Split Reductive Root Systems, Bruhat Cells, and Parabolics

By [F3] the split torus $T$ is isomorphic over $k$ to ${\mathbf{G}}_m^r$ with $r=\dim T$, and ${\mathbf{G}}_m^r=D_k({\mathbb{Z}}^r)$ by [F1]; the given isomorphism $T\cong D_k(M)$ and [F2] therefore identify $M$ with ${\mathbb{Z}}^r$. Applying [F2] to $M$ and to $\mathbb{Z}$ gives $X(T)={\mathrm{Hom}}_k(D_k(M),D_k(\mathbb{Z}))=\mathrm{Hom}(\mathbb{Z},M)\cong M$ and $X_{\ast }(T)={\mathrm{Hom}}_k(D_k(\mathbb{Z}),D_k(M))=\mathrm{Hom}(M,\mathbb{Z})=M^{\vee }$. Both are free abelian of rank $r=\dim T$; the identifications are induced by the anti-equivalence and are compatible with any change of the splitting isomorphism, which only renames $M$ by the induced automorphism.

For $\chi \in X(T)$ and $\lambda \in X_{\ast }(T)$ the composite $\chi \circ \lambda :{\mathbf{G}}_m\to {\mathbf{G}}_m$ is a homomorphism, and [F1] identifies ${\mathrm{Hom}}_k({\mathbf{G}}_m,{\mathbf{G}}_m)={\mathrm{Hom}}_k(D_k(\mathbb{Z}),D_k(\mathbb{Z}))=\mathrm{Hom}(\mathbb{Z},\mathbb{Z})=\mathbb{Z}$ by [F2]; so $⟨\chi ,\lambda ⟩:=\chi \circ \lambda$ is an integer and composition of homomorphisms is $\mathbb{Z}$-bilinear. Under the identifications of step 1.1, an element of $X(T)$ is an element $m\in M$ and an element of $X_{\ast }(T)$ is a homomorphism $\mu :M\to \mathbb{Z}$, and the corresponding composite is $\mu (m)$: $\chi \circ \lambda$ is evaluation of the character at the cocharacter. Therefore the map $X_{\ast }(T)\to X(T{)}^{\vee }$, $\lambda \mapsto (\chi \mapsto ⟨\chi ,\lambda ⟩)$, is exactly the identity $\mathrm{Hom}(M,\mathbb{Z})\to \mathrm{Hom}(M,\mathbb{Z})$ and hence an isomorphism, so the pairing is perfect and identifies $X_{\ast }(T)$ with $X(T{)}^{\vee }$.

Let $k^{\mathrm{\prime }}\supseteq k$ be a field extension. Base change of group algebras identifies $D_{k^{\mathrm{\prime }}}(M)=D_k(M{)}_{k^{\mathrm{\prime }}}$, and [F2] applied over the field $k^{\mathrm{\prime }}$ gives $X(T_{k^{\mathrm{\prime }}})=X(D_{k^{\mathrm{\prime }}}(M))=M$ and $X_{\ast }(T_{k^{\mathrm{\prime }}})=M^{\vee }$ with the same evaluation pairing, so the two identifications of step 1.1 commute with extension of scalars. For free lattices of finite rank the dual of a base change is the base change of the dual, so tensoring the perfect pairing of step 2.1 with $\mathbb{Q}$ exhibits $(X(T)\otimes \mathbb{Q}{)}^{\vee }=X_{\ast }(T)\otimes \mathbb{Q}$ and gives a perfect $\mathbb{Q}$-bilinear pairing of $X(T)\otimes \mathbb{Q}$ with $X_{\ast }(T)\otimes \mathbb{Q}$.

By [F3], $C_G(T)=T$ for every maximal torus, and the scheme centre lies in each such centralizer. Hence $Z(G)\subseteq T$. By [F4] the centre is of multiplicative type, proving (a). Its largest torus $Z(G{)}_t$ is central, smooth connected normal and solvable, so lies in $R(G)$. Conversely pass to algebraic closure. The unipotent radical of the smooth connected solvable group $R(G{)}_{k^{\mathrm{a}}}$ is characteristic there and normal in $G_{k^{\mathrm{a}}}$: conjugation preserves it on points by uniqueness, and smoothness makes this pointwise inclusion scheme-theoretic. Reductivity forces it to be trivial. Solvable splitting [F1] therefore makes $R(G{)}_{k^{\mathrm{a}}}$ a torus, so $R(G)$ is a torus. Rigidity [F5] makes it central, hence $R(G)\subseteq Z(G{)}_t$. Thus $R(G)=Z(G{)}_t$.

Put $G^{\mathrm{\prime }}=DG$. We prove $R(G)\cap G^{\mathrm{\prime }}$ finite without asserting smoothness of this intersection. Over algebraic closure take a faithful representation $W$ of $G$ and its central-torus weight decomposition $W={⨁}_{\chi }{W}_{\chi }$. Since $R(G)$ is central, each weight block is $G$-stable. Its determinant is a character of $G$, hence trivial on $G^{\mathrm{\prime }}$; on $R(G)$ it is ${\chi }^{d_{\chi }}$, where $d_{\chi }=\dim W_{\chi }>0$. Faithfulness makes the finitely many $\chi$ generate the character lattice of $R(G)$ by [F4]. The subgroup generated by their multiples $d_{\chi }\chi$ has finite index: a common positive multiple of all $d_{\chi }$ times the full lattice lies in it. Hence the subgroup killed by all these determinant characters is finite, and contains $R(G)\cap G^{\mathrm{\prime }}$. This proves finiteness over $k$, including infinitesimal kernels.

The centre is defined by commuting equations and commutes with field extension. By [F4], so does its largest torus, and therefore so does $R(G)$ by step1.1. For $Q=G\mathrm{/}R(G)$, a smooth connected normal solvable subgroup in $Q_{k^{\mathrm{a}}}$ pulls back to a smooth connected normal solvable subgroup of $G_{k^{\mathrm{a}}}$: its kernel is the smooth central torus $R(G{)}_{k^{\mathrm{a}}}$, and solvability is closed under extensions by pulling back derived series. Radical maximality forces the inverse image to equal $R(G{)}_{k^{\mathrm{a}}}$, so the subgroup in $Q_{k^{\mathrm{a}}}$ is trivial. Thus $Q$ is semisimple. Existence, affineness, smoothness and connectedness of this quotient follow from [F2]. This proves (b).

A finite central quotient of a smooth connected semisimple group is semisimple. Indeed over algebraic closure, the inverse image of its radical is an extension of a solvable group by a finite central commutative group, hence solvable. Its smooth connected reduced neutral component is normal in the semisimple source by [F2] and is therefore trivial. The finite quotient morphism preserves dimension, so the radical in the quotient has dimension zero and is trivial because smooth connected. Apply this to $G\mathrm{/}R(G)\to G\mathrm{/}Z(G)$, whose kernel $Z(G)\mathrm{/}R(G)$ is finite by [F4]. Thus $G\mathrm{/}Z(G)$ is semisimple, in particular reductive, and its centre is of multiplicative type by step1.1 applied to this quotient.

Since $R(G)\subseteq T$, its quotient $T\mathrm{/}R(G)$ is a torus in $G\mathrm{/}R(G)$. If a larger torus existed, its inverse image would be a smooth connected extension of tori with central kernel $R(G)$. Rigidity [F5] makes this extension commutative and of multiplicative type, hence a torus properly containing $T$, a contradiction. Thus $T\mathrm{/}R(G)$ is maximal. Consequently the semisimple rank is $\dim T-\dim R(G)=\mathrm{rk}G-\dim Z(G)$, since $Z(G)\mathrm{/}R(G)$ is finite. This proves (d).

Over algebraic closure the radical $R(G^{\mathrm{\prime }})$ is preserved by conjugation from $G$ by uniqueness, and this gives scheme normality because $G$ and $R(G^{\mathrm{\prime }})$ are smooth. Thus it lies in $R(G)\cap G^{\mathrm{\prime }}$, finite by step 1.2. Smooth connectedness then forces $R(G^{\mathrm{\prime }})=1$. The derived subgroup is smooth connected by [F2], so it is semisimple over $k$. Now $G\mathrm{/}R(G)$ is semisimple and perfect by [F6]. The closed normal image of $G^{\mathrm{\prime }}$ in this quotient has commutative quotient, because commutators lift fppf-locally to $G$ and land in $G^{\mathrm{\prime }}$. Perfectness makes that image the entire quotient. Therefore $G=R(G)G^{\mathrm{\prime }}$ and multiplication $R(G)\times G^{\mathrm{\prime }}\to G$ is a surjective homomorphism with kernel $\{(z,z^{-1}):z\in R(G)\cap G^{\mathrm{\prime }}\}$, finite by step 1.2. This proves all of (e).

Let $Z^{\mathrm{\prime }}$ be the inverse image in $G$ of $Z(G\mathrm{/}Z(G))$. It is normal, with central kernel $Z(G)$ and multiplicative-type quotient by step 3.1. The conjugation action of connected $G$ on $Z^{\mathrm{\prime }}$ is trivial by [F5], so $Z^{\mathrm{\prime }}\subseteq Z(G)$. Its quotient is therefore trivial: $G\mathrm{/}Z(G)$ has trivial scheme centre, proving (c).

By definition, semisimplicity means the geometric radical is trivial. Step2.1 identifies it with $R(G{)}_{k^{\mathrm{a}}}$, so this is equivalent to $R(G)=1$. By step1.1 and [F4] this occurs exactly when $Z(G)$ has dimension zero, equivalently is finite. Thus (f) holds. The distinction from the full neutral centre is essential: in characteristic $p$, ${\mathrm{S}\mathrm{L}}_p$ is semisimple with centre ${\mu }_p$, connected and nonreduced, while its largest central torus and radical are trivial.

On every $k$-algebra $R$, the functor of points of a group-scheme limit is the limit of the point functors. The Lie functor is the kernel of the reduction map from $R[\epsilon ]$ to $R$, and kernels commute with limits: a compatible tuple reduces to the identity exactly when each component does. Hence . Applying this to a kernel or a fiber product gives left exactness and the displayed fiber-product equality. For subgroup inclusions their fiber product is their scheme intersection, and the vector-space fiber product is the intersection of the Lie subspaces.

Put $\mathfrak{h}=\mathrm{Lie}(H)$. By [F1], $e^{\epsilon X}$ lies in the centralizer precisely when it commutes with $H(S)$ for every $k[\epsilon ]$-algebra $S$. This is equivalent to $X$ being invariant in the rational adjoint representation: invariance tested on any $k$-algebra $R$ gives $h{e}^{{\epsilon }^{\mathrm{\prime }}X}{h}^{-1}=e^{{\epsilon }^{\mathrm{\prime }}X}$ in $G(R[{\epsilon }^{\mathrm{\prime }}])$ for all $h\in H(R)$; taking $R=S$ and specializing ${\epsilon }^{\mathrm{\prime }}$ to the given nilpotent $\epsilon \in S$ proves the centralizer condition. Conversely take $S=R[\epsilon ]$ and $h\in H(R)$ to recover invariance. Thus $\mathrm{Lie}(C_G(H))={\mathfrak{g}}^H$. The tangent dimension criterion [F2] gives the stated dimension inequality and equality case.

Assume AC for part (b). Suppose $\mathrm{Lie}(H)=\mathrm{Lie}(G)$ with $H$ smooth. Since $H\subseteq G$ we have $\dim H\le \dim G$, and by [F2] $\dim G\le \dim \mathrm{Lie}(G)=\dim \mathrm{Lie}(H)=\dim H$, so equality holds throughout; thus $G$ is smooth and $\dim G=\dim H$. A closed subgroup of the connected smooth group $G$ of dimension $\dim G$ equals $G$ by [F2], so $H=G$. Now let $H_1,\dots ,H_n$ be smooth subgroups whose Lie algebras generate $\mathfrak{g}$, and let $H$ be the algebraic subgroup they generate, which is the scheme-theoretic closure of the union of finite product maps from the $H_i$. The product maps and their inverses are stable under multiplication and inversion, so their closure is a subgroup. These maps have geometrically reduced sources and are schematically dominant as a family onto that closure, hence the closure is geometrically reduced and therefore smooth as a finite-type group scheme; by functoriality of $\mathrm{Lie}$, $\mathrm{Lie}(H)$ is a Lie subalgebra of $\mathfrak{g}$ containing each $\mathrm{Lie}(H_i)$, hence containing the subalgebra they generate, which is all of $\mathfrak{g}$; so $\mathrm{Lie}(H)=\mathfrak{g}=\mathrm{Lie}(G)$ and the first part gives $H=G$.

For the normalizer, the same calculation gives $e^{\epsilon X}h{e}^{-\epsilon X}{h}^{-1}=e^{\epsilon (X-\mathrm{Ad}(h)X)}\mathrm{.}$ If $e^{\epsilon X}$ normalizes $H$, then for every $R$ and $h\in H(R)$ this commutator lies in $H(R[\epsilon ])$ and reduces to the identity; by [F1] this means $X-\mathrm{Ad}(h)X\in {\mathfrak{h}}_R$. Hence the class of $X$ in $\mathfrak{g}\mathrm{/}\mathfrak{h}$ is $H$-invariant. Conversely, if that class is invariant, take any $k[\epsilon ]$-algebra $S$ and $h\in H(S)$. The vector $X-\mathrm{Ad}(h)X$ lies in ${\mathfrak{h}}_S$, so the corresponding dual-number point of $H(S[{\epsilon }^{\mathrm{\prime }}])$ specializes to the displayed commutator in $H(S)$ under ${\epsilon }^{\mathrm{\prime }}\mapsto \epsilon$. This proves conjugation by $e^{\epsilon X}$ maps $H(S)$ into itself; applying the argument to $-X$ gives equality, so $e^{\epsilon X}$ normalizes $H$. Thus $\mathrm{Lie}(N_G(H))$ is the inverse image of $(\mathfrak{g}\mathrm{/}\mathfrak{h}{)}^H$, and its quotient by $\mathfrak{h}$ is that invariant space. This is a quotient of Lie algebras; it is not a claim about $\mathrm{Lie}(N_G(H)\mathrm{/}H)$ for nonsmooth $H$. These calculations prove all assertions.

Suppose first that $X=\mathrm{Spec}A$ is affine, with gradation $A={⨁}_{n\in \mathbb{Z}}{A}_n$ induced by the action, and let $\mathfrak{b}$ be the ideal generated by $\mathfrak{a}\cap A_0+{\sum }_{n<0}{A}_n$, so that $X(Z)=\mathrm{Spec}(A\mathrm{/}\mathfrak{b})$ in the sense of [F1]. For a $k$-algebra $R$ and a point $x\in X(R)$, identified with a homomorphism $x:A\to R$ by [F2], the orbit map corresponds to $A\to R[T,T^{-1}]$, ${\sum }_n{f}_n\mapsto {\sum }_{n}x(f_n)T^n$; the limit exists exactly when $x(A_n)=0$ for all $n<0$, and then the limit point is the composite $f\mapsto x(f_0)$ together with the point of $Z$. The limit lies in $Z(R)$ exactly when this composite kills $\mathfrak{a}$, that is, when $x(\mathfrak{a}\cap A_0)=0$; combining, $x$ satisfies the defining property of the concentrator functor exactly when $x$ kills $\mathfrak{b}$, i.e. exactly when $x$ factors through $A\mathrm{/}\mathfrak{b}$. Hence the affine concentrator represents the functor and the inclusion is a closed immersion.

Keep $X$ affine and suppose $X$ and $Z$ smooth. Work over an algebraic closure and let $x$ be a geometric point of $X(Z)$, with fixed limit $y$. Localize $A$ by $A_0\setminus {\mathfrak{m}}_0$, where ${\mathfrak{m}}_0$ is the ideal of $y$ in $A_0$. Evaluation at the fixed point $y$ kills every nonzero-degree piece, so $\mathfrak{m}={\mathfrak{m}}_0\oplus {⨁}_{n\mathrm{\ne }0}{A}_n$ is now a homogeneous maximal ideal. The local rings of $X$ and $Z$ at $y$ are regular, so [F3] shows that the local ring of $X(Z)$ at $y$ is regular. The quotient $A\mathrm{/}\mathfrak{b}$ has only nonnegative degrees, giving an ${\mathbf{A}}^1$-action that contracts $x$ to $y$. The nonregular locus of this finite-type scheme is closed and ${\mathbf{G}}_m$-stable; if it contained $x$, its closedness would also put $y$ in it, a contradiction. Hence every geometric point $x$ is regular and $X(Z)$ is smooth. In the affine case uniqueness as a smooth closed subscheme follows from reducedness and equality of geometric point sets.

Choose a finite cover by invariant affine opens $X_i$ and put $Z_i=Z\cap X_i$. The limit functor is local on the test scheme, because orbit extensions glue uniquely by separatedness. Its subfunctor consisting of extensions landing in $X_i$ is open: it is the condition that the limit morphism to $Z$ lands in $Z_i$. Indeed, an invariant closed complement $X\setminus X_i$ contains the limit of any orbit starting in it; hence a limit in $X_i$ forces the entire orbit extension into $X_i$, checked on geometric fibers. The affine schemes $X_i(Z_i)$ of step 1.1 therefore glue along these open subfunctors and cover the representing scheme $X(Z)$. On each such open the realization is a closed immersion into $X_i$, making it a local immersion. Also $p^{-1}(Z_i)=X_i(Z_i)$ is affine, so $p$ is affine. Smoothness follows locally from step 2.1, and representability supplies uniqueness of the scheme and both realization morphisms.

Base change: for a field extension $k^{\mathrm{\prime }}\mathrm{/}k$ the graded description of step 1.1 is stable under $A\mapsto A{\otimes }_k{k}^{\mathrm{\prime }}$ and limits of $R{\otimes }_k{k}^{\mathrm{\prime }}$-points are computed after this base change, so $X(Z{)}_{k^{\mathrm{\prime }}}=X_{k^{\mathrm{\prime }}}(Z_{k^{\mathrm{\prime }}})$ compatibly with $i$ and $p$; for a ${\mathbf{G}}_m$-equivariant morphism $f:X\to X^{\mathrm{\prime }}$ carrying $Z$ into $Z^{\mathrm{\prime }}$ the same functorial description shows $f(X(Z))\subseteq X^{\mathrm{\prime }}(Z^{\mathrm{\prime }})$, giving the morphism of pairs. For the conjugation action of ${\mathbf{G}}_m$ on a smooth affine group $G$ with cocharacter $\lambda$, let $H=G(\{g:{\lim }_{t\to 0}t\cdot g=e\})$. The defining condition is closed and its points form a subgroup of $G(R)$ for every $R$, because conjugation is a group homomorphism, $(gh{)}_t=tg{t}^{-1}th{t}^{-1}$, and the limit of a product is the product of the limits; hence $H$ is an algebraic subgroup. It is normal in $P_G(\lambda )$: for $g\in P_G(\lambda )(R)$ the limit $q={\lim }_{t\to 0}t\cdot g$ is defined, and for $h\in H(R)$ the conjugates $gh{g}^{-1}$ have limit $qe{q}^{-1}=e$, so $gH{g}^{-1}\subseteq H$, and symmetry gives $gH{g}^{-1}=H$. AC is used only through the geometric suppliers named in the deps.

Assume (a), $M=\mathfrak{m}M$. Localization is exact and commutes with the action, so $M_{\mathfrak{m}}=\mathfrak{m}{M}_{\mathfrak{m}}$; the ring $A_{\mathfrak{m}}$ is local with maximal ideal $\mathfrak{m}{A}_{\mathfrak{m}}$ and $M_{\mathfrak{m}}$ is finitely generated over it. If $M_{\mathfrak{m}}\mathrm{\ne }0$, choose a minimal generating tuple $x_1,\dots ,x_k$; then $x_k={\sum }_{i=1}^k{a}_i{x}_i$ with all $a_i\in \mathfrak{m}{A}_{\mathfrak{m}}$, whence $(1-a_k)x_k={\sum }_{i<k}{a}_i{x}_i$, and $1-a_k$ is a unit, contradicting minimality. So $M_{\mathfrak{m}}=0$ and (a) implies (b). Conversely assume (b) and let $x\in M$ be homogeneous. Since $x\mathrm{/}1=0$ in $M_{\mathfrak{m}}$, there is $a\in A\setminus \mathfrak{m}$ with $ax=0$. Write $a={\sum }_n{a}_n$; comparing homogeneous components of $ax$ gives $a_{n}x=0$ for every $n$. As $a\mathrm{\notin }\mathfrak{m}$ we have $a_0\mathrm{\notin }{\mathfrak{m}}_0$, so $a_0$ is a unit of the local ring $A_0$, and $a_{0}x=0$ forces $x=0$. A finitely generated graded module is generated by homogeneous elements, so $M=0$, which is (c); (c) trivially implies (a).

Let $N\subseteq M$ be finitely generated graded modules. The quotient $M\mathrm{/}N$ is a graded $A$-module, finitely generated because $A$ is Noetherian, and localization is exact, so $(M\mathrm{/}N{)}_{\mathfrak{m}}=M_{\mathfrak{m}}\mathrm{/}{N}_{\mathfrak{m}}$. Applying step 1.1 to $M\mathrm{/}N$ translates the three conditions: $M\mathrm{/}N=\mathfrak{m}(M\mathrm{/}N)$ is $M=N+\mathfrak{m}M$, while $M_{\mathfrak{m}}\mathrm{/}{N}_{\mathfrak{m}}=0$ is $M_{\mathfrak{m}}=N_{\mathfrak{m}}$ and $M\mathrm{/}N=0$ is $M=N$.

Assume AC for this and the next step, and suppose $A_{\mathfrak{m}}$ regular of dimension $d$ and $(A\mathrm{/}\mathfrak{a}{)}_{\mathfrak{m}}$ regular of dimension $d-r$, where $\mathfrak{a}\subseteq \mathfrak{m}$ because $\mathfrak{a}$ is graded and proper. Put $\mathfrak{n}=(\mathfrak{a}+\mathfrak{m})\mathrm{/}\mathfrak{a}$, the maximal ideal of $A\mathrm{/}\mathfrak{a}$, so that $(A\mathrm{/}\mathfrak{a}{)}_{\mathfrak{m}}$ is the localization of $A\mathrm{/}\mathfrak{a}$ at $\mathfrak{n}$. The exact sequence of $\kappa$-vector spaces $0\to (\mathfrak{a}+{\mathfrak{m}}^2)\mathrm{/}{\mathfrak{m}}^2\to \mathfrak{m}\mathrm{/}{\mathfrak{m}}^2\to \mathfrak{n}\mathrm{/}{\mathfrak{n}}^2\to 0$, together with ${\dim }_{\kappa }\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2=d$ and ${\dim }_{\kappa }\mathfrak{n}\mathrm{/}{\mathfrak{n}}^2=d-r$ from regularity [F2], gives ${\dim }_{\kappa }(\mathfrak{a}+{\mathfrak{m}}^2)\mathrm{/}{\mathfrak{m}}^2=r$. These spaces are spanned by images of homogeneous elements, so choose homogeneous $x_1,\dots ,x_r\in \mathfrak{a}$ whose images form a $\kappa$-basis of $(\mathfrak{a}+{\mathfrak{m}}^2)\mathrm{/}{\mathfrak{m}}^2$, and extend by homogeneous $x_{r+1},\dots ,x_d\in \mathfrak{m}$ whose images complete a $\kappa$-basis of $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$. Since $A_{\mathfrak{m}}$ is regular of dimension $d$, the images of $x_1,\dots ,x_d$ generate $\mathfrak{m}{A}_{\mathfrak{m}}$ minimally and form a regular system of parameters there; hence ${\mathfrak{a}}_0:=(x_1,\dots ,x_r)\subseteq \mathfrak{a}$ has $A_{\mathfrak{m}}\mathrm{/}{\mathfrak{a}}_0{A}_{\mathfrak{m}}$ regular of dimension $d-r$ by [F3], and it surjects onto the regular local ring $(A\mathrm{/}\mathfrak{a}{)}_{\mathfrak{m}}$ of the same dimension $d-r$. A surjective local homomorphism of regular local rings of equal dimension is an isomorphism: it induces a surjection of cotangent spaces of equal dimension, hence an isomorphism of associated graded rings, and its kernel lies in ${\bigcap }_i(\mathfrak{m}{A}_{\mathfrak{m}}{)}^i=0$ by Krull's intersection theorem [F3]. Therefore $\mathfrak{a}{A}_{\mathfrak{m}}={\mathfrak{a}}_0{A}_{\mathfrak{m}}$, and step 2.1 applied to the finitely generated graded modules ${\mathfrak{a}}_0\subseteq \mathfrak{a}$ gives $\mathfrak{a}={\mathfrak{a}}_0$. The same local argument with $\mathfrak{m}$ in place of $\mathfrak{a}$ shows $\mathfrak{m}{A}_{\mathfrak{m}}=(x_1,\dots ,x_d)A_{\mathfrak{m}}$, so step 2.1 gives $\mathfrak{m}=(x_1,\dots ,x_d)$.