Increasing Union of Subrings is Subring

Theorem

Let $R$ be a ring.

Let $S_0 \subseteq S_1 \subseteq S_2 \subseteq \ldots \subseteq S_i \subseteq \ldots$ be subrings of $R$.

Then the increasing union $S$:

$\ds S = \bigcup_{i \mathop \in \N} S_i$

is a subring of $R$.

Proof

Let $\ds S = \bigcup_{i \mathop \in \N} S_i$.

Clearly $0_R \in S$.

Let $a, b \in S$.

Then $\exists i, j \in \N: a \in S_i, b \in S_j$.

From the construction, we have that either of $S_i$ and $S_j$ contains the other.

Let $l = \max \set {i, j}$ so $a, b \in S_l$.

Then $a + \paren {-b} \in S_l$ and $a b \in S_l$, as $S_l$ is a subring.

Thus $a + \paren {-b} \in S$ and $a b \in S$.

By the Subring Test, $S$ is a subring of $R$.

$\blacksquare$

Sources

• 1970: B. Hartley and T.O. Hawkes: Rings, Modules and Linear Algebra ... (previous) ... (next): $\S 2.3$: Some properties of subrings and ideals: Lemma $2.13 \ \text{(ii)}$