Metamath Proof Explorer


Theorem 0ringmon1p

Description: There are no monic polynomials over a zero ring. (Contributed by Thierry Arnoux, 5-Feb-2025)

Ref Expression
Hypotheses 0ringmon1p.1 M = Monic 1p R
0ringmon1p.2 B = Base R
0ringmon1p.3 φ R Ring
0ringmon1p.4 φ B = 1
Assertion 0ringmon1p φ M =

Proof

Step Hyp Ref Expression
1 0ringmon1p.1 M = Monic 1p R
2 0ringmon1p.2 B = Base R
3 0ringmon1p.3 φ R Ring
4 0ringmon1p.4 φ B = 1
5 eqid Poly 1 R = Poly 1 R
6 eqid Base Poly 1 R = Base Poly 1 R
7 eqid 0 Poly 1 R = 0 Poly 1 R
8 eqid deg 1 R = deg 1 R
9 eqid 1 R = 1 R
10 5 6 7 8 1 9 ismon1p p M p Base Poly 1 R p 0 Poly 1 R coe 1 p deg 1 R p = 1 R
11 10 biimpi p M p Base Poly 1 R p 0 Poly 1 R coe 1 p deg 1 R p = 1 R
12 11 adantl φ p M p Base Poly 1 R p 0 Poly 1 R coe 1 p deg 1 R p = 1 R
13 12 simp3d φ p M coe 1 p deg 1 R p = 1 R
14 3 adantr φ p M R Ring
15 12 simp1d φ p M p Base Poly 1 R
16 12 simp2d φ p M p 0 Poly 1 R
17 eqid 0 R = 0 R
18 eqid coe 1 p = coe 1 p
19 8 5 7 6 17 18 deg1ldg R Ring p Base Poly 1 R p 0 Poly 1 R coe 1 p deg 1 R p 0 R
20 14 15 16 19 syl3anc φ p M coe 1 p deg 1 R p 0 R
21 2 17 9 0ring01eq R Ring B = 1 0 R = 1 R
22 3 4 21 syl2anc φ 0 R = 1 R
23 22 adantr φ p M 0 R = 1 R
24 20 23 neeqtrd φ p M coe 1 p deg 1 R p 1 R
25 24 neneqd φ p M ¬ coe 1 p deg 1 R p = 1 R
26 13 25 pm2.65da φ ¬ p M
27 26 eq0rdv φ M =