Metamath Proof Explorer


Theorem mat1rhm

Description: There is a ring homomorphism from a ring to the ring of matrices with dimension 1 over this ring. (Contributed by AV, 22-Dec-2019)

Ref Expression
Hypotheses mat1rhmval.k 𝐾 = ( Base ‘ 𝑅 )
mat1rhmval.a 𝐴 = ( { 𝐸 } Mat 𝑅 )
mat1rhmval.b 𝐵 = ( Base ‘ 𝐴 )
mat1rhmval.o 𝑂 = ⟨ 𝐸 , 𝐸
mat1rhmval.f 𝐹 = ( 𝑥𝐾 ↦ { ⟨ 𝑂 , 𝑥 ⟩ } )
Assertion mat1rhm ( ( 𝑅 ∈ Ring ∧ 𝐸𝑉 ) → 𝐹 ∈ ( 𝑅 RingHom 𝐴 ) )

Proof

Step Hyp Ref Expression
1 mat1rhmval.k 𝐾 = ( Base ‘ 𝑅 )
2 mat1rhmval.a 𝐴 = ( { 𝐸 } Mat 𝑅 )
3 mat1rhmval.b 𝐵 = ( Base ‘ 𝐴 )
4 mat1rhmval.o 𝑂 = ⟨ 𝐸 , 𝐸
5 mat1rhmval.f 𝐹 = ( 𝑥𝐾 ↦ { ⟨ 𝑂 , 𝑥 ⟩ } )
6 simpl ( ( 𝑅 ∈ Ring ∧ 𝐸𝑉 ) → 𝑅 ∈ Ring )
7 snfi { 𝐸 } ∈ Fin
8 2 matring ( ( { 𝐸 } ∈ Fin ∧ 𝑅 ∈ Ring ) → 𝐴 ∈ Ring )
9 7 6 8 sylancr ( ( 𝑅 ∈ Ring ∧ 𝐸𝑉 ) → 𝐴 ∈ Ring )
10 1 2 3 4 5 mat1ghm ( ( 𝑅 ∈ Ring ∧ 𝐸𝑉 ) → 𝐹 ∈ ( 𝑅 GrpHom 𝐴 ) )
11 eqid ( mulGrp ‘ 𝑅 ) = ( mulGrp ‘ 𝑅 )
12 eqid ( mulGrp ‘ 𝐴 ) = ( mulGrp ‘ 𝐴 )
13 1 2 3 4 5 11 12 mat1mhm ( ( 𝑅 ∈ Ring ∧ 𝐸𝑉 ) → 𝐹 ∈ ( ( mulGrp ‘ 𝑅 ) MndHom ( mulGrp ‘ 𝐴 ) ) )
14 10 13 jca ( ( 𝑅 ∈ Ring ∧ 𝐸𝑉 ) → ( 𝐹 ∈ ( 𝑅 GrpHom 𝐴 ) ∧ 𝐹 ∈ ( ( mulGrp ‘ 𝑅 ) MndHom ( mulGrp ‘ 𝐴 ) ) ) )
15 11 12 isrhm ( 𝐹 ∈ ( 𝑅 RingHom 𝐴 ) ↔ ( ( 𝑅 ∈ Ring ∧ 𝐴 ∈ Ring ) ∧ ( 𝐹 ∈ ( 𝑅 GrpHom 𝐴 ) ∧ 𝐹 ∈ ( ( mulGrp ‘ 𝑅 ) MndHom ( mulGrp ‘ 𝐴 ) ) ) ) )
16 6 9 14 15 syl21anbrc ( ( 𝑅 ∈ Ring ∧ 𝐸𝑉 ) → 𝐹 ∈ ( 𝑅 RingHom 𝐴 ) )