Sum of All Ring Products is Closed under Addition

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Theorem

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

Let $\left({S, +}\right)$ and $\left({T, +}\right)$ 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 $\left({S T, +}\right)$ is a closed subset of $\left({R, +}\right)$.


Proof

Let $x_1, x_2 \in S T$.

Then:

$\displaystyle x_1 = \sum_{i \mathop = 1}^j s_i \circ t_i, x_2 = \sum_{i \mathop = 1}^k s_i \circ t_i$

for some $s_i, t_i, j, k$, etc.

By renaming the indices, we can express $x_2$ as:

$\displaystyle x_2 = \sum_{i \mathop = j+1}^{j+k} s_i \circ t_i$

and hence:

$\displaystyle x_1 + x_2 = \sum_{i \mathop = 1}^j s_i \circ t_i + \sum_{i \mathop = j+1}^{j+k} s_i \circ t_i = \sum_{i \mathop = 1}^k s_i \circ t_i$

So $x_1 + x_2 \in S T$ and $\left({S T, +}\right)$ is shown to be closed.

$\blacksquare$


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