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 No typesetting found for |- S = ( IDLsrg ` R ) with typecode |-
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 Could not format S = ( IDLsrg ` R ) : No typesetting found for |- S = ( IDLsrg ` R ) with typecode |-
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