Metamath Proof Explorer


Theorem isfieldidl2

Description: Determine if a ring is a field based on its ideals. (Contributed by Jeff Madsen, 6-Jan-2011) (Revised by AV, 30-Jun-2026)

Ref Expression
Hypotheses isfieldidl.b B = Base R
isfieldidl.0 0 ˙ = 0 R
isfieldidl.i I = LIdeal R
Assertion isfieldidl2 R Field R CRing B 0 ˙ I = 0 ˙ B

Proof

Step Hyp Ref Expression
1 isfieldidl.b B = Base R
2 isfieldidl.0 0 ˙ = 0 R
3 isfieldidl.i I = LIdeal R
4 eqid 1 R = 1 R
5 1 2 3 4 isfieldidl R Field R CRing 0 ˙ 1 R I = 0 ˙ B
6 crngring R CRing R Ring
7 eqcom 0 ˙ = 1 R 1 R = 0 ˙
8 1 2 4 0ring01eqbi2 R Ring B = 0 ˙ 1 R = 0 ˙
9 7 8 bitr4id R Ring 0 ˙ = 1 R B = 0 ˙
10 6 9 syl R CRing 0 ˙ = 1 R B = 0 ˙
11 10 necon3bid R CRing 0 ˙ 1 R B 0 ˙
12 11 anbi1d R CRing 0 ˙ 1 R I = 0 ˙ B B 0 ˙ I = 0 ˙ B
13 12 pm5.32i R CRing 0 ˙ 1 R I = 0 ˙ B R CRing B 0 ˙ I = 0 ˙ B
14 3anass R CRing 0 ˙ 1 R I = 0 ˙ B R CRing 0 ˙ 1 R I = 0 ˙ B
15 3anass R CRing B 0 ˙ I = 0 ˙ B R CRing B 0 ˙ I = 0 ˙ B
16 13 14 15 3bitr4i R CRing 0 ˙ 1 R I = 0 ˙ B R CRing B 0 ˙ I = 0 ˙ B
17 5 16 bitri R Field R CRing B 0 ˙ I = 0 ˙ B