2
$\begingroup$

I've seen two definitions of graded modules.

Definition 1: Given a graded ring $R = \bigoplus_{n \in \mathbb{Z}}R^n$, an $R$-module $M$ is a graded module if $M$ can be expressed as a direct sum (as an abelian group) $$M = \bigoplus_{n \in \mathbb{Z}}M^n$$ such that $$R^kM^j \subseteq M^{k+j}.$$

Definition 2: A graded module is an indexed family $M = (M_p)_{p \in \mathbb{Z}}$ of $R$-modules for some ring $R$.

I assumed that these two definitions would be equivalent somehow (because they are two definitions of graded modules that I've seen in two different books), but I don't know how one would show that they are equivalent, because in the first case a graded module is, in fact, a module and in the second a graded module is just a family of modules.

For now let's just focus on proving that definition $2$ implies definition $1$. If $M$ is a graded module in the sense of definition $2$ then $M = (M_p)_{p \in \mathbb{Z}}$. Now I can firstly make $R$ into a graded ring in a natural way by defining $R^0 = R$ and $R^i = \{0_R\}$ for all $i \neq 0$. Then $R = \bigoplus_{n \in \mathbb{Z}}R^n$ and $R^nR^m \subseteq R^{n+m}$ making $R$ into a graded module. Furthermore if we define $M' = \bigoplus_{p \in \mathbb{Z}} M^p$ then we would end up with a graded module in the sense of definition $1$.

But this doesn't actually show that definition $2$ implies definition $1$, because it's not as if I've taken some graded $R$-module $N$ in the sense of definition $2$ and shown it's a graded $R$-module in the sense of definition $ 1$.

Are these two definitions in fact equivalent? If so how can I show that they are?

$\endgroup$
  • $\begingroup$ Mmmm it may not be equivalent in the purest sense, but I think you reasoning is enough: if you are given a graded module according to definition 2, then you have automatically defined a unique graded module in the sense of definition 1, and the other way round. $\endgroup$ – Dog_69 Jul 4 '19 at 0:46
  • $\begingroup$ Related: math.stackexchange.com/questions/2768795/… $\endgroup$ – Eric Wofsey Jul 4 '19 at 1:33
3
$\begingroup$

There are two essentially equivalent different ways of talking about "graded" algebraic structures. You can define a graded structure $S$ as a structure together with a direct sum decomposition $S=\bigoplus S_n$, or you can define a graded structure as a sequence of structures $(S_n)$ (in each case, there may be additional requirements about how the different $S_n$ interact with each other). While these are not literally equivalent definitions, they are interchangeable for all relevant purposes: given a graded structure in the first sense you can form the sequence $(S_n)$, and given a graded structure in the second sense you can form an external direct sum $\bigoplus S_n$ of the terms in the sequence, and these two constructions are inverse up to canonical isomorphism. (This sort of equivalence is analogous to how you could define a group as a pair $(G,\cdot)$ or as a quadruple $(G,\cdot,1,{}^{-1})$ and these definitions are not literally equivalent but are just different ways of encapsulating the same data.)

Besides this difference, in the case of graded modules, you can talk about either graded modules over an ordinary ring, or graded modules over a graded ring. The latter notion is more general: any ordinary ring can be considered as a graded ring which is nontrivial only in degree $0$, and then the two notions of graded modules over the ring are the same.

So, here's how your two definitions fit into this. Definition 1 is a definition of graded modules over graded rings using direct sum decompositions, and Definition 2 is a definition of graded modules over ordinary rings using sequences. So, Definition 2 is essentially equivalent to Definition 1 when you restrict to the special case that the graded ring comes from an ordinary ring (that is, it is nontrivial only in degree $0$). You have already described one direction of this equivalence in your question; to go the other direction, given a graded module $M=\bigoplus M_n$ by Definition 1 over a graded ring $R$ which is nontrivial only in degree $0$, the sequence $(M_n)$ is a graded $R$-module according to Definition 2, and this construction is inverse to your construction up to canonical isomorphism.

Finally, just to complete the story, let me give the definitions of graded rings and graded modules over graded rings using sequences rather than direct sum decompositions. We can define a graded ring as a sequence $(R_n)$ of abelian groups together with bilinear "multiplication" maps $R_m\times R_n\to R_{m+n}$ for each $m$ and $n$ such that $(rs)t=r(st)$ whenever $r\in R_i$, $s\in R_j$, $t\in R_k$ for any $i,j,k$, and an element $1\in R_0$ such that $1\cdot x=x\cdot 1=r$ for all $r\in R_n$ for any $n$.

[Given such an object, we can make $\bigoplus R_n$ a graded ring in the other sense, by extending the multiplication maps by bilinearity to give a single multiplication operation on all of $\bigoplus R_n$.]

Then, given such a graded ring $R=(R_n)$, a graded $R$-module can be defined as a sequence $(M_n)$ of abelian groups together with bilinear maps $R_m\times M_n\to M_{m+n}$ for each $m$ and $n$ such that $r(sm)=(rs)m$ whenever $r\in R_i,s\in R_j,m\in M_k$ for any $i,j,k$ and such that $1\cdot m=m$ for all $m\in M_n$ for any $n$.

[Again, given such an object $(M_n)$, we can then make $\bigoplus M_n$ a module over $\bigoplus R_n$ by extending the multiplication maps by bilinearity, and this gives a graded $\bigoplus R_n$-module as in your Definition 1.]

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for your detailed answer, it is really appreciated! $\endgroup$ – Perturbative Jul 4 '19 at 6:39
  • $\begingroup$ Is it then correct to say that while the two definitions aren't truly equivalent, there is a canonical way to pass from one definition to the other so in practice it doesn't matter which one we use? $\endgroup$ – Perturbative Jul 23 '19 at 21:35
  • $\begingroup$ Sure, that's a reasonable thing to say. $\endgroup$ – Eric Wofsey Jul 23 '19 at 21:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.