LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Subring criterion: $S \subseteq R$ is a subring if and only if $1_{R} \in S$ and $a - b \in S$ and $a b \in S$ for all $a, b \in S$ ; and an intersection of subrings is a subring

Statement

Let $R$ be a ring (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides) with zero $0_{R}$ and identity $1_{R}$ , and let $S \subseteq R$ . Then:

$S$ is a subring of $R$ (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication) if and only if $1_{R} \in S$ , and $a - b \in S$ and $a b \in S$ for all $a, b \in S$ ;
if $S$ is a nonempty set of subrings of $R$ , then $K = ⋂_{S \in S} S$ is a subring of $R$ . In particular the intersection of two subrings is a subring.

Facts & Assumptions

Given: A ring $R$ with zero $0_{R}$ and identity $1_{R}$ , and a subset $S \subseteq R$ ; $a - b$ abbreviates $a + (- b)$ (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides).

[L1]

A subring is a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication; it is then a ring with the same zero, identity and additive inverses as $R$ (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication).

[L2]

$(R, +, 0_{R})$ is an abelian group (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides, Group and abelian group).

[L3]

One-step subgroup test, written additively: a nonempty $T \subseteq R$ with $a - b \in T$ for all $a, b \in T$ is a subgroup of $(R, +, 0_{R})$ ; and a subgroup contains $0_{R}$ and is closed under addition and under additive inverses (One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $g h^{- 1} \in H$ for all $g, h \in H$ ; the identity and the inverses of $H$ are then those of $G$ , Subgroup).

[L4]

The intersection of a nonempty set of subgroups of a group is a subgroup (The intersection of a nonempty family of subgroups of $G$ is a subgroup of $G$ ).

Proof

technique · direct

1.1

Suppose $S$ is a subring. Then $1_{R} \in S$ by (T1); for $a, b \in S$ we have $- b \in S$ by (T3) and hence $a - b = a + (- b) \in S$ by (T2); and $a b \in S$ by (T4).

1.2

Conversely, suppose $1_{R} \in S$ and that $a - b \in S$ and $a b \in S$ for all $a, b \in S$ . Then $S$ is nonempty, so by the one-step test it is a subgroup of $(R, +, 0_{R})$ ; hence $0_{R} \in S$ , $S$ is closed under addition and $- x \in S$ for every $x \in S$ . Together with $1_{R} \in S$ and closure under multiplication, that is exactly (T1) to (T4), so $S$ is a subring.

L1L2L3

2.1

Steps 1.1 and 1.2 prove claim 1.

step 1.1step 1.2L1

2.2

Claim 2. Each $S \in S$ is a subgroup of $(R, +, 0_{R})$ by [L1] and [L3], so $K$ is a subgroup of $(R, +, 0_{R})$ by [L4]; in particular $K$ is closed under addition and under additive inverses. Also $1_{R} \in S$ for every $S \in S$ , so $1_{R} \in K$ ; and if $a, b \in K$ then $a b \in S$ for every $S \in S$ , so $a b \in K$ . Hence $K$ satisfies (T1) to (T4) and is a subring.

step 1.1L1L3L4

3.1

Claims 1 and 2 are established in steps 2.1 and 2.2.

step 2.1step 2.2∎

Remarks

Why the identity clause survives the economy. The one-step test compresses the three additive conditions into one, and nothing compresses (T1): a subset can satisfy $a - b \in S$ and $a b \in S$ and still miss $1_{R}$ altogether. The companion page's even integers are exactly that, which is why $1_{R} \in S$ is stated separately in claim 1 rather than folded in.
Nonemptiness comes for free here. One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $g h^{- 1} \in H$ for all $g, h \in H$ ; the identity and the inverses of $H$ are then those of $G$ requires the subset to be nonempty; in claim 1 that is supplied by $1_{R} \in S$ , so the criterion has no separate nonemptiness hypothesis.
Claim 2 is what would make "the subring generated by a set" meaningful. Intersections of subrings being subrings, the intersection of all subrings containing a given subset is the smallest one containing it. No such construction is used on this page; the claim is recorded because it costs one step and because The intersection of a nonempty family of subgroups of $G$ is a subgroup of $G$ already supplies the additive half.

Depends on

Used by

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Subring (Wikipedia) (standard reference, not scraped)
Subgroup test (Wikipedia) (standard reference, not scraped)
Thomas W. Judson, Abstract Algebra: Theory and Applications, §16.3: Rings (standard reference, not scraped)

Subring criterion: S⊆RS⊆R is a subring if and only if 1R∈S1R​∈S and a−b∈Sa−b∈S and ab∈Sab∈S for all a,b∈Sa,b∈S; and an intersection of subrings is a subring