The cusp retains a doubled tangent line

The polynomial $f=y^2-x^3\in k[x,y]$ has the unique monomial expansion $f=y^2-x^3$ by [F2], with homogeneous parts $f_2=y^2$ of degree $2$ and $f_3=-x^3$ of degree $3$ by [F3], and it generates the principal ideal $I=(f)$ by [F16]; the quotient ring $A=k[x,y]\mathrm{/}(f)$ defines the closed subscheme $C=\mathrm{Spec}A$ of ${\mathbb{A}}_k^2$, and $k[x]$ is a commutative polynomial ring by [F1].

The quotient ring $R=k[x,y]\mathrm{/}(y^2)$ is a free $k[x]$-module with basis $1,\bar{y}$: division by the monic polynomial $y^2\in k[x][y]$ by [F17] writes every $g\in k[x,y]$ uniquely as $g=q{y}^2+(a+by)$ with $a,b\in k[x]$ and ${\mathrm{deg}}_y(a+by)<2$, so the classes of $1$ and $y$ form a $k[x]$-basis of $R$ and $a+b\bar{y}=0$ in $R$ holds exactly when $a=b=0$.

The multiplicity of the equation $f$ at the origin is $2$: the translated expansion $f(0+t)=t_y^2-t_x^3$ is $f_2+f_3$ with $f_2=t_y^2\mathrm{\ne }0$ and no part of degree $0$ or $1$, so the least $j$ with $f_j\mathrm{\ne }0$ is $j=2$ by [F4]; equivalently the $\mathfrak{m}$-adic order of $f$ in the local ring of ${\mathbb{A}}_k^2$ at the origin is $2$.

The origin is a $k$-rational point of $C$: the evaluation map $\epsilon :k[x,y]\to k$ at $(0,0)$ has kernel $(x,y)$, which is maximal by [F9]; since $f$ has no constant term, $f\in (x,y)=\ker \epsilon$ by [F16], so [F10] factors $\epsilon$ through a surjective $k$-algebra homomorphism $A\to k$, and [F11] exhibits it as a $k$-rational point $P=0$ of $C$.

The scheme-theoretic tangent cone: $f$ lies in $\mathfrak{q}=(x,y)$ because it has no constant term, and $I=(f)$, so [F5] applies with $P=k[x,y]$ and gives ${\mathrm{Cone}}_0(C)\cong \mathrm{Spec}(k[x,y]\mathrm{/}{\mathrm{in}}_{\mathfrak{q}}(I))$ with ${\mathrm{in}}_{\mathfrak{q}}(I)=(f_{\min })=(y^2)$ in the principal case; by [F6] this spectrum is the scheme-theoretic tangent cone, so ${\mathrm{Cone}}_0(C)\cong \mathrm{Spec}(k[x,y]\mathrm{/}(y^2))$, the closed subscheme of ${\mathbb{A}}_k^2$ defined by $y^2=0$.

The nilradical of $R$ is the principal ideal $(\bar{y})$: for $a,b\in k[x]$ and $n\ge 1$, the binomial theorem [F18] gives $(a+b\bar{y}{)}^n={\sum }_k\left(\genfrac{}{}{0px}{}{n}{k}\right)a^{n-k}{b}^k{\bar{y}}^k=a^n+n{a}^{n-1}b\bar{y}$ in $R$, because ${\bar{y}}^k=0$ for $k\ge 2$; if this is zero, then $a^n=0$ and $n{a}^{n-1}b=0$ by the uniqueness of the basis representation in step 1.2, and since $k[x]$ is an integral domain by [F14] and [F15], $a^n=0$ forces $a=0$; hence every nilpotent element of $R$ lies in $(\bar{y})$, while $(\bar{y})\subseteq \mathrm{Nil}(R)$ because ${\bar{y}}^2=0$, so $\mathrm{Nil}(R)=(\bar{y})$ by [F12].

The tangent space is two-dimensional: since $I=(f)$ is generated by the single equation $f=y^2-x^3$, the Jacobian matrix of the list $(f)$ at $P$ is the $1\times 2$ matrix $(-3x^2,2y)$ evaluated at $(0,0)$, namely $(0,0)$, by [F7]; its kernel is all of $k^2$, so [F8] gives a canonical isomorphism $T_{0}C\cong \ker (0:k^2\to k)=k^2$, of dimension $2$. The two entries $-3x^2$ and $2y$ vanish at the origin in every characteristic, so the computation does not use the nonvanishing of $2$ or $3$.

The degree-one part of the cone ideal $(y^2)$ is zero: every nonzero element of $(y^2)$ has the form $y^{2}h$ with $h\mathrm{\ne }0$, and each of its monomials has $y$-exponent at least $2$, hence total degree at least $2$; so no nonzero linear form lies in $(y^2)$, and $(y^2)\cap P_1=0$ for the degree-one space $P_1$ of [F2, F3].

The reduced cone is the $x$-axis: by [F13] the reduction of ${\mathrm{Cone}}_0(C)=\mathrm{Spec}R$ is $\mathrm{Spec}(R\mathrm{/}\sqrt{(0)})=\mathrm{Spec}(R\mathrm{/}\mathrm{Nil}(R))=\mathrm{Spec}(R\mathrm{/}(\bar{y}))$ by [F12] and step 2.4; the preimage of $(\bar{y})$ under the quotient map $k[x,y]\to R$ is $(y)$, because $g$ maps into $(\bar{y})$ exactly when $g=q{y}^2+by$ with $b\in k[x]$ by the basis of step 1.2, that is exactly when $g\in (y)$; hence $R\mathrm{/}(\bar{y})\cong k[x,y]\mathrm{/}(y)$ and the reduced cone is the line $\{y=0\}$, the $x$-axis of ${\mathbb{A}}_k^2$; under the coordinate-velocity identification $T_{0}C\cong k^2$ of step 3.1 this line is the set $\{(c,0):c\in k\}$, whose $k$-linear span is the one-dimensional subspace $\{(c,0)\}$, properly contained in $k^2$; the same preimage computation gives $\sqrt{(y^2)}=(y)$ in $k[x,y]$ by [F19], so $y$ lies in the radical of the cone ideal but not in the cone ideal, as step 3.2 records.

Conclusion: the tangent cone is the doubled line $\mathrm{Spec}(k[x,y]\mathrm{/}(y^2))$, nonreduced with $\bar{y}\mathrm{\ne }0$ and ${\bar{y}}^2=0$, its reduction is the $x$-axis, and that line spans only a one-dimensional subspace of the two-dimensional tangent space $T_{0}C\cong k^2$; the multiplicity of the equation at the origin is $2$ by step 2.1, so the origin is a multiple point of the plane cubic. The tangent cone therefore records the doubling that the tangent space does not see, while its reduction has directions along only one line. The full graded cone does determine the tangent space: its degree-one piece is ${\mathfrak{m}}_0\mathrm{/}{\mathfrak{m}}_0^2$ by [F6], and here has the independent classes of $x,y$ by step 3.2, whose dual is $T_{0}C\cong k^2$. The computation retains nilpotents and performs no reduction of the cone or of the curve.

Boundary and scope dispositions: the curve and the cone are nonempty, since $P=0$ is a $k$-rational point of $C$ by step 2.2 and the cone is a spectrum over $k$; the zero case appears in the Jacobian matrix $(0,0)$ of step 3.1, whose kernel is all of $k^2$, in the zero degree-one part $(y^2)\cap P_1=0$ of step 3.2, and in the vanishing ${\bar{y}}^2=0$ of step 2.4; there is one equation, one Jacobian row and one cone equation $y^2$ of degree two (step 1.1 and step 2.3); the degenerate nonreduced case is exactly the cone $k[x,y]\mathrm{/}(y^2)$ with nilpotent nonzero $\bar{y}$, retained and not removed by any reduction in step 2.3, its reduction being computed separately in step 4.1, and the multiplicity is $2$ by step 2.1; the endpoint in the exponent is the least interesting case $y^2$ itself, and the characteristic hypothesis $\mathrm{char}k\mathrm{\ne }2,3$ comes from the source's standing assumption, no step using the invertibility of $2$ or $3$, as noted in step 3.1; no Axiom of Choice or dependent choice is used, since the equation, the Jacobian, the initial form, the basis and the nilpotent computation are all exhibited explicitly; and no biconditional is asserted or used, the only two-sided claim being the computation of the nilradical in step 2.4, proved by two inclusions.