Sum of Ring Products is Subring of Commutative Ring

Theorem
Let $\struct {R, +, \circ}$ be a commutative ring.

Let $\struct {S, +, \circ}$ and $\struct {T, +, \circ}$ be subrings of $\struct {R, +, \circ}$.

Let $S T$ be defined as:
 * $\ds S T = \set {\sum_{i \mathop = 1}^n s_i \circ t_i: s_1 \in S, t_i \in T, i \in \closedint 1 n}$

Then $S T$ is a subring of $\struct {R, +, \circ}$.

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

Let $x_1, x_2 \in S T$.

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

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

Then:

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