Metamath Proof Explorer


Theorem rspecbas

Description: The prime ideals form the base of the spectrum of a ring. (Contributed by Thierry Arnoux, 2-Jun-2024)

Ref Expression
Hypothesis rspecbas.1 𝑆 = ( Spec ‘ 𝑅 )
Assertion rspecbas ( 𝑅 ∈ Ring → ( PrmIdeal ‘ 𝑅 ) = ( Base ‘ 𝑆 ) )

Proof

Step Hyp Ref Expression
1 rspecbas.1 𝑆 = ( Spec ‘ 𝑅 )
2 prmidlssidl ( 𝑅 ∈ Ring → ( PrmIdeal ‘ 𝑅 ) ⊆ ( LIdeal ‘ 𝑅 ) )
3 eqid ( IDLsrg ‘ 𝑅 ) = ( IDLsrg ‘ 𝑅 )
4 eqid ( LIdeal ‘ 𝑅 ) = ( LIdeal ‘ 𝑅 )
5 3 4 idlsrgbas ( 𝑅 ∈ Ring → ( LIdeal ‘ 𝑅 ) = ( Base ‘ ( IDLsrg ‘ 𝑅 ) ) )
6 2 5 sseqtrd ( 𝑅 ∈ Ring → ( PrmIdeal ‘ 𝑅 ) ⊆ ( Base ‘ ( IDLsrg ‘ 𝑅 ) ) )
7 eqid ( ( IDLsrg ‘ 𝑅 ) ↾s ( PrmIdeal ‘ 𝑅 ) ) = ( ( IDLsrg ‘ 𝑅 ) ↾s ( PrmIdeal ‘ 𝑅 ) )
8 eqid ( Base ‘ ( IDLsrg ‘ 𝑅 ) ) = ( Base ‘ ( IDLsrg ‘ 𝑅 ) )
9 7 8 ressbas2 ( ( PrmIdeal ‘ 𝑅 ) ⊆ ( Base ‘ ( IDLsrg ‘ 𝑅 ) ) → ( PrmIdeal ‘ 𝑅 ) = ( Base ‘ ( ( IDLsrg ‘ 𝑅 ) ↾s ( PrmIdeal ‘ 𝑅 ) ) ) )
10 6 9 syl ( 𝑅 ∈ Ring → ( PrmIdeal ‘ 𝑅 ) = ( Base ‘ ( ( IDLsrg ‘ 𝑅 ) ↾s ( PrmIdeal ‘ 𝑅 ) ) ) )
11 rspecval ( 𝑅 ∈ Ring → ( Spec ‘ 𝑅 ) = ( ( IDLsrg ‘ 𝑅 ) ↾s ( PrmIdeal ‘ 𝑅 ) ) )
12 1 11 syl5eq ( 𝑅 ∈ Ring → 𝑆 = ( ( IDLsrg ‘ 𝑅 ) ↾s ( PrmIdeal ‘ 𝑅 ) ) )
13 12 fveq2d ( 𝑅 ∈ Ring → ( Base ‘ 𝑆 ) = ( Base ‘ ( ( IDLsrg ‘ 𝑅 ) ↾s ( PrmIdeal ‘ 𝑅 ) ) ) )
14 10 13 eqtr4d ( 𝑅 ∈ Ring → ( PrmIdeal ‘ 𝑅 ) = ( Base ‘ 𝑆 ) )