# External Direct Sum of Rings is Ring

## 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.

Then their (external) direct product:

$\displaystyle \left({R, +, \circ}\right) = \prod_{k \mathop = 1}^n \left({R_k, +_k, \circ_k}\right)$

is a ring.

## Proof

Consider the structures $\left({R_1, +_1}\right), \left({R_2, +_2}\right), \ldots, \left({R_n, +_n}\right)$.

By the definition of a ring, these are all groups.

From External Direct Product of Groups is Group we have that the their external direct product:

$\displaystyle \left({R, +}\right) = \prod_{k \mathop = 1}^n \left({R_k, +_k}\right)$

is a group.

Similarly, consider the structures $\left({R_1, \circ_1}\right), \left({R_2, \circ_2}\right), \ldots, \left({R_n, \circ_n}\right)$.

By the definition of a ring, these are all semigroups.

From External Direct Product of Semigroups we have that the their external direct product:

$\displaystyle \left({R, \circ}\right) = \prod_{k \mathop = 1}^n \left({R_k, \circ_k}\right)$

is a semigroup.

Finally we note that from External Direct Product Distributivity, $\circ$ as defined here is distributive over $+$.

Hence the result, by definition of ring.

$\blacksquare$