Direct sum of graded modules is *isomorphic* to a graded module?

Question

Let $R$ be a graded ring and let $M_1,\ldots,M_m$ be graded $R$-modules. We can form the direct sum $N=\bigoplus_{i=1}^m M_i$, which according to my textbook is also a graded $R$-module using the grading $N_t:=(M_1)_t\oplus\cdots\oplus (M_m)_t$ for $t\geq0$. It is clear to me that $\bigoplus_{t\geq0}N_t$ is a graded $R$-module, but we only have $N\cong\bigoplus_{t\geq0} N_t$ rather than the equality $N=\bigoplus_{t\geq0} N_t$. So strictly speaking, $N$ is isomorphic to a graded $R$-module and therefore inherits a grading from this isomorphism? Is this the correct idea? My book makes no mention of this subtlety so I just wanted to check to make sure I'm interpreting this correctly.

Answer 1

There are different conventions of what exactly a “graded $R$-module $M$” over a graded ring $R$ is, for precisely such reasons. There are basically two flavours of definition:

• A single $R$-module $M$ together with a family $(M_t)_t$ of abelian subgroups such that $M = ⨁_t M_t$ (as abelian groups) and $R_s M_t ⊆ M_{s + t}$ for all $s, t$.

• A family $(M_t)_t$ of abelian groups together with multiplication maps $$ α_{s, t} \colon R_s × M_t \to M_{s + t} $$ satisfying certain compatibility conditions. (Each $α_{s, t}$ must be additive in both components; each $α_{0, t}$ must have the element $1_R$ of $R_0$ act on $M_t$ as the identity; both ways of going from $R_s × R_t × M_u$ to $M_{s + t + u}$ must agree.)

One might think about the first kind of definition as an “internal” one, and as the second kind as an “external” one. Both definitions lead to essentially the same kinds of objects and the same theory, but have different advantages and disadvantages.

The book by you refer to (“Using Algebraic Geometry” by Cox, Little and O’Shea) uses the first, “internal” definition. (Namely Definition 3.2 on page 266.) This definition can lead to some slight set-theoretic issues stemming – for example – from the set-theoretic difference between internal direct sums (as used in this definition of a graded module) and external direct sums (as often used for the construction of new modules).

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However, I need to point out that this problem doesn’t occur in the given example. In the definition of $N$ as $N = ⨁_{i = 1}^m M_i$ we use the exernal direct sum; in other words, $$ N = \{ (x_1, \dotsc, x_n) \mid x_i ∈ M_i \} \,. $$ Each $(M_i)_t$ is an abelian subgroup of $M_i$, whence the external direct sum $$ N_t ≔ (M_1)_t ⊕ \dotsb ⊕ (M_m)_t = \{ (x_1, \dotsc, x_m) \mid x_i ∈ (M_i)_t \} $$ is again an actual subgroup of $N$. We therefore have the internal (!) direct sum decomposition $$ N = \bigoplus_t N_t \,. $$ We can therefore apply the internal definition of a graded module without problem.

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It is nevertheless true that such set-theoretic problems can occur; which is why some people prefer to work with the external definition, or to just ignore such set-theoretic subtleties altogether.

Answer 2

When we say "a module $M$ is graded if it admits a direct sum decomposition $M=\bigoplus_t M_t$", what it actually means is: there is a choice of submodule $M_t$ for each $t$, such that the induced canonical morphism $\bigoplus_t M_t\to M$ is an isomorphism. So there is no need to fret about strict equalities, they will basically never happen anyway.

If all $M_i$ are graded, then choosing $N_t = (M_1)_t\oplus\dots\oplus (M_m)_t$ does induce a canonical isomorphism $\bigoplus_t N_t\to N$, so this is indeed a grading on $N$.