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