Finite normalization alone does not make a curve regular

The point $p=(0,0)$ is a singular point of the cusp: at $p$ the local ring $R={\mathcal{O}}_{Z,p}$ has dimension one and embedding dimension two, hence is not regular by [F1]. Therefore $Z$ is not regular.

The morphism $\nu :{\mathbf{A}}_k^1\to Z$, $t\mapsto (t^2,t^3)$, is finite: the image $k[t^2,t^3]\subseteq k[t]$ is a $k$-subalgebra over which $k[t]$ is generated as a module by $1$ and $t$ (because $t^2$ and $t^3$ lie in the subalgebra). It is birational: the induced map of fraction fields is $k(t^2,t^3)\hookrightarrow k(t)$, an equality since $t=t^3\mathrm{/}{t}^2$, and $\nu$ is an isomorphism away from the origin with inverse $(x,y)\mapsto y\mathrm{/}x$. It is bijective on scheme points: it is an isomorphism on the complement of the origin by the displayed inverse, and its origin fibre has coordinate ring $k[t]\mathrm{/}(t^2,t^3)$, supported at the single point $t=0$. Normality in [F3] follows directly from unique factorization: a reduced fraction $a\mathrm{/}b$ satisfying a monic integral equation has $b\mid a^n$ after denominators are cleared, so coprimality makes $b$ a unit. Since ${\mathbf{A}}_k^1$ is therefore normal, $\nu$ is a finite normalization of $Z$; by the uniqueness in [F2] the normalization of $Z$ is finite.

On the blowup chart $y=xt$, the strict transform has equation $t^2-x=0$, hence coordinate ring $k[t]$. On the other chart $x=ys$, its equation is $1-y{s}^3=0$, which makes $s$ and $y$ invertible; this portion lies in the overlap with the first chart. Thus the whole strict transform is the regular affine line and its map to $Z$ is $t\mapsto (t^2,t^3)$, the normalization of step 1.2 (Affine blowup standard charts and overlaps).

The curve $Z$ therefore has finite normalization by step 1.2 and is not regular by step 1.1, so the false claim is refuted by the explicit witness $(Z,\nu )$. Moreover $\nu$ is a finite birational morphism which is not an isomorphism over the singular point: if it were an isomorphism at the origin, then $Z$ would be regular at the origin, contradicting step 1.1. Thus finite normalization is not a substitute for the blowup procedure: the regularization theorem requires point blowups, and step 2.1 shows explicitly that one point blowup makes its strict transform regular (Regularization of a one-dimensional integral curve with finite normalization by point blowups, Blowing up a non-regular point strictly increases the finite normalization subalgebra).