Metamath Proof Explorer


Theorem irrednzr

Description: A ring with an irreducible element cannot be the zero ring. (Contributed by Thierry Arnoux, 18-May-2025)

Ref Expression
Hypotheses irrednzr.1
|- I = ( Irred ` R )
irrednzr.2
|- ( ph -> R e. Ring )
irrednzr.3
|- ( ph -> X e. I )
Assertion irrednzr
|- ( ph -> R e. NzRing )

Proof

Step Hyp Ref Expression
1 irrednzr.1
 |-  I = ( Irred ` R )
2 irrednzr.2
 |-  ( ph -> R e. Ring )
3 irrednzr.3
 |-  ( ph -> X e. I )
4 eqid
 |-  ( Base ` R ) = ( Base ` R )
5 1 4 irredcl
 |-  ( X e. I -> X e. ( Base ` R ) )
6 3 5 syl
 |-  ( ph -> X e. ( Base ` R ) )
7 eqid
 |-  ( 0g ` R ) = ( 0g ` R )
8 1 7 irredn0
 |-  ( ( R e. Ring /\ X e. I ) -> X =/= ( 0g ` R ) )
9 2 3 8 syl2anc
 |-  ( ph -> X =/= ( 0g ` R ) )
10 6 9 eldifsnd
 |-  ( ph -> X e. ( ( Base ` R ) \ { ( 0g ` R ) } ) )
11 7 4 ringelnzr
 |-  ( ( R e. Ring /\ X e. ( ( Base ` R ) \ { ( 0g ` R ) } ) ) -> R e. NzRing )
12 2 10 11 syl2anc
 |-  ( ph -> R e. NzRing )