Smoothness of the source cannot be dropped in the extension of rational maps

Geometric integrality over every field extension. Let $K\mathrm{/}k$ be any field extension. In $K(x)$, the order valuation $v_{x+1}$ gives $v_{x+1}(x^2(x+1))=1$ because $x$ is a unit at the prime $(x+1)$ of $K[x]$; a square has even valuation, so $x^2(x+1)$ is not a square in $K(x)$. Since $\mathrm{char}K\mathrm{\ne }2$, the quadratic $y^2-x^2(x+1)$ has no root and is irreducible in $K(x)[y]$. It is monic, hence primitive, over the UFD $K[x]$, so Gauss's lemma makes it irreducible in $K[x,y]$ [F7]. The ring $K[x,y]$ is a UFD, so this irreducible polynomial is prime. Thus $K[x,y]\mathrm{/}(y^2-x^2(x+1))$ is a nonzero domain and its spectrum is integral Integral schemes. This holds for every extension $K\mathrm{/}k$, in particular for $\bar{k}\mathrm{/}k$, so $X$ is geometrically integral.

The point $o$ is closed since $A\mathrm{/}\mathfrak{m}=k$. Moreover $A\mathrm{/}(x)\cong k[y]\mathrm{/}(y^2)$ has the unique prime $(y)$, so $D(x)=X\setminus \{o\}$. The local ring at $o$ has dimension one: every open neighbourhood of the closed point $o$ contains the generic point of the integral curve $X$ and hence has chain dimension one, and [F8] identifies that local dimension with $\dim {\mathcal{O}}_{X,o}$. The maximal ideal modulo its square has basis given by the classes of $x,y$, since $y^2-x^2(x+1)\in (x,y{)}^2$. Therefore $\mathrm{edim}{\mathcal{O}}_{X,o}=2\mathrm{\ne }1=\dim {\mathcal{O}}_{X,o}$, so ${\mathcal{O}}_{X,o}$ is not regular and $X$ is not smooth at $o$ embedding dimension and regular local ring and [F9].

The principal open $U:=D(x)$ is smooth. Set $t=y\mathrm{/}x\in A_x$; the relation gives $x=t^2-1$ and $y=t(t^2-1)$, so $A_x\cong k[t,(t^2-1{)}^{-1}]\cong k[t,s]\mathrm{/}(s(t^2-1)-1)$. The derivative with respect to $s$ is the unit $t^2-1$, so this is standard smooth over $k$ by [F9]. Since $D(x)$ is the complement of the singular point $o$, the smooth locus of $X$ is exactly $U$.

The rational map. The regular function $y\mathrm{/}x$ defines a $k$-morphism ${\phi }_U:U\to {\mathbb{P}}^1$ into $U_0$ by [F4]. Let $\phi :X⇢{\mathbb{P}}^1$ be represented by $(U,{\phi }_U)$. The target is proper and separated over $k$ by [F5], so it satisfies the target hypotheses of [F2].

Curve and dimension. For $K=k$, the injection $k[x]\hookrightarrow A$ follows because a nonzero polynomial in $x$ cannot be divisible by the degree-two polynomial in $y$. Hence $x$ is transcendental over $k$ and $y$ is algebraic over $k(x)$, so ${\mathrm{trdeg}}_k\mathrm{Frac}(A)=1$. By [F8], $\dim A=1$ and $\mathrm{Spec}A$ has chain dimension one. The ring $A$ is finite type over $k$, its affine structure morphism is separated [F10], and it is nonempty and geometrically integral by step 1.1. Thus $X$ is a curve over $k$ [F1, F8].

The branch parameterization is a morphism. The ring map $A\to k[t]$, $x\mapsto t^2-1$, $y\mapsto t(t^2-1)$ is well defined because $t^2(t^2-1{)}^2-(t^2-1{)}^3-(t^2-1{)}^2=0$, and it induces $\nu$. Both $t=1$ and $t=-1$ map to $o$; on $t^2-1\mathrm{\ne }0$, ${\phi }_U\circ \nu$ is the chart-coordinate map $\tau :{\mathbf{A}}^1\to {\mathbb{P}}^1$ given by $\widehat{t}=t$.

Suppose a morphism $\psi :X\to {\mathbb{P}}^1$ extends $\phi$. By rational-map equivalence, $\psi {\mathrm{\mid }}_U$ and ${\phi }_U$ agree on a nonempty open of the integral scheme $U$, hence on a dense open. Since $U$ is reduced and ${\mathbb{P}}^1$ is separated, [F3] gives $\psi {\mathrm{\mid }}_U={\phi }_U$. Thus $\psi \circ \nu$ and $\tau$ agree on ${\mathbf{A}}^1\setminus \{1,-1\}$, a dense open of the reduced scheme ${\mathbf{A}}^1$; [F3] gives $\psi \circ \nu =\tau$ everywhere.

Evaluating at $1$ and $-1$ gives $\psi (o)=\tau (1)$ and $\psi (o)=\tau (-1)$. The chart-coordinate points $\widehat{t}=1$ and $\widehat{t}=-1$ are distinct because $\mathrm{char}k\mathrm{\ne }2$, a contradiction. Therefore $\phi$ has no extension.

Any two extensions of $\phi$ agree on the dense open $U$ of the reduced scheme $X$; since ${\mathbb{P}}^1$ is separated, [F3] makes them equal. This disproves existence while preserving conditional uniqueness. It is an integral singular curve with a proper target, so the smoothness hypothesis in [F6] cannot be dropped. Choice availability is accounted for here: AC [A1] supplies DC [A2], the premise used by [F6] for the smooth-source comparison; this availability note is not part of the explicit two-branch contradiction.