Normalizing the cuspidal plane curve

The substitution kills $P$: $\sigma (y{)}^2-\sigma (x{)}^3=t^6-t^6=0$, so by [F1] there is a unique $k$-algebra homomorphism $\phi \text{\hspace{0.17em}⁣}:A\to k[t]$ with $\phi (\bar{x})=t^2$ and $\phi (\bar{y})=t^3$.

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 $\phi (a+by)=a(t^2)+t^{3}b(t^2)$ separates into an even part $a(t^2)$ and an odd part $t^{3}b(t^2)$, so $\phi (a+by)=0$ forces both to vanish. If $c\in k[x]$ is nonzero then $c(t^2)\mathrm{\ne }0$, because the term of highest degree $2\mathrm{deg}c$ has coefficient equal to the leading coefficient of $c$; hence $a=0$, and then $t^{3}b(t^2)=0$ with $k[t]$ a domain gives $b(t^2)=0$, hence $b=0$. So $A$ is a domain isomorphic to $k[t^2,t^3]$. 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 $\bar{y}\mathrm{/}\bar{x}=t^3\mathrm{/}{t}^2=t$, so $k(t)\subseteq \mathrm{Frac}(A)$ and, since $A\subseteq k[t]$, equality $\mathrm{Frac}(A)=k(t)$ holds. Moreover $t^2=\bar{x}\in A$, so $t$ is integral over $A$, and every even power of $t$ belongs to $A$ as a power of $t^2$, and every odd power belongs to $At$; hence $k[t]=A+At$ is 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 therefore lies in $k[t]$ because $k[t]$ is integrally closed [F3]; conversely every element of $k[t]=A+At$ is integral over $A$. Hence the integral closure of $A$ in its fraction field is exactly $k[t]$. By [F4] the normalization of the cusp is $\nu \text{\hspace{0.17em}⁣}:{\mathbf{A}}^1\to X$ induced by $A\hookrightarrow k[t]$, namely $\nu (t)=(t^2,t^3)$, a finite birational map.

The map $\nu$ is bijective. If $u=0$ for a point $(u,v)$ of the cusp, then $v^2=u^3=0$, so $v=0$ and the unique parameter is $t=0$; if $u\mathrm{\ne }0$, then $t=v\mathrm{/}u$ satisfies $t^2=v^2\mathrm{/}{u}^2=u^3\mathrm{/}{u}^2=u$ and $t^3=v\cdot v^2\mathrm{/}{u}^3=v$, so $(u,v)$ is the image of the unique parameter $v\mathrm{/}u$. Hence every point of $X$ has exactly one preimage.

The pullback $\phi \text{\hspace{0.17em}⁣}:k[t^2,t^3]\hookrightarrow k[t]$ is not surjective: its image consists of sums of monomials $t^n$ with $n\in ⟨2,3⟩=\{0,2,3,4,\dots \}$, and $t$ has exponent $1$, so $t$ is not in the image. By the anti-equivalence [F4] the map $\nu$ is not an isomorphism, although it is bijective. Its fibre over the cusp is the singleton $\{0\}$, because $t^2=0$ forces $t=0$ in the field $k$; hence the cusp is unibranch [F6].

The conductor is the maximal ideal $\mathfrak{m}=(t^2,t^3)$ of $A$: indeed $t^{2}k[t]\subseteq A$ and $t^{3}k[t]\subseteq A$ because every exponent $\ge 2$ lies in $⟨2,3⟩$, so $\mathfrak{m}\subseteq \mathfrak{c}$; conversely an $a\in A$ outside $\mathfrak{m}$ has nonzero constant coefficient $c$. All its other monomials have exponent at least two, so $at$ has a nonzero coefficient $c$ at exponent one and cannot belong to $A$, so $\mathfrak{c}\subseteq \mathfrak{m}$. By [F5] $V(\mathfrak{c})$ is the non-normal locus, and $V(t^2,t^3)$ is the single point $(0,0)$; hence the cusp is the unique non-normal point of $X$.

Summing up: the normalization of the cusp is $\nu \text{\hspace{0.17em}⁣}:{\mathbf{A}}^1\to X$, $t\mapsto (t^2,t^3)$, a finite birational bijection which is not an isomorphism; the cusp is its unique non-normal point, with singleton fibre $\{0\}$, so it is unibranch but not normal; and the conductor of the extension is the maximal ideal $(t^2,t^3)$ of $k[t^2,t^3]$.