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

1.1L1

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).

1.2L1L2L3

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.

2.1step 1.1step 1.2L1

Steps 1.1 and 1.2 prove claim 1.

2.2step 1.1L1L3L4

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.

3.1step 2.1step 2.2∎

Claims 1 and 2 are established in steps 2.1 and 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.

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

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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