ in the domain $k[x,y]$, so $b=0$ because $p(h)\mathrm{\ne }0$ (step 1.1); hence $B$ is torsion-free over $k[t]$. By [F4] $B$ is flat over $k[t]$, so $f$ is flat at every point by the affine-local criterion [F2], which rests on the germwise definition of flatness [F1].

Finite presentation. The $k[t]$-algebra $k[t,x,y]$ is a polynomial algebra, hence finitely presented by [F10], and $B$ is its quotient by the ideal generated by the single element $F$, hence $B$ is a finitely presented $k[t]$-algebra; therefore $f$ is locally of finite presentation by [F10].

We show that $h$ is irreducible in $k[x,y]$, so that $(h)$ is a prime of $k[x,y]$ contained in $(x,y)$. Suppose $h=pq$ with $p,q\in k[x,y]$ nonunits; view $p,q$ as polynomials in $y$ over $k[x]$. Since the coefficient of $y^2$ in $h$ is $1$, the $y$-degrees of $p$ and $q$ add to two and their leading coefficients multiply to $1$, hence are units of $k[x]$, i.e. nonzero constants. If one factor had $y$-degree zero it would be a nonunit of $k[x]$ contributing that nonunit to the leading coefficient of the other factor, impossible; so after absorbing constants $h=(y-\alpha )(y-\beta )$ with $\alpha ,\beta \in k[x]$. Then $\alpha +\beta =0$ and $\alpha \beta =-x^2(x+1)$, so ${\beta }^2=x^2(x+1)$; but ${\beta }^2$ has even degree in $x$ while $x^2(x+1)$ has degree three, a contradiction. Hence $h$ is irreducible, so $(h)$ is a prime of $k[x,y]$ by [F13], and it lies in $(x,y)$ because $h\in (x,y{)}^2$. Its degree-two initial form at the origin is $y^2-x^2=(y-x)(y+x)$; the factors are distinct because $\mathrm{char}k\mathrm{\ne }2$, so the origin is an ordinary node by [F14].

Cotangent and dimension of the special fibre at the origin. Put $S:=k[x,y{]}_{(x,y)}$ and $A_{\mathfrak{m}}=S\mathrm{/}(h)S$ (step 2.1). The ambient local ring $S$ has cotangent space with $k$-basis the classes of $x,y$: every element of $S$ is $f\mathrm{/}g$ with $g(0)\mathrm{\ne }0$, so it is congruent modulo $(x,y)S$ to the constant term $f(0)\mathrm{/}g(0)$, and congruent modulo $(x,y{)}^{2}S$ to $f(0)\mathrm{/}g(0)$ plus the linear part of $f\mathrm{/}g$. Hence $\mathrm{edim}S=2$ and by [F5], $\dim S\le 2$; the chain $(0)⊊(x)S⊊(x,y)S$ shows $\dim S\ge 2$, so $\dim S=2$.

In $A=k[x,y]\mathrm{/}(h)$ we have $h=y^2-x^2(x+1)\in {\mathfrak{m}}^2$; hence $(h)+(x,y{)}^2=(x,y{)}^2$ and $A\mathrm{/}(x,y{)}^2\cong k[x,y]\mathrm{/}(x,y{)}^2$, so $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ has $k$-basis the classes of $x,y$ and $\mathrm{edim}{A}_{\mathfrak{m}}=2$ by [F6].

Dimension of the fibre at the origin. The chain $(0)⊊\mathfrak{m}{A}_{\mathfrak{m}}$ shows $\dim A_{\mathfrak{m}}\ge 1$, since $A_{\mathfrak{m}}$ is a domain by step 2.4 and $x$ is a nonzero element of its maximal ideal. If $\dim A_{\mathfrak{m}}\ge 2$, then, since $A_{\mathfrak{m}}$ is a local domain by step 2.4, there is a strict chain $(0)⊊{\bar{P}}_0⊊{\bar{P}}_1$ in $A_{\mathfrak{m}}$. Lifting to $k[x,y]$ gives primes $(h)⊊P_0⊊P_1\subseteq (x,y)$. Their localizations yield the strict chain $(0)⊊(h)S⊊(P_0)S⊊(P_1)S$ in $S=k[x,y{]}_{(x,y)}$, of length three, contradicting $\dim S=2$ from step 3.1. Hence $\dim A_{\mathfrak{m}}=1$.

Non-regularity and failure of smoothness. By steps 3.2 and 4.1 the local ring $A_{\mathfrak{m}}$ is a nonzero Noetherian local ring with $\dim A_{\mathfrak{m}}=1\mathrm{\ne }2=\mathrm{edim}{A}_{\mathfrak{m}}$, so it is not regular by [F6]. If the fibre were geometrically regular at the image of $\mathfrak{q}$, then by the case $K=\kappa (\mathfrak{p})=k$ of [F7] the localisation $A_{\mathfrak{m}}$ would be regular; it is not, so the fibre is not geometrically regular at the image of $\mathfrak{q}$. By [F8] the morphism $f$ is not smooth at $\mathfrak{q}$, even though by step 2.2 it is flat at $\mathfrak{q}$ and by step 2.3 locally of finite presentation there. Taking $K=k$ in the quantifier of [F7] is legitimate because $k$ is a field extension of $\kappa ((t))=k$; no further choice is made. The Axiom of Choice [F12] is assumed in the Statement and is used exactly through the bound [F5] in step 3.1. [F5, F7, F8, F12, step 2.2, step 3.2, step 4.1] $\mathrm{\square }$