Graded balanced tensor product and homogeneous Hom

Total internal degree on the balanced tensor product. Let $A$ be a graded $k$-algebra, $M$ a graded right $A$-module and $N$ a graded left $A$-module (Associative graded algebras, bimodules, and internal shifts). First use the canonical decomposition of the ordinary tensor product of the underlying abelian groups $M{\otimes }_{\mathbb{Z}}N\cong {⨁}_{i,j\in \mathbb{Z}}{M}_i{\otimes }_{\mathbb{Z}}{N}_j\mathrm{.}$ Its inverse sends $m\otimes n$ to the finite sum ${\sum }_{i,j}{m}_i\otimes n_j$, where $m={\sum }_i{m}_i$ and $n={\sum }_j{n}_j$ are their homogeneous decompositions; the two maps are inverse because they agree with the identity on homogeneous elementary tensors. Give the summand $M_i{\otimes }_{\mathbb{Z}}{N}_j$ total degree $i+j$, and write $G_d:={⨁}_{i+j=d}{M}_i{\otimes }_{\mathbb{Z}}{N}_j$.

The balanced tensor product $M{\otimes }_{A}N=(M{\otimes }_{\mathbb{Z}}N)\mathrm{/}{\mathcal{R}}_A$ (The tensor product $M{\otimes }_{R}N$ from the additive group underlying the free $\mathbb{Z}$-module on $M\times N$, elementary tensors, and finite tensor sums), where ${\mathcal{R}}_A$ is generated by the balance relations $(ma)\otimes n-m\otimes (an)$. These relations are generated by the homogeneous ones: for $m={\sum }_i{m}_i$, $a={\sum }_{\mathrm{\ell }}{a}_{\mathrm{\ell }}$ and $n={\sum }_j{n}_j$, the relation is the finite sum of $(m_i{a}_{\mathrm{\ell }})\otimes n_j-m_i\otimes (a_{\mathrm{\ell }}{n}_j)$ over $i,\mathrm{\ell },j$. Each such homogeneous relation belongs to $G_{i+\mathrm{\ell }+j}$, since the two module actions preserve total degree. Thus ${\mathcal{R}}_A={⨁}_d({\mathcal{R}}_A\cap G_d)$, and

$M{\otimes }_{A}N={⨁}_{d\in \mathbb{Z}}(M{\otimes }_{A}N{)}_d,(M{\otimes }_{A}N{)}_d:=(G_d+{\mathcal{R}}_A)\mathrm{/}{\mathcal{R}}_A,$

is a graded abelian group. For homogeneous $m\in M_i$ and $n\in N_j$ the elementary tensor $m\otimes n$ has total internal degree $i+j$; every tensor is a finite sum of homogeneous elementary tensors, and the balancing relations used in the quotient are homogeneous. This grading is the only grading used below, and it introduces no sign.

Outer actions. If $M$ is a graded $(B,A)$-bimodule and $N$ a graded left $A$-module, then the action $b(m\otimes n):=(bm)\otimes n$ of A commuting outer scalar action descends to a tensor product makes $M{\otimes }_{A}N$ a graded left $B$-module: the action is homogeneous because homogeneous $b\in B_l$ sends $m\otimes n$ of degree $i+j$ to an element of degree $l+i+j$. Symmetrically, if $N$ is a graded $(A,C)$-bimodule, the action $(m\otimes n)c:=m\otimes (nc)$ makes $M{\otimes }_{A}N$ a graded right $C$-module, and the two outer actions commute when both are present.

Homogeneous Hom. For graded left $A$-modules $P,Q$, with ${\mathrm{Hom}}_A^{\mathrm{u}\mathrm{n}\mathrm{g}\mathrm{r}}(P,Q)$ the abelian group of all $A$-linear maps $P\to Q$, put

${\mathrm{Hom}}_{A,d}(P,Q):=\{f\in {\mathrm{Hom}}_A^{\mathrm{u}\mathrm{n}\mathrm{g}\mathrm{r}}(P,Q)\mid f(P_i)\subseteq Q_{i+d}\text{ for all }i\in \mathbb{Z}\},{\mathrm{HOM}}_A(P,Q):={⨁}_{d\in \mathbb{Z}}{\mathrm{Hom}}_{A,d}(P,Q)\mathrm{.}$

Each ${\mathrm{Hom}}_{A,d}(P,Q)$ is a subgroup of ${\mathrm{Hom}}_A^{\mathrm{u}\mathrm{n}\mathrm{g}\mathrm{r}}(P,Q)$, and the sum is direct: if $f={\sum }_d{f}_d={\sum }_d{f}_d^{\mathrm{\prime }}$ with $f_d,f_d^{\mathrm{\prime }}\in {\mathrm{Hom}}_{A,d}$, then for $p\in P_i$ the element $f(p)$ has homogeneous components $f_d(p)\in Q_{i+d}$, and comparison of the graded components in $Q$ gives $f_d(p)=f_d^{\mathrm{\prime }}(p)$ for every $d$ and $p$, so $f_d=f_d^{\mathrm{\prime }}$. Thus ${\mathrm{HOM}}_A(P,Q)$ is a graded abelian group whose degree-$d$ part is ${\mathrm{Hom}}_{A,d}(P,Q)$, and the degree-zero morphisms are ${\mathrm{Hom}}_{A,0}(P,Q)$. By definition ${\mathrm{HOM}}_A(P,Q)$ consists of the finite sums of homogeneous $A$-linear maps, so ${\mathrm{HOM}}_A(P,Q)\subseteq {\mathrm{Hom}}_A^{\mathrm{u}\mathrm{n}\mathrm{g}\mathrm{r}}(P,Q)$ with equality only when every $A$-linear map has bounded spread in degree.

The associative left action. If $M$ is a graded $(B,A)$-bimodule and $Y$ a graded left $B$-module, then ${\mathrm{HOM}}_B(M,Y)$ carries a left $A$-module structure

$(a\cdot f)(m):=f(ma)(a\in A,m\in M),$

which is homogeneous: for $a\in A_i$ and $f\in {\mathrm{Hom}}_{B,d}(M,Y)$ one has $ma\in M_{l+i}$ for $m\in M_l$, hence $(a\cdot f)(m)\in Y_{l+i+d}$ and $a\cdot f\in {\mathrm{Hom}}_{B,d+i}(M,Y)$. The action is associative and unital by the module axioms of $M$, so ${\mathrm{HOM}}_B(M,Y)$ is a graded left $A$-module and an object of ${\mathrm{GrMod}}_0(A)$. The same formula defines an action on the full ungraded $\mathrm{Hom}$ group, where it is denoted by the same symbol.

${\mathrm{HOM}}_A(P,Q)$ need not be all of ${\mathrm{Hom}}_A^{\mathrm{u}\mathrm{n}\mathrm{g}\mathrm{r}}(P,Q)$. Let $k$ be a field, $A=k[x]$ with $\mathrm{deg}x=1$, and $Q=A$; let $P={⨁}_{j\ge 0}A\{-j\}$ with generator $u_j:=1_A\in A\{-j\}$, an element of degree $-j$, and define the $A$-linear map

$f:P⟶A,f(u_j):=x^{2j}(j\ge 0)\mathrm{.}$

For each $j$ the map $f_j$ that sends $u_j$ to $x^{2j}$ and every other $u_l$ to $0$ is $A$-linear and homogeneous of degree $3j$: it kills all pieces $A\{-l{\}}_d$ with $l\mathrm{\ne }j$, and on $A\{-j{\}}_d=A_{d+j}$ it sends $a{u}_j$ to $a{x}^{2j}\in A_{d+3j}$. The pointwise sum ${\sum }_j{f}_j$ equals $f$, because every element of $P$ has finite support, but it is not a finite sum: if $f={\sum }_{t\in T}{g}_t$ with $T$ finite and $g_t$ homogeneous of degree $t$, then evaluating at $u_j$ gives $x^{2j}={\sum }_t{g}_t(u_j)$ with $g_t(u_j)\in A_{t-j}$, and since $x^{2j}$ is a nonzero homogeneous element of $A$ of degree $2j$ the only contribution comes from $t-j=2j$. Hence $g_{3j}(u_j)=x^{2j}\mathrm{\ne }0$ for every $j\ge 0$, so $T$ would have to contain the infinitely many distinct degrees $3j$, contradicting finiteness. Therefore $f\mathrm{\notin }{\mathrm{HOM}}_A(P,A)$, and the inclusion ${\mathrm{HOM}}_A(P,Q)\subseteq {\mathrm{Hom}}_A^{\mathrm{u}\mathrm{n}\mathrm{g}\mathrm{r}}(P,Q)$ is proper in general.