External Direct Product Distributivity

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Theorem

Let $\left({R_1, +_1, \circ_1}\right), \left({R_2, +_2, \circ_2}\right), \ldots, \left({R_n, +_n, \circ_n}\right)$ be ringoids.

Let their external direct product be $\displaystyle \left({R, +, \circ}\right) = \prod_{k \mathop = 1}^n \left({R_k, +_k, \circ_k}\right)$.

Then the operation $\circ$ is distributive over $+$.


Proof

As all the $\left({R_k, +_k, \circ_k}\right)$ are ringoids, $\circ_k$ distributes over $+_k$ for all $k$.


Let $x, y, z \in R$.

Then:

\(\displaystyle x \circ \left({y + z}\right)\) \(=\) \(\displaystyle \left({x_1, x_2, \ldots, x_n}\right) \circ \left({\left({y_1, y_2, \ldots, y_n}\right) + \left({z_1, z_2, \ldots, z_n}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({x_1, x_2, \ldots, x_n}\right) \circ \left({y_1 + z_1, y_2 + z_2, \ldots, y_n + z_n}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({x_1 \circ \left({y_1 + z_1}\right), x_2 \circ \left({y_2 + z_2}\right), \ldots, x_n \circ \left({y_n + z_n}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({\left({x_1 \circ z_1}\right) + \left({y_1 \circ z_1}\right), \left({x_2 \circ z_2}\right) + \left({y_2 \circ z_2}\right), \ldots, \left({x_n \circ z_n}\right) + \left({y_n \circ z_n}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({\left({x_1 \circ z_1}\right), \left({x_2 \circ z_2}\right), \ldots, \left({x_n \circ z_n}\right)}\right) + \left({\left({y_1 \circ z_1}\right) + \left({y_2 \circ z_2}\right) + \ldots + \left({y_n \circ z_n}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({\left({x_1, x_2, \ldots, x_n}\right) \circ \left({z_1, z_2, \ldots, z_n}\right)}\right) + \left({\left({x_1, x_2, \ldots, x_n}\right) \circ \left({z_1, z_2, \ldots, z_n}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({x \circ y}\right) + \left({x \circ z}\right)\)


In the same way:

$\left({y + z}\right) \circ x = \left({y \circ x}\right) + \left({z \circ x}\right)$

$\blacksquare$