All initial forms define the tangent cone

Noetherian affine scheme. The finite-generation theorem [F4] says every ideal of $P$ is finitely generated. The left regular modules ${}_{P}P$ and ${}_{A}A$ have scalar action by multiplication [F23]; hence [F7] and [F8] identify their submodules with left ideals, which are ideals because both rings are commutative. Thus every submodule of ${}_{P}P$ is finitely generated; [F6] makes this module Noetherian and [F5] makes $P$ a Noetherian ring. Put $A=P\mathrm{/}I$ and let $\pi :P\to A$ be the quotient map [F10, F11]. For an ideal $K\subseteq A$, its preimage $J={\pi }^{-1}(K)$ is an ideal of $P$: additivity and closure under multiplication follow by applying the homomorphism $\pi$ and the ideal property of $K$. Since $P$ is Noetherian, [F7, F8] identify $J$ with a submodule of ${}_{P}P$, and [F6] gives a finite generating list $p_1,\dots ,p_s$ for $J$. Every $a\in K$ has a representative $p\in P$ by [F11], and then $p\in J$; writing $p={\sum }_i{c}_i{p}_i$ shows $a={\sum }_i\pi (c_i)\pi (p_i)$. Each $\pi (p_i)$ lies in $K$, so they generate $K$. Thus every ideal of $A$ is finitely generated. By [F7, F8, F23], every submodule of ${}_{A}A$ is an ideal; it too is finitely generated, so [F6] and [F5] show that $A$ is Noetherian. The affine cover $X=\mathrm{Spec}A$ itself makes $X$ locally Noetherian by [F22]. Each argument fixes one ideal or submodule and establishes existence of a finite generating list for it; no simultaneous family of lists is chosen, and no choice principle is used.

The polynomial associated graded ring. Let $P_d$ be the homogeneous polynomials of total degree $d$. By [F16, F17], every polynomial is a finite sum of homogeneous parts. For every $r\ge 0$, ${\mathfrak{q}}^r={⨁}_{d\ge r}{P}_d$: a product of $r$ elements of $\mathfrak{q}$ has no term of degree below $r$, while every monomial of degree at least $r$ factors as $r$ variables times a monomial; the finite-sum description of ideal products [F9] gives the two inclusions. Thus the degree-$d$ map $P_d\to {\mathfrak{q}}^d\mathrm{/}{\mathfrak{q}}^{d+1}$ is an isomorphism. Since multiplication in the associated graded ring is induced from $P$ [F3], these maps identify ${\mathrm{gr}}_{\mathfrak{q}}P$ with $P$ as graded rings. In degree zero this is the identification $P_0=k$.

The principal case. Suppose $I=(f)$ for $0\mathrm{\ne }f\in P$. By [F20, F21], $P$ is a domain. If $0\mathrm{\ne }h\in I$, [F19] writes $h=gf$; since $h\mathrm{\ne }0$, $g\mathrm{\ne }0$. Let $d$ and $e$ be the lowest degrees of $f$ and $g$. The lowest-degree part of $gf$ is $g_{\min }{f}_{\min }$: all other products of homogeneous parts have degree greater than $d+e$, and this product is nonzero in the domain $P$. Thus every lowest form of a nonzero element of $I$ lies in $(f_{\min })$, while $f_{\min }$ is itself one of those forms. By [F18], ${\mathrm{in}}_{\mathfrak{q}}(I)=(f_{\min })$.

The rational stalk. Evaluation at the origin sends every $t_i$ to zero and each constant to itself. By the unique monomial expansion [F16], its kernel in $P$ is $\mathfrak{q}$. As $I\subseteq \mathfrak{q}$, evaluation factors through $A$ with kernel $\mathfrak{m}=\mathfrak{q}\mathrm{/}I$ and quotient $A\mathrm{/}\mathfrak{m}\cong k$; [F12] makes $\mathfrak{m}$ maximal, and [F13] makes it prime. The stalk formula [F14] gives ${\mathcal{O}}_{X,x}\cong A_{\mathfrak{m}}$, whose unique maximal ideal is $\mathfrak{n}=\mathfrak{m}{A}_{\mathfrak{m}}$ by [F15]. By [F1] and step 1.1, ${\mathrm{Cone}}_x(X)=\mathrm{Spec}({\mathrm{gr}}_{\mathfrak{n}}{A}_{\mathfrak{m}})$.

Quotient filtration and its kernel. The quotient operations [F10, F11] give ${\mathfrak{m}}^r=({\mathfrak{q}}^r+I)\mathrm{/}I$ for every $r$, by induction on $r$. Projection therefore defines a graded ring map $\varphi :{\mathrm{gr}}_{\mathfrak{q}}P\to {\mathrm{gr}}_{\mathfrak{m}}A$, which is surjective in each degree because every class in ${\mathfrak{m}}^r\mathrm{/}{\mathfrak{m}}^{r+1}$ is represented by an element of ${\mathfrak{q}}^r$. Use step 1.2 to identify its source with $P$. For a homogeneous $h\in P_d$, its class lies in the degree-$d$ kernel exactly when $h\in I+{\mathfrak{q}}^{d+1}$. Write $h=g+u$ with $g\in I$ and $u\in {\mathfrak{q}}^{d+1}$. If $h\mathrm{\ne }0$, then $g=h-u$ has no terms below degree $d$ and its degree-$d$ part is $h$, so $g_{\min }=h$. Conversely, if $0\mathrm{\ne }g\in I$ has lowest part $g_{\min }$ of degree $d$, then $g-g_{\min }\in {\mathfrak{q}}^{d+1}$, so $g_{\min }$ lies in the degree-$d$ kernel. The kernel is a homogeneous ideal; it contains every lowest form, and each of its homogeneous elements is itself a lowest form. It is therefore exactly ${\mathrm{in}}_{\mathfrak{q}}(I)$, including when $I=0$.

The local tangent cone. The localization maps of [F2] give an isomorphism in every degree from ${\mathrm{gr}}_{\mathfrak{m}}A$ to ${\mathrm{gr}}_{\mathfrak{n}}{A}_{\mathfrak{m}}$. They preserve multiplication because each degree map is induced by the ring map $A\to A_{\mathfrak{m}}$ and the products in both associated graded rings are induced by ring multiplication [F3]. Thus step 2.2 yields the canonical graded $k$-algebra isomorphism $P\mathrm{/}{\mathrm{in}}_{\mathfrak{q}}(I)\cong {\mathrm{gr}}_{{\mathfrak{m}}_x}{\mathcal{O}}_{X,x}$, sending each variable to its degree-one initial class. By [F1] and step 2.1, its spectrum is the scheme-theoretic tangent cone at $x$.

A generating list need not suffice, and boundary cases. For $P=k[X,Y,Z]$, set $f_1=XY$, $f_2=XZ+Z(Y^2-Z^2)$, and $I=(f_1,f_2)\subseteq (X,Y,Z)$, so the origin belongs to $X=\mathrm{Spec}(P\mathrm{/}I)$. The initial forms of this generating list are $XY$ and $XZ$, which generate an ideal contained in $(X)$. But $h=YZ(Y^2-Z^2)=Y{f}_2-Z{f}_1$ is a nonzero homogeneous element of $I$, so it is its own lowest form and belongs to ${\mathrm{in}}_{(X,Y,Z)}(I)$. Since $(XY,XZ)\subseteq (X)$ while $h\mathrm{\notin }(X)$ (its image modulo $(X)$ is the same nonzero polynomial), the two displayed initial forms do not generate the initial ideal. This witness works in every characteristic because the distinct monomials $Y^{3}Z$ and $Y{Z}^3$ cannot cancel. For $n=0$, $\mathfrak{q}=0$ forces $I=0$ and $P=A=k$, so the graded ring is $k$ in degree zero and the initial ideal is zero; the nonzero principal case is excluded. For $n=1$, each ${\mathfrak{q}}^r\mathrm{/}{\mathfrak{q}}^{r+1}$ is generated by $t_1^r$, consistent with steps 1.2 and 2.2. Degree zero has $P_0=k$ and $I\subseteq \mathfrak{q}$ contains no nonzero constant; degree one is covered by the same kernel calculation with ${\mathfrak{q}}^2$. If $I=0$, the set of nonzero elements of $I$ is empty and its generated initial ideal is zero. The condition $I\subseteq \mathfrak{q}$ implies $I\mathrm{\ne }P$, so the origin exists and $X$ is not empty. The lemma asserts no iff, so there are no converse directions to check.