Localisation at a homogeneous element is graded, with graded kernels and dehomogenised degree-zero parts

For $r\in R$ written in its finite decomposition $r={\sum }_n{r}_n$ with $r_n\in R_n$, and any $m\ge 0$, one has $r\mathrm{/}{t}^m={\sum }_n{r}_n\mathrm{/}{t}^m$ with $r_n\mathrm{/}{t}^m\in (R_t{)}_{n-m\delta }$, so every element of $R_t$ is a finite sum of elements of the sets $(R_t{)}_n$; each $(R_t{)}_n$ is an additive subgroup, because a sum of two fractions with numerators in $R_{n+m\delta }$ can be written over the common denominator $t^m$ with numerator again in $R_{n+m\delta }$, and likewise for additive inverses.

Let $j\in J$; by [L3] $j={\sum }_i{c}_i{a}_i$ for finitely many $c_i\in R$, and decomposing $c_i={\sum }_n{c}_{i,n}$ with $c_{i,n}\in R_n$, the $N$-th homogeneous component of $j$ is ${\sum }_i{c}_{i,N-d_i}{a}_i$ (over those $i$ with $N\ge d_i$), a combination of the generators and hence an element of $J$; so every homogeneous component of every element of $J$ lies in $J$.

If $z_n=r_n\mathrm{/}{t}^m\in (R_t{)}_n$ are finitely many elements with pairwise distinct degrees $n$ and ${\sum }_n{z}_n=0$, then $\left({\sum }_n{r}_n\right)\mathrm{/}{t}^m=0$, so $t^N{\sum }_n{r}_n=0$ for some $N\ge 0$ by [L2]; the elements $t^N{r}_n\in R_{n+m\delta +N\delta }$ are homogeneous of pairwise distinct degrees, so each $t^N{r}_n=0$ by [L1], whence $z_n=r_n{t}^N\mathrm{/}{t}^{m+N}=0$; therefore the sum of the $(R_t{)}_n$ is direct and every element of $R_t$ has a unique decomposition into homogeneous pieces.

Hence $R\mathrm{/}J$ is nonnegatively graded with $(R\mathrm{/}J{)}_n:=(R_n+J)\mathrm{/}J$: every class $r+J$ is the finite sum ${\sum }_n(r_n+J)$ of such classes, and if ${\sum }_n(r_n+J)=0$ with $r_n\in R_n$ then ${\sum }_n{r}_n=j\in J$ and step 1.2 gives $r_n\in J$ for every $n$, so each $r_n+J=0$; moreover $(R_i+J)\mathrm{/}J\cdot (R_j+J)\mathrm{/}J\subseteq (R_{i+j}+J)\mathrm{/}J$, and the quotient map carries $R_n$ into $(R\mathrm{/}J{)}_n$.

The elements of the extended ideal $J{R}_t$ are exactly the fractions $j\mathrm{/}{t}^m$ with $j\in J$ and $m\ge 0$: one inclusion is $j\mathrm{/}{t}^m=(1\mathrm{/}{t}^m)(j\mathrm{/}1)\in J{R}_t$, and conversely every element of $J{R}_t$ is ${\sum }_i{y}_i(a_i\mathrm{/}1)$ by [L3], which after writing the finitely many $y_i\in R_t$ over a common denominator $t^m$ becomes $\left({\sum }_i{c}_i{a}_i\right)\mathrm{/}{t}^m$ with ${\sum }_i{c}_i{a}_i\in J$.

The grading is multiplicative and ${\lambda }_t$ is degree-preserving: $(r\mathrm{/}{t}^m)(s\mathrm{/}{t}^{m^{\mathrm{\prime }}})=rs\mathrm{/}{t}^{m+m^{\mathrm{\prime }}}$ with $rs\in R_{(i+m\delta )+(j+m^{\mathrm{\prime }}\delta )}$ whenever $r\in R_{i+m\delta }$ and $s\in R_{j+m^{\mathrm{\prime }}\delta }$, so $(R_t{)}_i(R_t{)}_j\subseteq (R_t{)}_{i+j}$, and ${\lambda }_t(r)=r\mathrm{/}1=r\mathrm{/}{t}^0\in (R_t{)}_n$ for $r\in R_n$; together with steps 1.1 and 2.1 this gives the graded $R$-module structure and the direct sum decomposition of 1.

Let $\varphi :R\to T$ satisfy $\varphi (R_n)\subseteq T_n$ and let $r={\sum }_n{r}_n$ with $r_n\in R_n$; then $\varphi (r)={\sum }_n\varphi (r_n)$ with $\varphi (r_n)\in T_n$ of pairwise distinct degrees $n$, so $\varphi (r)=0$ if and only if $\varphi (r_n)=0$ for every $n$, i.e. if and only if every homogeneous component of $r$ lies in $\ker \varphi$; hence $\ker \varphi ={\sum }_n(\ker \varphi \cap R_n)$, a direct sum because the $R_n$ are independent in $R$, which is claim 2.

Assume $\delta =1$ and let $x\in (J{R}_t{)}_0$; by step 2.3 write $x=j\mathrm{/}{t}^m$ with $j\in J$, and replacing $j$ by its homogeneous component of degree $m$, which lies in $J$ by step 1.2 and contributes exactly the degree-zero part of $j\mathrm{/}{t}^m$ by step 2.1, we may suppose $j$ homogeneous of degree $m$; by [L3] write $j={\sum }_i{c}_i{a}_i$ with $c_i\in R_{m-d_i}$ (the degree-$(m-d_i)$ components of arbitrary coefficients), so $x={\sum }_i(c_i\mathrm{/}{t}^{m-d_i})(a_i\mathrm{/}{t}^{d_i})$ with $c_i\mathrm{/}{t}^{m-d_i}\in (R_t{)}_0$ and $a_i\mathrm{/}{t}^{d_i}\in (R_t{)}_0$; conversely $a_i\mathrm{/}{t}^{d_i}\in (J{R}_t{)}_0$ because $a_i\in J$ and $\mathrm{deg}(a_i\mathrm{/}{t}^{d_i})=d_i-d_i=0$. Hence $(J{R}_t{)}_0=(a_1\mathrm{/}{t}^{d_1},\dots ,a_k\mathrm{/}{t}^{d_k})$ as ideals of $(R_t{)}_0$.

Assume $\delta =1$ and $J=(a_1,\dots ,a_k)$; by [L4] the surjection $R_t\to (R\mathrm{/}J{)}_t$ has kernel $J{R}_t$, and it carries homogeneous elements to homogeneous elements of the same degree, because a degree-$n$ element of $R_t$ can be written $r\mathrm{/}{t}^m$ with $r\in R_{n+m\delta }$ homogeneous and the quotient map is degree-preserving by step 2.2; hence the induced map $(R_t{)}_0\to ((R\mathrm{/}J{)}_t{)}_0$ of degree-zero parts is surjective with kernel $(J{R}_t{)}_0$, and step 3.3 identifies this kernel, giving $((R\mathrm{/}J{)}_t{)}_0\cong (R_t{)}_0\mathrm{/}(a_1\mathrm{/}{t}^{d_1},\dots ,a_k\mathrm{/}{t}^{d_k})$.

Claims 1 to 5 are proved: the localisation at a homogeneous element is graded with the displayed degree pieces, degree-preserving ring maps have graded kernels, the quotient by an ideal generated by homogeneous elements is graded, and for a degree-one $t$ the degree-zero parts of $J{R}_t$ and of $(R\mathrm{/}J{)}_t$ are the displayed dehomogenised ideals and quotients; no choice principle is used, the argument working with the explicit fraction calculus of $R_t$.