Metamath Proof Explorer


Theorem ascl0

Description: The scalar 0 embedded into a left module corresponds to the 0 of the left module if the left module is also a ring. (Contributed by AV, 31-Jul-2019)

Ref Expression
Hypotheses ascl0.a A = algSc W
ascl0.f F = Scalar W
ascl0.l φ W LMod
ascl0.r φ W Ring
Assertion ascl0 φ A 0 F = 0 W

Proof

Step Hyp Ref Expression
1 ascl0.a A = algSc W
2 ascl0.f F = Scalar W
3 ascl0.l φ W LMod
4 ascl0.r φ W Ring
5 2 lmodfgrp W LMod F Grp
6 3 5 syl φ F Grp
7 eqid Base F = Base F
8 eqid 0 F = 0 F
9 7 8 grpidcl F Grp 0 F Base F
10 6 9 syl φ 0 F Base F
11 eqid W = W
12 eqid 1 W = 1 W
13 1 2 7 11 12 asclval 0 F Base F A 0 F = 0 F W 1 W
14 10 13 syl φ A 0 F = 0 F W 1 W
15 eqid Base W = Base W
16 15 12 ringidcl W Ring 1 W Base W
17 4 16 syl φ 1 W Base W
18 eqid 0 W = 0 W
19 15 2 11 8 18 lmod0vs W LMod 1 W Base W 0 F W 1 W = 0 W
20 3 17 19 syl2anc φ 0 F W 1 W = 0 W
21 14 20 eqtrd φ A 0 F = 0 W