Ideal of External Direct Sum of Rings

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 rings.

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

For each $$k \in \left[{1 \,. \, . \, n}\right]$$, let:


 * $$R'_k = \left\{{\left({x_1, \ldots, x_n}\right) \in R: \forall j \ne k: x_j = 0}\right\}$$

Then:
 * $$\forall k \in \left[{1 \, . \, . \, n}\right]: R'_k$$ is an ideal of $$R$$.

Proof
Let $$y = \left({y_1, y_2, \ldots, y_n}\right) \in R$$.

Let $$x = \left({x_1, x_2, \ldots, x_n}\right) \in R_k$$

By definition of external direct product, we have:
 * $$x \circ y = \left({x_1 \circ y_1, x_2 \circ y_2, \ldots, x_n \circ y_n}\right)$$
 * $$y \circ x = \left({y_1 \circ x_1, y_2 \circ x_2, \ldots, y_n \circ x_n}\right)$$

But we have, $$\forall j \ne k: x_j \circ y_j = 0 = y_j \circ x_j$$ as $$x_j = 0$$.

So it follows that $$x \circ y \in R_k$$ and $$y \circ x \in R_k$$

Hence $$R'_k$$ is an ideal of $$R$$.