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=BaseR
scmatrhmval.a A=NMatR
scmatrhmval.o 1˙=1A
scmatrhmval.t ˙=A
scmatrhmval.f F=xKx˙1˙
scmatrhmval.c C=NScMatR
Assertion scmatrhmcl NFinRRingXKFXC

Proof

Step Hyp Ref Expression
1 scmatrhmval.k K=BaseR
2 scmatrhmval.a A=NMatR
3 scmatrhmval.o 1˙=1A
4 scmatrhmval.t ˙=A
5 scmatrhmval.f F=xKx˙1˙
6 scmatrhmval.c C=NScMatR
7 1 2 3 4 5 scmatrhmval RRingXKFX=X˙1˙
8 7 3adant1 NFinRRingXKFX=X˙1˙
9 3simpa NFinRRingXKNFinRRing
10 simp3 NFinRRingXKXK
11 2 matring NFinRRingARing
12 11 3adant3 NFinRRingXKARing
13 eqid BaseA=BaseA
14 13 3 ringidcl ARing1˙BaseA
15 12 14 syl NFinRRingXK1˙BaseA
16 1 2 13 4 matvscl NFinRRingXK1˙BaseAX˙1˙BaseA
17 9 10 15 16 syl12anc NFinRRingXKX˙1˙BaseA
18 oveq1 c=Xc˙1˙=X˙1˙
19 18 eqeq2d c=XX˙1˙=c˙1˙X˙1˙=X˙1˙
20 19 adantl NFinRRingXKc=XX˙1˙=c˙1˙X˙1˙=X˙1˙
21 eqidd NFinRRingXKX˙1˙=X˙1˙
22 10 20 21 rspcedvd NFinRRingXKcKX˙1˙=c˙1˙
23 1 2 13 3 4 6 scmatel NFinRRingX˙1˙CX˙1˙BaseAcKX˙1˙=c˙1˙
24 23 3adant3 NFinRRingXKX˙1˙CX˙1˙BaseAcKX˙1˙=c˙1˙
25 17 22 24 mpbir2and NFinRRingXKX˙1˙C
26 8 25 eqeltrd NFinRRingXKFXC