Differentials of a polynomial ring and of a cuspidal hypersurface

The quotient formula. With $f=y^2-x^3$ one has $\mathrm{\partial }f\mathrm{/}\mathrm{\partial }x=-3x^2$ and $\mathrm{\partial }f\mathrm{/}\mathrm{\partial }y=2y$ on the monomial basis [F1], so the Jacobian matrix of the single equation is the row $(-3x^2,2y)$ and ${\mathrm{\Omega }}_{A\mathrm{/}k}\cong A^2\mathrm{/}A\cdot (-3x^2,2y)=(A\mathrm{d}x\oplus A\mathrm{d}y)\mathrm{/}(2y\mathrm{d}y-3x^2\mathrm{d}x)$, exactly as displayed; no injectivity of the conormal map is used or claimed [F1].

The curve has dimension one. The subring $k[x]\subseteq A$ is a polynomial ring, and $A$ is a finite $k[x]$-module with basis $1,y$ because $y^2=x^3\in k[x]$, so the extension is integral and injective and $A\mathrm{\ne }0$; hence $\dim A=\dim k[x]=1$ by [F2] and [F3].

The fibre at the origin. The maximal ideal $\mathfrak{m}=(x,y)A$ corresponds to the origin, with residue field $\kappa (\mathfrak{m})=k$ because $A\mathrm{/}\mathfrak{m}=k$; tensoring the presentation of step 1.1 with $A\to k$ gives ${\mathrm{\Omega }}_{A\mathrm{/}k}{\otimes }_{A}k\cong k^2\mathrm{/}k\cdot (0,0)=k^2$, as both coefficients $2y$ and $3x^2$ vanish at the origin even when $2=0$ or $3=0$ in $k$. Thus the cotangent fibre at the origin has dimension two, equal to the number of variables, while $A$ has dimension one by step 1.2.