I'm quite confused about the definition of $\Bbb ZG$ modules and $\Bbb ZG$ rings.

  • Is the definition of $\Bbb ZG$ module equivalent to "free $\Bbb Z$-module on $G$"?
  • Given $G$, are the underlying set of $\Bbb ZG$ module and $\Bbb ZG$ ring exactly the same?
  • Given $G$, are the addition operation of $\Bbb ZG$ module and $\Bbb ZG$ ring exactly the same?
  • When $G$ is finite, is the below picture(from Dummit and Foote) the definition of $\Bbb ZG$ ring? And is the addition operation defined below also the addition operation of $\Bbb ZG$ module when $G$ is finite?enter image description here
  • Finally, is (iv) in the below picture the standard definition of $\Bbb ZG$ ring when $G$ is infinite? Why does the $x^{-1}$ appear? How to think about it? enter image description here

Let me try to clarify matters, in the order you addressed them.

  1. NO, a $\mathbb{Z}[G]$-module emphatically does NOT mean the free $\mathbb{Z}$-module $\mathbb{Z}^{(G)}$ defined over set $G$, even when $G$ were endowed with a group structure.
  2. To delineate the conceptual approach, one first realises that given any arbitrary ring $A$ (whether commutative or not, doesn't matter) and any arbitrary monoid $M$ (whether finite or not, finiteness is not a prerequisite), to these two objects a new ring can be associated in a canonical (natural) manner. The associated ring is called the monoid ring of $M$ over $A$ (equivalently, with coefficients in $A$), and denoted by $A[M]$. This construction will of course apply in particular for monoids that are actually groups, and the corresponding object will be correspondingly called group ring. The notation is inspired by the traditional notation for polynomial rings, $A[X]$ et al, as these rings are themselves standard examples of monoid rings (over free commutative monoids).
  3. For an already given and fixed pair of ring $A$ and monoid $M$, we refer to modules whose ring of scalars is the monoid ring $A[M]$ as $A[M]$-modules, in accordance with the general terminology: if $T$ is a module over the ring $A$, then we call it an $A$-module. Therefore, if you are given an $A[M]$-module $T$, the internal addition of $T$ is NOT the same as the internal addition of the monoid ring $A[M]$.
  4. The quoted paragraph serves indeed as a description of group rings (yet no more, as it does not rigorously define ''formal sums'' with a proper grounding in set-theory, it just gives you a ''feel'' of what the elements look like and how you perform the algebraic operations on them). The definition goes explicitly as follows: for given ring $A$ and monoid $M$ consider first the direct product of additive groups $(A, +)_{t \in M}$, namely

$$A^M= \prod_{t \in M} A$$ Then, for arbitrary $u \in A[M]$ define the support $\mathrm{Supp}\ u=\{t \in M\ |\ u_t \neq 0_A\}$ and introduce the subset $$A^{(M)}=\{u \in A^M\ |\ \mathrm{Supp}\ u\ \mathrm{finite} \}$$

which is easily seen to be a subgroup of the direct product $A^M$, namely the direct sum of the additive groups $(A, +)_{t \in M}$. At this point you thus already have an internal addition on the set $A^{(M)}$, conferring an abelian group structure.

Furthermore, define the multiplicative operation $\cdot : A^{(M)} \times A^{(M)} \to A^{(M)}$ such that for any $u, v \in A^{(M)}$ and any $t \in M$:

$$(uv)_t=\sum_{r, s \in M \\ rs=t} u_rv_s \tag{1}$$

what we would call multiplication by convolution, as it intertwines the internal multiplication of the ring $A$ with that of the monoid $M$.

A routine check will establish that this multiplication on $A^{(M)}$ will in conjunction with the addition described above yield a ring structure on $A^{(M)}$. This is the monoid ring, and from this point on you can denote it by $A[M]$, to highlight the fact that you are considering the entire ring structure on it, and not just the additive group direct sum $A^{(M)}$ (nevertheless, as support sets they are one and the same thing, $A[M]=A^{(M)}$, it is merely a matter of what algebraic structure one is focusing on).

  1. In the special case you are examining the internal multiplication on a group ring $A[G]$ (so $G$ is assumed to be a group this time, not merely a monoid), fixing elements $u, v \in A[G]$ and $t \in G$, let us refer to definition (1): notice there that in order to obtain the $t$-component of the product $uv$ you must perform a summation over indices which are ordered pairs of elements $(r,s)$ of $G$, subject to the condition $rs=t$. As in a group all elements are symmetrizable, this set of ordered pairs - your index set for the summation - can actually be written more simply as $\{(r, r^{-1}t)\}_{r \in G}$ (bijectively paramaterized by $r \in G$) or equivalently as $\{(ts^{-1}, s)_{s \in G}$ (bijectively parameterized by $s$).
  2. The way to justify the introduction of this convolutive multiplication is to think about how you multiply ordinary one-indeterminate integer polynomials, for instance: in order to obtain the coefficient at degree $n$ for a product $fg$ you have to evaluate

$$\sum_{0 \leqslant k \leqslant n} f_kg_{n-k}$$ (where by $f_k$ I mean the coefficient at degree $k$ in $f$). In other words, you are considering all pairs $(k, l)$ of natural numbers such that $k+l=n$ and for every such pair you are contributing with the term $f_kg_l$ to the sum considered over all such pairs. That is because the one indeterminate polynomial ring $A[X]$ is actually canonically (naturally) isomorphic to the monoid ring $A[\mathbb{N}]$!

  • $\begingroup$ Thanks! Keep reading the answer. A little question I'm thinking about, is $\Bbb ZG=\oplus_{i\in G}\Bbb Z$? (ignoring the operations, just care about the sets itself) $\endgroup$ – Eric Apr 6 at 12:48
  • 1
    $\begingroup$ Yes of course. The support set of $A[M]$ is obtained as the support set of $A^{(M)}=\bigoplus_{t \in M} A$. In general that is how you define the direct sum of a family of modules (over a given ring), $\bigoplus_{i \in I} M_i \subseteq \prod_{i \in I} M_i$ containing only those families of finite support. $\endgroup$ – ΑΘΩ Apr 6 at 13:45

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