A smooth Weierstrass surface with a type II cusp fiber

The weighted projective bundle is projective over $B$, and the hypersurface is closed in it, so $\pi$ is projective and therefore proper. The coordinate $z=0$ defines the distinguished section. Since $g=u^{12}-v^{12}$ is a degree-twelve section of $\mathcal{O}(12)=L^6$, the equation is globally well-defined in the line-bundle twists of [F1]. The polynomial $u^{12}-v^{12}$ has twelve distinct roots on ${\mathbb{P}}^1$, none at $[1:0]$, because its affine roots are the twelve distinct twelfth roots of unity and its derivative $12u^{11}$ is nonzero at each.

On the chart $z\mathrm{\ne }0$, use local coordinates $x=X\mathrm{/}{z}^2$, $y=Y\mathrm{/}{z}^3$ and a base parameter $t$. The equation is $F=y^2-x^3-g(t)=0$. A singular point of the total hypersurface would satisfy $F=0$, $2y=0$, $-3x^2=0$, and $-g^{\mathrm{\prime }}(t)=0$, hence $x=y=0$, $g(t)=g^{\mathrm{\prime }}(t)=0$, contradicting the simple-root property in step 1.1. At $z=0$ the equation forces $x,y\mathrm{\ne }0$; the stabilizer is trivial since ${\lambda }^2={\lambda }^3=1$ implies $\lambda =1$, and the derivative in the $y$ direction is $2y\mathrm{\ne }0$. Thus the total hypersurface is smooth by [F2].

By [F6], each fiber over $t\in B$ is isomorphic to the plane cubic $Y^{2}Z=X^3+g(t)Z^3$. For $a=g(t)\mathrm{\ne }0$, the partial derivatives $-3X^2,2YZ,Y^2-3a{Z}^2$ have no common zero on this cubic, so it is smooth. It is geometrically integral by a direct factorization test. Over the algebraic closure $K=\bar{\mathbb{C}}$, the polynomial $X^3+a$ has three distinct roots because $a\mathrm{\ne }0$ and its derivative $3X^2$ has no common root with it. If a reduced fraction $r=p\mathrm{/}q\in K(X)$ had $r^2=X^3+a$, then at any root $\alpha$ the multiplicity of $X-\alpha$ in $r^2$ would be $2({\mathrm{ord}}_{\alpha }p-{\mathrm{ord}}_{\alpha }q)$, an even integer, whereas its multiplicity in $X^3+a$ is one. Thus $X^3+a$ is not a square, so the quadratic $Y^2-(X^3+a)$ is irreducible in $K(X)[Y]$. A factorization in $K[X,Y]$ would give one there; a factor depending only on $X$ would divide the leading coefficient $1$, so no such nonunit factor exists. A homogeneous factor of its projective closure that dehomogenizes to a constant would be a power of $Z$, but $Z$ does not divide $Y^{2}Z-X^3-a{Z}^3$. Thus there is no extra component at infinity, and the plane cubic is geometrically integral. It is genus one by [F4]. For $a=0$, the affine equation $Y^2=X^3$ is irreducible because $X^3$ is not a square in $\mathbb{C}(X)$; its only singular point is $(0,0)$, with local equation $Y^2-X^3=0$, an ordinary cusp, and the point at infinity is smooth. Thus exactly the twelve simple zeros in step 1.1 give integral cuspidal fibers.

The smooth projective total surface $Y$ is connected. By [F7], each connected component is an irreducible projective surface. Properness makes its image in $B$ closed; irreducibility of the component makes that image irreducible, so it is either a point or all of $B$. It cannot be a point because a surface cannot lie in a one-dimensional fiber. Thus every component surjects onto $B$. If there were at least two components, every fiber would be a disjoint union of nonempty closed intersections with these components, contradicting the connectedness of the integral fibers from step 2.2.

The morphism is flat. At a point $y$ over a closed point $t_0\in B$, let $R={\mathcal{O}}_{B,t_0}$, a discrete valuation ring with parameter $s=t-t_0$, let $S$ be the corresponding Noetherian local ring of the smooth ambient space, and let $M={\mathcal{O}}_{Y,y}$, a finite $S$-module. The fiber module $M\mathrm{/}sM$ is a vector space over $R\mathrm{/}(s)$ and is flat. The total space is smooth, so $M$ is a regular local domain. By step 3.1, $Y$ is integral and $\pi$ is dominant; hence the pullback of $s$ is nonzero in the function field of $Y$ and remains nonzero in every local ring $M$. Therefore multiplication by $s$ is injective, equivalently $(s){\otimes }_{R}M\to M$ is injective. Applying [F3] with $I=(s)$ proves $M$ flat over $R$ at every such point. Over the generic point of $B$, the base local ring is a field, so flatness is automatic. Thus $\pi$ is flat.

The section and flat proper family make this a genus-one fibration with section: over $K=\overline{\mathbb{C}(B)}$, [F6] identifies the geometric generic fiber with the plane cubic $Y^{2}Z=X^3+g{Z}^3$. Here $g\mathrm{\ne }0$. The derivative and nonsquare arguments of step 2.2 apply over this algebraically closed characteristic-zero field: $X^3+g$ has distinct roots, and the odd order at each root prevents it from being a square in $K(X)$. Thus the geometric generic fiber is smooth, proper and geometrically integral; [F4], applied over $K$, gives ${\dim }_K{H}^1=1$. Thus the definitions Flat genus-one fibration and Elliptic fibration with a section apply. No fiber is multiple: each smooth fiber is integral and each singular fiber is an irreducible reduced cusp, so its fiber divisor has multiplicity one. There is no reducible fiber component, hence no smooth rational vertical $(-1)$-curve; the surface is relatively minimal. At a simple zero $t_0$ of $g$, the analytic coordinate $s=g(t)$ is valid because $g^{\mathrm{\prime }}(t_0)\mathrm{\ne }0$, and the transverse local equation is exactly $y^2=x^3+s$. The fiber at $s=0$ is the irreducible cusp from step 2.2, so [F5] identifies it as Kodaira type II. This also confirms the standard valuation row $(\mathrm{ord}f,\mathrm{ord}g,\mathrm{ord}\mathrm{\Delta })=(\ge 1,1,2)$ for $f=0$, $g=s$, and $\mathrm{\Delta }=27s^2$.