Normalizing the nodal plane curve

The substitution kills $P$: $\sigma (y{)}^2-\sigma (x{)}^2(\sigma (x)+1)=t^2(t^2-1{)}^2-(t^2-1{)}^2{t}^2=0$, because $\sigma (x)+1=t^2$. By [F1] there is a unique $k$-algebra homomorphism $\phi \text{\hspace{0.17em}⁣}:A\to k[t]$ with $\phi (\bar{x})=t^2-1$ and $\phi (\bar{y})=t(t^2-1)$.

The map $\phi$ is injective. By [F2] every element of $A$ is uniquely $a(x)+b(x)y$ with $a,b\in k[x]$, and $\phi (a+by)=a(t^2-1)+t(t^2-1)b(t^2-1)$ separates into an even part $a(t^2-1)$ and an odd part $t(t^2-1)b(t^2-1)$, so $\phi (a+by)=0$ forces both parts to vanish. If $c\in k[x]$ is nonzero then $c(t^2-1)\mathrm{\ne }0$: the term of highest degree $2\mathrm{deg}c$ has coefficient equal to the leading coefficient of $c$ by [F2]. Hence $a=0$, and since $k[t]$ is a domain and $t^2-1\mathrm{\ne }0$ we get $b(t^2-1)=0$, hence $b=0$. So $\phi$ is injective and $A$ is a domain, isomorphic to $k[t^2-1,t(t^2-1)]$. If a polynomial $Q(x,y)$ vanishes on $X$, its substitution vanishes at every $t\in k$ under the displayed parametrization. The field $k$ is infinite, so this substituted polynomial is zero. The kernel just computed is $(P)$, hence $I(X)=(P)$, justifying the coordinate-ring identification in the Given.

In $\mathrm{Frac}(A)$ one has $\bar{y}\mathrm{/}\bar{x}=(t^3-t)\mathrm{/}(t^2-1)=t$, so $k(t)\subseteq \mathrm{Frac}(A)$, and $A\subseteq k[t]$ gives the reverse inclusion; hence $\mathrm{Frac}(A)=k(t)$. Moreover $t^2=\bar{x}+1\in A$, so $t$ is integral over $A$, and since $t^n=t^{n-2}(\bar{x}+1)$ for $n\ge 2$ we get $k[t]=A+At$, a finite $A$-module.

If $z\in k(t)=\mathrm{Frac}(A)$ is integral over $A$, then a monic equation for $z$ over $A$ has coefficients in $k[t]$, so $z$ is integral over $k[t]$ and hence lies in $k[t]$ because $k[t]$ is integrally closed [F3]. Conversely every element of $k[t]=A+At$ is integral over $A$ by step 3.1. Therefore the integral closure of $A$ in $\mathrm{Frac}(A)$ is exactly $k[t]$.

By [F4] the normalization of $X$ is the affine variety with coordinate ring $k[t]$, namely ${\mathbf{A}}^1$, with the structure morphism $\nu \text{\hspace{0.17em}⁣}:{\mathbf{A}}^1\to X$ induced by $A\hookrightarrow k[t]$; by [F4] applied to the parametrization this is $\nu (t)=(t^2-1,t(t^2-1))$, a finite birational map, and it is surjective.

The fibre of $\nu$ over the node $(0,0)$ consists of the parameters with $t^2-1=0$, namely $t=1$ and $t=-1$; these are distinct because $\mathrm{char}k\mathrm{\ne }2$, and both map to the origin because $t(t^2-1)=0$. Hence the node has a two-point normalization fibre, so it is not unibranch [F5]; since normal points are unibranch [F5], the node is not normal.

Summing up, the normalization of the nodal curve is the finite birational surjection $\nu \text{\hspace{0.17em}⁣}:{\mathbf{A}}^1\to X$, $t\mapsto (t^2-1,t(t^2-1))$, whose fibre over the node is the two-point set $\{1,-1\}$; the node is therefore not unibranch and not normal.