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Let $R$ be a commutative ring with identity $1$. The definition of a (unital) R-module is given as an abelian group $M$ together with an operation $R \times M \to M$, usually just written as $rv$ when $r \in R$ and $v \in M$. The scaling operation satisfies the following conditions: \begin{align} 1v = v ~~~\text{ for all }~~~~ v \in M \\ (rs)v = r(sv)~~~\text{for all }r,s \in R \text{ and all }~v \in M \\ (r+s)v = rv + sv \text{ for all } r,s \in R \text{ and all }~ v\in M \\ r(v+w) = rv + rw~~\text{ for all }r \in R ~~\text{ and }v,w \in M \end{align} It is then stated that "If $M$ is a R-module, so is the set of finite formal linear combinations $L(M)$ of elements of $M$, where in $L(M)$ you are not allowed to simplify, so for instance, $r(v)$ and $1(rv)$ are considered distinct elements if $r \neq 1$."

Is the idea here to endow $L(M)$ with the same operation as $M$ making $L(M)$ an abelian group (as it is currently just a set) and to define the scaling operation $R \times L(M) \to L(M)$ in the same way as we do $R \times M \to M$?

Alex
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1 Answers1

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You won't give $L(M)$ the same addition structure as $M$. An arbitrary element of $L(M)$ has the form $\ell = \sum_{i = 1}^n r_i [m_i]$, where $m_i\in M$, or if you prefer $\ell = \sum_{m\in M} r_m [m]$, but writing in this second way means you must demand that all but finitely many $r_m$ are equal to $0$. Then to add two elements $ \ell = \sum_{m\in M} r_m [m]$ and $\ell' = \sum_{m\in M} r_m' [m]$, you simply add the coefficients of each $[m]$: no performing the addition in $M$ at all! $$ \ell + \ell' = \sum_{m\in M} r_m [m]+\sum_{m\in M} r_m' [m] := \sum_{m\in M} (r_m + r'_m) [m] $$ (one should check that all but finitely many of the $r_m + r'_m$ are equal to $0$ to ensure that the sum is a well-defined element of $L(M)$. And now, the action of $R$ on $L(M)$ is just scaling on the coefficients.

This construction does not depend on $M$ having a $R$-module structure; you can form $L(S)$ for any set $S$ in the same way, and this will have the same addition operation. In fact, if $\#S = \#M$, then $L(M)\cong L(S)$ (see if you can find the isomorphism).

Stahl
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  • Thanks for your answer. What is the notation $[\cdot]$ representing? Why not just $l = \sum_{i=1}^{n}r_im_i$? – Alex Nov 14 '16 at 19:26
  • I use $[m]$ to indicate the difference between $[m]$ and $m$. That is, to distinguish the sum $\sum_{i = 1}^n r_i m_i$ within $M$ from the formal sum in $L(M)$. – Stahl Nov 14 '16 at 19:29
  • What is the difference? I thought $L(M)$ was the set of finite linear combinations, does that not just refer to the addition operation of $M$. – Alex Nov 14 '16 at 19:30
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    $\sum_{i = 1}^n r_i m_i$ combines in $M$. For example, if $m\neq -m$, then $m + (-m) = m + (-1)m = (-1 + 1)m = 0$ in $M$, but $[m] + [-m]\neq 0$ in $L(M)$ (you can't pull the $-1$ out from inside the bracket and simplify: $[m]$ and $[-m]$ can only be added formally because $m\neq -m$). – Stahl Nov 14 '16 at 19:32
  • By added formally you mean you can just state $[m] + [-m]$ without further simplification? – Alex Nov 14 '16 at 19:34
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    Yes, $[m] + [-m]$ is just that: $[m] + [-m]$. Think about addition in $L(M)$ like polynomial addition, where different elements of $m$ give different powers or different variables. If you have the polynomial $x^3 + x^2$, there's no way to simplify that (forgetting for a moment that polynomials have a multiplication structure as well). Or perhaps a better example, the polynomial in two variables $x + y$ cannot be simplified at all. Actually, if you only look at addition, $k[x,y]\cong L({1, x, y, x^2, y^2, \dots})$ as $k$-vector spaces or $k$-modules. – Stahl Nov 14 '16 at 19:47
  • What is the overall structure of $L(M)$ then, with this operation of addition as you described it, is it an an abelian group (as I thought it has to be by definition of R-module)? – Alex Nov 14 '16 at 19:53
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    It is an abelian group with the addition I described in the answer, and the $R$-module structure is multiplication on the coefficients. – Stahl Nov 14 '16 at 19:55
  • If you have a chance, check out my other post. – Alex Nov 14 '16 at 21:02