Sum of All Ring Products is Associative

Theorem

Let $\left({R, +, \circ}\right)$ be a ring.

Let $\left({S, +}\right), \left({T, +}\right), \left({U, +}\right)$ be additive subgroups of $\left({R, +, \circ}\right)$.

Let $S T$ be defined as:

$\displaystyle S T = \left\{{\sum_{i \mathop = 1}^n s_i \circ t_i: s_1 \in S, t_i \in T, i \in \left[{1 \,.\,.\,n}\right]}\right\}$

Then:

$\left({S T}\right) U = S \left({T U}\right)$

We have by definition that $S T$ is made up of all finite sums of elements of the form $s \circ t$ where $s \in S, t \in T$.

Therefore, so are $\left({S T}\right) U$ and $S \left({T U}\right)$.

Let $z \in \left({S T}\right) U$.

Then $z$ is a finite sum of elements in the form $x \circ u$ where $x \in ST$ and $u \in U$.

So $x$ is a finite sum of elements in the form $s \circ t$ where $s \in S, t \in T$.

Therefore $z$ is a finite sum of elements in the form $\left({s \circ t}\right) \circ u$ where $s \in S, t \in T, u \in U$.

As $\left({R, +, \circ}\right)$ os a ring, $\circ$ is associative.

So $z$ is a finite sum of elements in the form $s \circ \left({t \circ u}\right)$ where $s \in S, t \in T, u \in U$.

So these elements all belong to $S \left({T U}\right)$.

Since $S \left({T U}\right)$ is closed under addition, $z \in S \left({T U}\right)$.

So:

$\left({S T}\right) U \subseteq S \left({T U}\right)$

By a similar argument in the other direction, $S \left({T U}\right) \subseteq \left({S T}\right) U $

and so by definition of set equality:

$\left({S T}\right) U = S \left({T U}\right)$

$\blacksquare$