Metamath Proof Explorer


Theorem idlsrgmulrss1

Description: In a commutative ring, the product of two ideals is a subset of the first one. (Contributed by Thierry Arnoux, 16-Jun-2024)

Ref Expression
Hypotheses idlsrgmulrss1.1 ⊢ S = IDLsrg ⁡ R
idlsrgmulrss1.2 ⊢ B = LIdeal ⁡ R
idlsrgmulrss1.3 ⊢ ⊗ ˙ = ⋅ S
idlsrgmulrss1.4 ⊢ · ˙ = ⋅ R
idlsrgmulrss1.5 ⊢ φ → R ∈ CRing
idlsrgmulrss1.6 ⊢ φ → I ∈ B
idlsrgmulrss1.7 ⊢ φ → J ∈ B
Assertion idlsrgmulrss1 ⊢ φ → I ⊗ ˙ J ⊆ I

Proof

Step Hyp Ref Expression
1 idlsrgmulrss1.1 ⊢ S = IDLsrg ⁡ R
2 idlsrgmulrss1.2 ⊢ B = LIdeal ⁡ R
3 idlsrgmulrss1.3 ⊢ ⊗ ˙ = ⋅ S
4 idlsrgmulrss1.4 ⊢ · ˙ = ⋅ R
5 idlsrgmulrss1.5 ⊢ φ → R ∈ CRing
6 idlsrgmulrss1.6 ⊢ φ → I ∈ B
7 idlsrgmulrss1.7 ⊢ φ → J ∈ B
8 eqid ⊢ mulGrp R = mulGrp R
9 eqid ⊢ LSSum ⁡ mulGrp R = LSSum ⁡ mulGrp R
10 1 2 3 8 9 5 6 7 idlsrgmulrval ⊢ φ → I ⊗ ˙ J = RSpan ⁡ R ⁡ I LSSum ⁡ mulGrp R J
11 crngring ⊢ R ∈ CRing → R ∈ Ring
12 rlmlmod ⊢ R ∈ Ring → ringLMod ⁡ R ∈ LMod
13 5 11 12 3syl ⊢ φ → ringLMod ⁡ R ∈ LMod
14 eqid ⊢ Base R = Base R
15 14 2 lidlss ⊢ I ∈ B → I ⊆ Base R
16 6 15 syl ⊢ φ → I ⊆ Base R
17 14 2 lidlss ⊢ J ∈ B → J ⊆ Base R
18 7 17 syl ⊢ φ → J ⊆ Base R
19 6 2 eleqtrdi ⊢ φ → I ∈ LIdeal ⁡ R
20 14 8 9 5 18 19 ringlsmss1 ⊢ φ → I LSSum ⁡ mulGrp R J ⊆ I
21 rlmbas ⊢ Base R = Base ringLMod ⁡ R
22 rspval ⊢ RSpan ⁡ R = LSpan ⁡ ringLMod ⁡ R
23 21 22 lspss ⊢ ringLMod ⁡ R ∈ LMod ∧ I ⊆ Base R ∧ I LSSum ⁡ mulGrp R J ⊆ I → RSpan ⁡ R ⁡ I LSSum ⁡ mulGrp R J ⊆ RSpan ⁡ R ⁡ I
24 13 16 20 23 syl3anc ⊢ φ → RSpan ⁡ R ⁡ I LSSum ⁡ mulGrp R J ⊆ RSpan ⁡ R ⁡ I
25 5 11 syl ⊢ φ → R ∈ Ring
26 eqid ⊢ RSpan ⁡ R = RSpan ⁡ R
27 26 2 rspidlid ⊢ R ∈ Ring ∧ I ∈ B → RSpan ⁡ R ⁡ I = I
28 25 6 27 syl2anc ⊢ φ → RSpan ⁡ R ⁡ I = I
29 24 28 sseqtrd ⊢ φ → RSpan ⁡ R ⁡ I LSSum ⁡ mulGrp R J ⊆ I
30 10 29 eqsstrd ⊢ φ → I ⊗ ˙ J ⊆ I