Metamath Proof Explorer


Theorem matbas

Description: The matrix ring has the same base set as its underlying group. (Contributed by Stefan O'Rear, 4-Sep-2015)

Ref Expression
Hypotheses matbas.a A = N Mat R
matbas.g G = R freeLMod N × N
Assertion matbas N Fin R V Base G = Base A

Proof

Step Hyp Ref Expression
1 matbas.a A = N Mat R
2 matbas.g G = R freeLMod N × N
3 baseid Base = Slot Base ndx
4 basendxnmulrndx Base ndx ndx
5 3 4 setsnid Base G = Base G sSet ndx R maMul N N N
6 eqid R maMul N N N = R maMul N N N
7 1 2 6 matval N Fin R V A = G sSet ndx R maMul N N N
8 7 fveq2d N Fin R V Base A = Base G sSet ndx R maMul N N N
9 5 8 eqtr4id N Fin R V Base G = Base A