Graded Nakayama and the Hesselink regularity comparison

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 $\mathfrak{m}=(x_1,\dots ,x_d)$.

Let $\mathfrak{b}$ be the ideal generated by $L={\mathfrak{a}}_0+{\sum }_{n<0}{A}_n$ (with ${\mathfrak{a}}_0=\mathfrak{a}\cap A_0$) and let ${\mathfrak{b}}_0$ be the ideal generated by those $x_i$ lying in $L$, that is, by the $x_i$ with $\mathrm{deg}{x}_i<0$, together with $x_1,\dots ,x_r$ of degree $\le 0$. Then ${\mathfrak{b}}_0$ is generated by a subset of the regular system of parameters $x_1,\dots ,x_d$ of $A_{\mathfrak{m}}$, so $(A\mathrm{/}{\mathfrak{b}}_0{)}_{\mathfrak{m}}$ is regular by [F3]. Every element of $L$ lies in ${\mathfrak{b}}_0+\mathfrak{m}\mathfrak{b}$: if $b\in A_n$ with $n<0$, then $b\in \mathfrak{m}=(x_1,\dots ,x_d)$ by step 3.1, so $b={\sum }_i{a}_i{x}_i$ with $a_i$ homogeneous, and for each $i$ either $\mathrm{deg}{x}_i<0$, whence $x_i\in {\mathfrak{b}}_0$ and $a_i{x}_i\in {\mathfrak{b}}_0$, or $\mathrm{deg}{x}_i\ge 0$, whence $\mathrm{deg}{a}_i<0$, so $a_i\in {\sum }_{n<0}{A}_n\subseteq \mathfrak{b}$ and $a_i{x}_i\in \mathfrak{m}\mathfrak{b}$; if $b\in \mathfrak{a}\cap A_0$, then $b\in \mathfrak{a}=(x_1,\dots ,x_r)$ and $b={\sum }_{i\le r}{x}_i{b}_i$ with $b_i$ homogeneous and for each $i$ either $\mathrm{deg}{x}_i\le 0$, whence $x_i\in {\mathfrak{b}}_0$, or $\mathrm{deg}{x}_i>0$, whence $\mathrm{deg}{b}_i<0$ and $b_i\in \mathfrak{b}$, so $x_i{b}_i\in {\mathfrak{b}}_0\cup \mathfrak{m}\mathfrak{b}$. Since $L$ generates $\mathfrak{b}$, this gives $\mathfrak{b}\subseteq {\mathfrak{b}}_0+\mathfrak{m}\mathfrak{b}$, that is, $\mathfrak{b}\mathrm{/}{\mathfrak{b}}_0=\mathfrak{m}(\mathfrak{b}\mathrm{/}{\mathfrak{b}}_0)$; step 2.1 applied to ${\mathfrak{b}}_0\subseteq \mathfrak{b}$ yields $\mathfrak{b}={\mathfrak{b}}_0$. Hence $(A\mathrm{/}\mathfrak{b}{)}_{\mathfrak{m}}=(A\mathrm{/}{\mathfrak{b}}_0{)}_{\mathfrak{m}}$ is regular.