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Metamath Proof Explorer

Theorem znzrh

Description: The ZZ ring homomorphism of Z/nZ is inherited from the quotient ring it is based on. (Contributed by Mario Carneiro, 14-Jun-2015) (Revised by AV, 13-Jun-2019)

Ref Expression
Hypotheses znval2.s 
|- S = ( RSpan ` ZZring )
znval2.u 
|- U = ( ZZring /s ( ZZring ~QG ( S ` { N } ) ) )
znval2.y 
|- Y = ( Z/nZ ` N )
Assertion znzrh 
|- ( N e. NN0 -> ( ZRHom ` U ) = ( ZRHom ` Y ) )

Proof

Step Hyp Ref Expression
1 znval2.s 
 |-  S = ( RSpan ` ZZring )
2 znval2.u 
 |-  U = ( ZZring /s ( ZZring ~QG ( S ` { N } ) ) )
3 znval2.y 
 |-  Y = ( Z/nZ ` N )
4 eqidd 
 |-  ( N e. NN0 -> ( Base ` U ) = ( Base ` U ) )
5 1 2 3 znbas2 
 |-  ( N e. NN0 -> ( Base ` U ) = ( Base ` Y ) )
6 1 2 3 znadd 
 |-  ( N e. NN0 -> ( +g ` U ) = ( +g ` Y ) )
7 6 oveqdr 
 |-  ( ( N e. NN0 /\ ( x e. ( Base ` U ) /\ y e. ( Base ` U ) ) ) -> ( x ( +g ` U ) y ) = ( x ( +g ` Y ) y ) )
8 1 2 3 znmul 
 |-  ( N e. NN0 -> ( .r ` U ) = ( .r ` Y ) )
9 8 oveqdr 
 |-  ( ( N e. NN0 /\ ( x e. ( Base ` U ) /\ y e. ( Base ` U ) ) ) -> ( x ( .r ` U ) y ) = ( x ( .r ` Y ) y ) )
10 4 5 7 9 zrhpropd 
 |-  ( N e. NN0 -> ( ZRHom ` U ) = ( ZRHom ` Y ) )