Sum of Ring Products is Subring of Commutative Ring

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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:

\(\ds x_1 \circ x_2\) \(=\) \(\ds \paren {\sum_{i \mathop = 1}^m s_i \circ t_i} \circ \paren {\sum_{j \mathop = 1}^n s'_j \circ t'_j}\)
\(\ds \) \(=\) \(\ds \sum_{\substack {1 \mathop \le i \mathop \le m \\ 1 \mathop \le j \mathop \le n} } \paren {s_i \circ s'_j} \circ \paren {t_i \circ t'_j}\) 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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