Sum of Ring Products is Subring of Commutative Ring

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Theorem

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

Let $\left({S, +, \circ}\right)$ and $\left({T, +, \circ}\right)$ be subrings 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 $S T$ is a subring of $\left({R, +, \circ}\right)$.


Proof

From Sum of All Ring Products is Additive Subgroup we have that $\left({S T, +}\right)$ is an additive subgroup of $R$.


Let $x_1, x_2 \in S T$.

Then:

$\displaystyle x_1 = \sum_{i \mathop = 1}^m s_i \circ t_i, x_2 = \sum_{i \mathop = 1}^n s_j \circ t_j$

for some $s_i, t_i, s_j, t_j, m, n$, etc.


Then:

\(\displaystyle x_1 \circ x_2\) \(=\) \(\displaystyle \left({\sum_{i \mathop = 1}^m s_i \circ t_i}\right) \circ \left({\sum_{j \mathop = 1}^n s'_j \circ t'_j}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le n} } \left({s_i \circ s'_j}\right) \circ \left({t_i \circ t'_j}\right)\) as $\circ$ is commutative

So $x_1 \circ x_2 \in S T$ and the result follows from the Subring Test.

$\blacksquare$


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