We construct $Q_{k}$ by replacing $y^{3}$ by $x$ in the polynomial

Since $j=3[j/3]+r_{j}$, $r_{j}\in\{0,1,2\}$, it follows that

We now check that there is no polynomial $Q(x,y)$ of degree $k$ so that $Q(y^{3},y)=(y+1)^{3k}$. Indeed, if $Q(x,y)=\sum_{j+l\leq k}c_{jl}x^{j}y^{l}$ then

does not contain the monomial $y^{3k-1}$.

Note that $\overline{X}=\{xt^{2}=y^{3}\}=X\cup\{a\}$, where $a=[0:1:0]$, so the germ $(\overline{X},a)$ is irreducible. Proposition 3.1 implies that $\eta\,|_{{}_{X}}$ has an extension in ${\mathcal{L}}_{1/3}({\mathbb{C}}^{2})$. ∎