Pulling a divisor back along the cusp normalization

$A$ is a subring of the domain $k[t]$ by [F1], so $A$ is a domain, and $x=t^2\mathrm{\ne }0$; $A$ is a finitely generated $k$-algebra by construction and is Noetherian by [F3]; $k[t]$ is Noetherian by [F2], and the inclusion $A\hookrightarrow k[t]$ induces the morphism $\nu :\mathrm{Spec}k[t]\to C$ with ${\nu }^{\mathrm{\#}}(x)=t^2$ and ${\nu }^{\mathrm{\#}}(y)=t^3$ by [F4], where $C=\mathrm{Spec}A$.

Since $A$ is a subring of the domain $k[t]$ by [F1] and $0\mathrm{\ne }x=t^2\in A$, multiplication by $x$ on $A$ and on every localisation of $A$ is injective, so $x$ is a regular section in the sense of [F6]; hence $D=V(x)$ is the effective Cartier divisor on $C$ with ideal sheaf $x{\mathcal{O}}_C$ and local equation $x$ on the whole of $C$, in particular a Cartier divisor by [F5].

The polynomial ring $k[t]$ is an integrally closed domain by [F7]; every prime localisation $k[t{]}_{\mathfrak{p}}$ is therefore integrally closed by [F11]; since $k[t]$ is Noetherian by [F2] and of dimension $\dim k[t]=1$ by [F9], the affine line $\mathrm{Spec}k[t]$ is a normal Noetherian integral scheme of dimension one in the sense of [F10].

As $A$ is a domain and Noetherian, $C=\mathrm{Spec}A$ is an integral Noetherian scheme by [F10].

The element $t^2={\nu }^{\mathrm{\#}}(x)$ is a nonzero element of the domain $k[t]$, hence a nonzerodivisor, so the pulled-back regular equation ${\nu }^{\mathrm{\#}}(x)$ is regular; by the effective case of [F17] the datum $x\mathrm{/}1$ for $D$ of step 1.2 is $\nu$-admissible and ${\nu }^{\ast }D$ is the effective Cartier divisor on $\mathrm{Spec}k[t]$ cut out by the global equation $t^2$.

The fraction field $\mathrm{Frac}(A)$ equals $k(t)$, because $t=y\mathrm{/}x$ is a fraction of elements of $A$ and conversely $A\subseteq k[t]$; the element $t$ is integral over $A$ since $t^2=x\in A$, so $k[t]=A+At$ is a finite $A$-module; every $z=a+bt$ with $a,b\in A$ satisfies $z^2-2az+(a^2-b^{2}x)=0$, so $k[t]$ is integral over $A$, and every element of $k(t)$ integral over $A$ is integral over $k[t]$ and hence lies in $k[t]$ because $k[t]$ is integrally closed by [F7]; therefore $k[t]$ is the integral closure of $A$ in $k(t)$ by [F12], and $\nu$ is the normalization of $C$, the source being normal by step 1.3.

Since $\dim k[t]=1$ by [F9] and height is computed by [F13], a prime $(\mathrm{\ne }(0))$ of $k[t]$ has height one exactly when it is nonzero, and $(0)$ has height zero; by [F8] every nonzero prime of $k[t]$ is principal, say $(g)$ with $g$ a prime element of $k[t]$. Hence the prime divisors of $\mathrm{Spec}k[t]$, which by [F10] are its height-one integral closed subschemes, are exactly the closed points $V(g)$ for prime elements $g\in k[t]$.

The element $t$ is prime in $k[t]$ because $k[t]\mathrm{/}(t)\cong k$ is a field, so $[0]=V(t)$ is a prime divisor by step 2.4; the local ring $k[t{]}_{(t)}$ is a discrete valuation ring by [F15], since $k[t]$ is a Noetherian integrally closed domain by [F2] and [F7] and $(t)$ has height one by step 2.4, and its maximal ideal is the extension $(t)k[t{]}_{(t)}$ by [F14], generated by the image of $t$; consequently $v_{[0]}(t)=1$ for the normalised valuation of [F16].

Let $Z=V(g)$ be a prime divisor of $\mathrm{Spec}k[t]$ with $(g)\mathrm{\ne }(t)$, as in step 2.4. If $t^2\in (g)$ then $g\mid t^2$, and since the prime element $g$ divides the product $t\cdot t$ it divides $t$, so $(t)\subseteq (g)$; both are height-one primes by step 2.4, so $(t)=(g)$, contradicting $Z\mathrm{\ne }[0]$. Hence $t^2\mathrm{\notin }(g)$, and $t^2$ lies outside the maximal ideal $(g)k[t{]}_{(g)}$ of the local ring, so it is a unit there by [F14] and $v_Z(t^2)=0$, that is, ${\mathrm{ord}}_Z(t^2)=0$ by [F16].

By [F18] applied to the morphism $\nu$ and the divisor $D$ of step 1.2, whose pullback is defined by step 2.2, there is a canonical isomorphism ${\mathcal{O}}_{\mathrm{Spec}k[t]}({\nu }^{\ast }D)\cong {\nu }^{\ast }{\mathcal{O}}_C(D)$ matching the constant sections $1$.

The element $t^2$ is a global meromorphic unit of the integral scheme $\mathrm{Spec}k[t]$, and its order at the prime divisor $[0]$ is ${\mathrm{ord}}_{[0]}(t^2)=v_{[0]}(t^2)=2v_{[0]}(t)=2$ by [F16] and step 3.1.

In summary, $D=V(x)$ is an effective Cartier divisor on the cusp $C$ by step 1.2; the pullback ${\nu }^{\ast }D$ is defined and is cut out by the global equation $t^2$ on the normal affine line $\mathrm{Spec}k[t]$ by steps 1.3 and 2.2; the equation $t^2$ has order $2$ at the origin and order $0$ at every other prime divisor by steps 3.2 and 4.1; $k[t]$ is the integral closure of $A$ by step 2.3, so that $\nu$ is the normalization of the cusp; and the associated invertible sheaves agree by step 3.3.