Conditions for Internal Ring Direct Sum

Theorem
Let $$\left({R, +, \circ}\right)$$ be a ring.

Let $$\left \langle {\left({S_k, +, \circ}\right)} \right \rangle$$ be a sequence of subrings of $$R$$.

Then $$R$$ is the ring direct sum of $$\left \langle {S_k} \right \rangle_{1 \le k \le n}$$ iff:


 * $$(1) \quad R = S_1 + S_2 + \cdots + S_n$$


 * $$(2) \quad \left \langle {\left({S_k, +}\right)} \right \rangle_{1 \le k \le n}$$ is a sequence of independent subgroups of $$\left({R, +}\right)$$


 * $$(3) \quad \forall k \in \left[{1 \, . \, . \, n}\right]: S_k$$ is an ideal of $$R$$.

Proof
Let $$S$$ be the cartesian product of $$\left \langle {\left({S_k, +}\right)} \right \rangle$$

First note that $$\phi$$ is a group homomorphism from $$\left({S, +}\right)$$ to $$\left({R, +}\right)$$, as:
 * $$\sum_{j=1}^n \left({x_j + y_j}\right) = \sum_{j=1}^n x_j + \sum_{j=1}^n y_j$$ from Associativity on Indexing Set.

So $$\phi$$ is a ring homomorphism iff:
 * $$\left({\sum_{j=1}^n x_j}\right) \circ \left({\sum_{j=1}^n y_j}\right) = \sum_{j=1}^n \left({x_j \circ y_j}\right)$$

Now, let:
 * $$\left({S, +, \circ}\right)$$ be the cartesian product of $$\left \langle {\left({S_k, +, \circ}\right)} \right \rangle$$
 * $$\phi: S \to R$$ be the mapping defined as:
 * $$\phi\left({\left({x_1, x_2, \ldots, x_n}\right)}\right) = x_1 + x_2 + \cdots x_n$$

Clearly $$\phi$$ is a surjection iff $$(1)$$ holds.

By Internal Direct Product Generated by Subgroups, $$\phi$$ is a ring isomorphism iff $$(1)$$ and $$(2)$$ hold.

Let $$\phi: S \to R$$ be a ring isomorphism.

By Canonical Injection from Ideal of External Direct Product of Rings, $$\operatorname{in}_k \left({S_k}\right)$$ is an ideal of $$S$$.

So $$\phi \left({\operatorname{in}_k \left({S_k}\right)}\right)$$ is an ideal of $$R$$.

But $$\phi$$ and $$\operatorname{pr}_k$$ coincide on $$\operatorname{in}_k \left({S_k}\right)$$.

So:
 * $$\phi \left({\operatorname{in}_k \left({S_k}\right)}\right) = \operatorname{pr}_k \left({\operatorname{in}_k \left({S_k}\right)}\right) = S_k$$

and so $$(3)$$ holds.

Now suppose $$(3)$$ holds, and the $$S_k$$ are all ideals of $$R$$.

By $$(2)$$ and Condition for Subgroups to be Independent, we have:
 * $$i \ne j \implies S_i \cap S_j = \left\{{0_R}\right\}$$

So for all $$\left({x_1, x_2, \ldots, x_n}\right), \left({y_1, y_2, \ldots, y_n}\right) \in S$$:

$$ $$ $$ $$

because as $$S_i, S_j$$ are ideals, we have:
 * $$x_i \circ j_j = S_i \cap S_j = \left\{{0}\right\}$$

Hence the three conditions are sufficient for $$\phi$$ to be a ring isomorphism.