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 eqtrid ⊢ 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