An elliptic curve with good reduction and an elliptic curve with bad reduction

The constant family is an abelian scheme: $E_0$ is a smooth projective curve of genus one with a rational point, base change preserves smoothness, properness and the group law (Base change and products of abelian schemes), and its special fibre is $E_0$; this proves the good-reduction clause.

Suppose $E_t$ had an abelian scheme model $A\to \mathrm{Spec}R$. By [F1] multiplication by $2$ on $A$ is finite etale of rank $4$, so $A[2]$ is finite etale over $R$; [F1] then makes reduction $A[2](R)\to A[2](k)$ a bijection, and $A[2](k)=\mathbb{Z}\mathrm{/}2\times \mathbb{Z}\mathrm{/}2$ because the special fibre is an elliptic curve in characteristic not $2$. Hence $A[2](K)=E_t[2](K)$ would have four elements, while the two-torsion of $E_t$ is $\{O\}$ together with the roots in $K$ of $x^3+t=0$ by [F2]; the polynomial $x^3+t$ has no root in $K=k(\text{\hspace{0.17em}⁣}(t)\text{\hspace{0.17em}⁣})$, since a root would satisfy $3v(x)=v(t)=1$, impossible for an integer valuation. This contradiction shows that no abelian scheme model exists, so $E_t$ has bad reduction.

The contrast is sharp: over the totally ramified extension $k(\text{\hspace{0.17em}⁣}(s)\text{\hspace{0.17em}⁣})$ with $s^6=t$, the substitution $x=s^2{x}^{\mathrm{\prime }}$, $y=s^3{y}^{\mathrm{\prime }}$ identifies $E_t{\otimes }_{k(\text{\hspace{0.17em}⁣}(t)\text{\hspace{0.17em}⁣})}k(\text{\hspace{0.17em}⁣}(s)\text{\hspace{0.17em}⁣})$ with $Y^{2}Z=X^3+Z^3$, the base change from $k$ of an elliptic curve of good reduction, so $E_t$ has bad reduction over $k[\text{\hspace{0.17em}⁣}[t]\text{\hspace{0.17em}⁣}]$ but acquires good reduction after a finite ramified extension. In step 2.1 the properness of a hypothetical model is what extends each $K$-point of $A[2]$ uniquely to $R$; no claim that the singular chosen equation alone excludes a different smooth model is made.