Metamath Proof Explorer


Theorem subrgsubm

Description: A subring is a submonoid of the multiplicative monoid. (Contributed by Mario Carneiro, 15-Jun-2015)

Ref Expression
Hypothesis subrgsubm.1 𝑀 = ( mulGrp ‘ 𝑅 )
Assertion subrgsubm ( 𝐴 ∈ ( SubRing ‘ 𝑅 ) → 𝐴 ∈ ( SubMnd ‘ 𝑀 ) )

Proof

Step Hyp Ref Expression
1 subrgsubm.1 𝑀 = ( mulGrp ‘ 𝑅 )
2 eqid ( Base ‘ 𝑅 ) = ( Base ‘ 𝑅 )
3 2 subrgss ( 𝐴 ∈ ( SubRing ‘ 𝑅 ) → 𝐴 ⊆ ( Base ‘ 𝑅 ) )
4 eqid ( 1r𝑅 ) = ( 1r𝑅 )
5 4 subrg1cl ( 𝐴 ∈ ( SubRing ‘ 𝑅 ) → ( 1r𝑅 ) ∈ 𝐴 )
6 subrgrcl ( 𝐴 ∈ ( SubRing ‘ 𝑅 ) → 𝑅 ∈ Ring )
7 eqid ( 𝑅s 𝐴 ) = ( 𝑅s 𝐴 )
8 7 1 mgpress ( ( 𝑅 ∈ Ring ∧ 𝐴 ∈ ( SubRing ‘ 𝑅 ) ) → ( 𝑀s 𝐴 ) = ( mulGrp ‘ ( 𝑅s 𝐴 ) ) )
9 6 8 mpancom ( 𝐴 ∈ ( SubRing ‘ 𝑅 ) → ( 𝑀s 𝐴 ) = ( mulGrp ‘ ( 𝑅s 𝐴 ) ) )
10 7 subrgring ( 𝐴 ∈ ( SubRing ‘ 𝑅 ) → ( 𝑅s 𝐴 ) ∈ Ring )
11 eqid ( mulGrp ‘ ( 𝑅s 𝐴 ) ) = ( mulGrp ‘ ( 𝑅s 𝐴 ) )
12 11 ringmgp ( ( 𝑅s 𝐴 ) ∈ Ring → ( mulGrp ‘ ( 𝑅s 𝐴 ) ) ∈ Mnd )
13 10 12 syl ( 𝐴 ∈ ( SubRing ‘ 𝑅 ) → ( mulGrp ‘ ( 𝑅s 𝐴 ) ) ∈ Mnd )
14 9 13 eqeltrd ( 𝐴 ∈ ( SubRing ‘ 𝑅 ) → ( 𝑀s 𝐴 ) ∈ Mnd )
15 1 ringmgp ( 𝑅 ∈ Ring → 𝑀 ∈ Mnd )
16 1 2 mgpbas ( Base ‘ 𝑅 ) = ( Base ‘ 𝑀 )
17 1 4 ringidval ( 1r𝑅 ) = ( 0g𝑀 )
18 eqid ( 𝑀s 𝐴 ) = ( 𝑀s 𝐴 )
19 16 17 18 issubm2 ( 𝑀 ∈ Mnd → ( 𝐴 ∈ ( SubMnd ‘ 𝑀 ) ↔ ( 𝐴 ⊆ ( Base ‘ 𝑅 ) ∧ ( 1r𝑅 ) ∈ 𝐴 ∧ ( 𝑀s 𝐴 ) ∈ Mnd ) ) )
20 6 15 19 3syl ( 𝐴 ∈ ( SubRing ‘ 𝑅 ) → ( 𝐴 ∈ ( SubMnd ‘ 𝑀 ) ↔ ( 𝐴 ⊆ ( Base ‘ 𝑅 ) ∧ ( 1r𝑅 ) ∈ 𝐴 ∧ ( 𝑀s 𝐴 ) ∈ Mnd ) ) )
21 3 5 14 20 mpbir3and ( 𝐴 ∈ ( SubRing ‘ 𝑅 ) → 𝐴 ∈ ( SubMnd ‘ 𝑀 ) )