Metamath Proof Explorer


Theorem gsumvsmul1

Description: Pull a scalar multiplication out of a sum of vectors. This theorem properly generalizes gsummulc1 , since every ring is a left module over itself. (Contributed by Thierry Arnoux, 12-Jun-2023)

Ref Expression
Hypotheses gsumvsmul1.b B = Base R
gsumvsmul1.s S = Scalar R
gsumvsmul1.k K = Base S
gsumvsmul1.z 0 ˙ = 0 S
gsumvsmul1.t · ˙ = R
gsumvsmul1.r φ R LMod
gsumvsmul1.1 φ S CMnd
gsumvsmul1.a φ A V
gsumvsmul1.x φ Y B
gsumvsmul1.y φ k A X K
gsumvsmul1.n φ finSupp 0 ˙ k A X
Assertion gsumvsmul1 φ R k A X · ˙ Y = S k A X · ˙ Y

Proof

Step Hyp Ref Expression
1 gsumvsmul1.b B = Base R
2 gsumvsmul1.s S = Scalar R
3 gsumvsmul1.k K = Base S
4 gsumvsmul1.z 0 ˙ = 0 S
5 gsumvsmul1.t · ˙ = R
6 gsumvsmul1.r φ R LMod
7 gsumvsmul1.1 φ S CMnd
8 gsumvsmul1.a φ A V
9 gsumvsmul1.x φ Y B
10 gsumvsmul1.y φ k A X K
11 gsumvsmul1.n φ finSupp 0 ˙ k A X
12 lmodcmn R LMod R CMnd
13 cmnmnd R CMnd R Mnd
14 6 12 13 3syl φ R Mnd
15 1 2 5 3 lmodvslmhm R LMod Y B x K x · ˙ Y S GrpHom R
16 6 9 15 syl2anc φ x K x · ˙ Y S GrpHom R
17 ghmmhm x K x · ˙ Y S GrpHom R x K x · ˙ Y S MndHom R
18 16 17 syl φ x K x · ˙ Y S MndHom R
19 oveq1 x = X x · ˙ Y = X · ˙ Y
20 oveq1 x = S k A X x · ˙ Y = S k A X · ˙ Y
21 3 4 7 14 8 18 10 11 19 20 gsummhm2 φ R k A X · ˙ Y = S k A X · ˙ Y