Subring Generated by Subset

There are multiple equivalent definitions of the (sub)ring generated by a subset. In this article, we state these definitions and prove their equivalence.

The first definition is intuitively “applying the operations repeatedly to the original subset.” This makes even more sense when 0,1 are represented as unary operators and not constants. The definition is:

Definition 1 (Ring Generated by Subset.) Let $X \subseteq R$ . The ring generated by $X$ is the subring $S$ of $R$ formed by the elements of $X$ , then 0 and 1, and then all sums, products, and negatives of any elements in $S$ .

Another definition is:

Definition 2 (Ring Generated by Subset.) Equivalently, $S$ is the intersection of all subrings containing $X$ . (This intersection is necessarily a ring.)

Let’s prove the equivalence of these now. First:

Lemma. The arbitrary intersection of any collection of subrings of $R$ is a subring of $R$ .

Proof. Let the intersection be $I = \cap_{a}I_{a}$ . Since each $I_{a}$ is a subring of $R$ , $0,1 \in I_{a}$ , so $0,1 \in I$ . We have:

Addition: If $a,b \in I$ , then $a,b \in I_{a}$ for all $a$ , hence $a + b \in I_{a}$ for all $a$ and $a + b \in I$ .

The rest is similar. This is true of any operation on any algebra (in the universal algebraic sense): for any elements $a_{i}$ , if an operation $f$ on the $a_{i}$ exists in every set in a collection, then $f\left( a_{i} \right)$ exists in the intersection too (by definition of intersection.) Thus, we can state more generally that the intersection of subalgebras of an equational class (at least equational class, I think it would be true of any algebra of any signature too) is itself a subalgebra. $\blacksquare$

Now, let’s add some formalism to Definition 1, so that we can prove things using it.

The ring generated by $X$ is produced as follows:

$S_{1} = X \cup \left\{ 0,1 \right\}$
$S_{i} = \left\{ \text{all sums, products, negatives of } S_{i - 1} \right\}$ ; clearly, $S_{i - 1} \subseteq S_{i}$
$S = \left\{ r \in R \middle| r \in S_{i} \text{ for some } i \right\}$ , or equivalently, $S = \cup_{i}S_{i}$

If $a,b \in S$ , then $a \in S_{i},b \in S_{j}$ for some $i,j$ , so $a,b \in S_{\max(i,j)}$ and $a + b \in S_{\max(i,j)} \rightarrow a + b \in S$ ; similarly we have closure on the other operations so $S$ is a subring. We now show that $S$ is equal to the intersection of all rings containing $X$ .

Let $T$ be any subring containing $X$ . $T$ must include (by closure) each $S_{i}$ , and therefore must include $S$ . But this is true of any such $T$ , so $S$ is part of the intersection of all of them. Conversely, one such $T$ is $S$ itself, so the intersection is part of $S$ . But that then implies what we want. $\blacksquare$

Written some time ago, when I was first studying ring theory.

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