External Direct Sum of Rings is Ring

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Theorem

Let $\struct {R_1, +_1, \circ_1}, \struct {R_2, +_2, \circ_2}, \ldots, \struct {R_n, +_n, \circ_n}$ be rings.


Then their (external) direct product:

$\ds \struct {R, +, \circ} = \prod_{k \mathop = 1}^n \struct {R_k, +_k, \circ_k}$

is a ring.


Proof

Consider the structures $\struct {R_1, +_1}, \struct {R_2, +_2}, \ldots, \struct {R_n, +_n}$.

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:

$\ds \struct {R, +} = \prod_{k \mathop = 1}^n \struct {R_k, +_k}$

is a group.


Similarly, consider the structures $\struct {R_1, \circ_1}, \struct {R_2, \circ_2}, \ldots, \struct {R_n, \circ_n}$.

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

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

$\ds \struct {R, \circ} = \prod_{k \mathop = 1}^n \struct {R_k, \circ_k}$

is a semigroup.


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


Hence the result, by definition of ring.

$\blacksquare$


Sources

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