Metamath Proof Explorer


Theorem scmatrhmcl

Description: The value of the ring homomorphism F is a scalar matrix. (Contributed by AV, 22-Dec-2019)

Ref Expression
Hypotheses scmatrhmval.k K = Base R
scmatrhmval.a A = N Mat R
scmatrhmval.o 1 ˙ = 1 A
scmatrhmval.t ˙ = A
scmatrhmval.f F = x K x ˙ 1 ˙
scmatrhmval.c C = N ScMat R
Assertion scmatrhmcl N Fin R Ring X K F X C

Proof

Step Hyp Ref Expression
1 scmatrhmval.k K = Base R
2 scmatrhmval.a A = N Mat R
3 scmatrhmval.o 1 ˙ = 1 A
4 scmatrhmval.t ˙ = A
5 scmatrhmval.f F = x K x ˙ 1 ˙
6 scmatrhmval.c C = N ScMat R
7 1 2 3 4 5 scmatrhmval R Ring X K F X = X ˙ 1 ˙
8 7 3adant1 N Fin R Ring X K F X = X ˙ 1 ˙
9 3simpa N Fin R Ring X K N Fin R Ring
10 simp3 N Fin R Ring X K X K
11 2 matring N Fin R Ring A Ring
12 11 3adant3 N Fin R Ring X K A Ring
13 eqid Base A = Base A
14 13 3 ringidcl A Ring 1 ˙ Base A
15 12 14 syl N Fin R Ring X K 1 ˙ Base A
16 1 2 13 4 matvscl N Fin R Ring X K 1 ˙ Base A X ˙ 1 ˙ Base A
17 9 10 15 16 syl12anc N Fin R Ring X K X ˙ 1 ˙ Base A
18 oveq1 c = X c ˙ 1 ˙ = X ˙ 1 ˙
19 18 eqeq2d c = X X ˙ 1 ˙ = c ˙ 1 ˙ X ˙ 1 ˙ = X ˙ 1 ˙
20 19 adantl N Fin R Ring X K c = X X ˙ 1 ˙ = c ˙ 1 ˙ X ˙ 1 ˙ = X ˙ 1 ˙
21 eqidd N Fin R Ring X K X ˙ 1 ˙ = X ˙ 1 ˙
22 10 20 21 rspcedvd N Fin R Ring X K c K X ˙ 1 ˙ = c ˙ 1 ˙
23 1 2 13 3 4 6 scmatel N Fin R Ring X ˙ 1 ˙ C X ˙ 1 ˙ Base A c K X ˙ 1 ˙ = c ˙ 1 ˙
24 23 3adant3 N Fin R Ring X K X ˙ 1 ˙ C X ˙ 1 ˙ Base A c K X ˙ 1 ˙ = c ˙ 1 ˙
25 17 22 24 mpbir2and N Fin R Ring X K X ˙ 1 ˙ C
26 8 25 eqeltrd N Fin R Ring X K F X C