Metamath Proof Explorer


Theorem esplyfval2

Description: When K is out-of-bounds, the K -th elementary symmetric polynomial is zero. (Contributed by Thierry Arnoux, 25-Jan-2026)

Ref Expression
Hypotheses esplyfval2.d 𝐷 = { ∈ ( ℕ0m 𝐼 ) ∣ finSupp 0 }
esplyfval2.i ( 𝜑𝐼 ∈ Fin )
esplyfval2.r ( 𝜑𝑅 ∈ Ring )
esplyfval2.k ( 𝜑𝐾 ∈ ( ℕ0 ∖ ( 0 ... ( ♯ ‘ 𝐼 ) ) ) )
esplyfval2.z 𝑍 = ( 0g ‘ ( 𝐼 mPoly 𝑅 ) )
Assertion esplyfval2 ( 𝜑 → ( ( 𝐼 eSymPoly 𝑅 ) ‘ 𝐾 ) = 𝑍 )

Proof

Step Hyp Ref Expression
1 esplyfval2.d 𝐷 = { ∈ ( ℕ0m 𝐼 ) ∣ finSupp 0 }
2 esplyfval2.i ( 𝜑𝐼 ∈ Fin )
3 esplyfval2.r ( 𝜑𝑅 ∈ Ring )
4 esplyfval2.k ( 𝜑𝐾 ∈ ( ℕ0 ∖ ( 0 ... ( ♯ ‘ 𝐼 ) ) ) )
5 esplyfval2.z 𝑍 = ( 0g ‘ ( 𝐼 mPoly 𝑅 ) )
6 2 adantr ( ( 𝜑𝑐 ∈ 𝒫 𝐼 ) → 𝐼 ∈ Fin )
7 elpwi ( 𝑐 ∈ 𝒫 𝐼𝑐𝐼 )
8 7 adantl ( ( 𝜑𝑐 ∈ 𝒫 𝐼 ) → 𝑐𝐼 )
9 6 8 ssfid ( ( 𝜑𝑐 ∈ 𝒫 𝐼 ) → 𝑐 ∈ Fin )
10 hashcl ( 𝑐 ∈ Fin → ( ♯ ‘ 𝑐 ) ∈ ℕ0 )
11 9 10 syl ( ( 𝜑𝑐 ∈ 𝒫 𝐼 ) → ( ♯ ‘ 𝑐 ) ∈ ℕ0 )
12 11 nn0red ( ( 𝜑𝑐 ∈ 𝒫 𝐼 ) → ( ♯ ‘ 𝑐 ) ∈ ℝ )
13 hashcl ( 𝐼 ∈ Fin → ( ♯ ‘ 𝐼 ) ∈ ℕ0 )
14 2 13 syl ( 𝜑 → ( ♯ ‘ 𝐼 ) ∈ ℕ0 )
15 14 nn0red ( 𝜑 → ( ♯ ‘ 𝐼 ) ∈ ℝ )
16 15 adantr ( ( 𝜑𝑐 ∈ 𝒫 𝐼 ) → ( ♯ ‘ 𝐼 ) ∈ ℝ )
17 4 eldifad ( 𝜑𝐾 ∈ ℕ0 )
18 17 nn0red ( 𝜑𝐾 ∈ ℝ )
19 18 adantr ( ( 𝜑𝑐 ∈ 𝒫 𝐼 ) → 𝐾 ∈ ℝ )
20 hashss ( ( 𝐼 ∈ Fin ∧ 𝑐𝐼 ) → ( ♯ ‘ 𝑐 ) ≤ ( ♯ ‘ 𝐼 ) )
21 6 8 20 syl2anc ( ( 𝜑𝑐 ∈ 𝒫 𝐼 ) → ( ♯ ‘ 𝑐 ) ≤ ( ♯ ‘ 𝐼 ) )
22 14 nn0zd ( 𝜑 → ( ♯ ‘ 𝐼 ) ∈ ℤ )
23 nn0diffz0 ( ( ♯ ‘ 𝐼 ) ∈ ℕ0 → ( ℕ0 ∖ ( 0 ... ( ♯ ‘ 𝐼 ) ) ) = ( ℤ ‘ ( ( ♯ ‘ 𝐼 ) + 1 ) ) )
24 14 23 syl ( 𝜑 → ( ℕ0 ∖ ( 0 ... ( ♯ ‘ 𝐼 ) ) ) = ( ℤ ‘ ( ( ♯ ‘ 𝐼 ) + 1 ) ) )
25 4 24 eleqtrd ( 𝜑𝐾 ∈ ( ℤ ‘ ( ( ♯ ‘ 𝐼 ) + 1 ) ) )
26 eluzp1l ( ( ( ♯ ‘ 𝐼 ) ∈ ℤ ∧ 𝐾 ∈ ( ℤ ‘ ( ( ♯ ‘ 𝐼 ) + 1 ) ) ) → ( ♯ ‘ 𝐼 ) < 𝐾 )
27 22 25 26 syl2anc ( 𝜑 → ( ♯ ‘ 𝐼 ) < 𝐾 )
28 27 adantr ( ( 𝜑𝑐 ∈ 𝒫 𝐼 ) → ( ♯ ‘ 𝐼 ) < 𝐾 )
29 12 16 19 21 28 lelttrd ( ( 𝜑𝑐 ∈ 𝒫 𝐼 ) → ( ♯ ‘ 𝑐 ) < 𝐾 )
30 12 29 ltned ( ( 𝜑𝑐 ∈ 𝒫 𝐼 ) → ( ♯ ‘ 𝑐 ) ≠ 𝐾 )
31 30 neneqd ( ( 𝜑𝑐 ∈ 𝒫 𝐼 ) → ¬ ( ♯ ‘ 𝑐 ) = 𝐾 )
32 31 ralrimiva ( 𝜑 → ∀ 𝑐 ∈ 𝒫 𝐼 ¬ ( ♯ ‘ 𝑐 ) = 𝐾 )
33 rabeq0 ( { 𝑐 ∈ 𝒫 𝐼 ∣ ( ♯ ‘ 𝑐 ) = 𝐾 } = ∅ ↔ ∀ 𝑐 ∈ 𝒫 𝐼 ¬ ( ♯ ‘ 𝑐 ) = 𝐾 )
34 32 33 sylibr ( 𝜑 → { 𝑐 ∈ 𝒫 𝐼 ∣ ( ♯ ‘ 𝑐 ) = 𝐾 } = ∅ )
35 34 imaeq2d ( 𝜑 → ( ( 𝟭 ‘ 𝐼 ) “ { 𝑐 ∈ 𝒫 𝐼 ∣ ( ♯ ‘ 𝑐 ) = 𝐾 } ) = ( ( 𝟭 ‘ 𝐼 ) “ ∅ ) )
36 ima0 ( ( 𝟭 ‘ 𝐼 ) “ ∅ ) = ∅
37 35 36 eqtrdi ( 𝜑 → ( ( 𝟭 ‘ 𝐼 ) “ { 𝑐 ∈ 𝒫 𝐼 ∣ ( ♯ ‘ 𝑐 ) = 𝐾 } ) = ∅ )
38 37 fveq2d ( 𝜑 → ( ( 𝟭 ‘ 𝐷 ) ‘ ( ( 𝟭 ‘ 𝐼 ) “ { 𝑐 ∈ 𝒫 𝐼 ∣ ( ♯ ‘ 𝑐 ) = 𝐾 } ) ) = ( ( 𝟭 ‘ 𝐷 ) ‘ ∅ ) )
39 ovex ( ℕ0m 𝐼 ) ∈ V
40 1 39 rabex2 𝐷 ∈ V
41 indconst0 ( 𝐷 ∈ V → ( ( 𝟭 ‘ 𝐷 ) ‘ ∅ ) = ( 𝐷 × { 0 } ) )
42 40 41 mp1i ( 𝜑 → ( ( 𝟭 ‘ 𝐷 ) ‘ ∅ ) = ( 𝐷 × { 0 } ) )
43 38 42 eqtrd ( 𝜑 → ( ( 𝟭 ‘ 𝐷 ) ‘ ( ( 𝟭 ‘ 𝐼 ) “ { 𝑐 ∈ 𝒫 𝐼 ∣ ( ♯ ‘ 𝑐 ) = 𝐾 } ) ) = ( 𝐷 × { 0 } ) )
44 43 coeq2d ( 𝜑 → ( ( ℤRHom ‘ 𝑅 ) ∘ ( ( 𝟭 ‘ 𝐷 ) ‘ ( ( 𝟭 ‘ 𝐼 ) “ { 𝑐 ∈ 𝒫 𝐼 ∣ ( ♯ ‘ 𝑐 ) = 𝐾 } ) ) ) = ( ( ℤRHom ‘ 𝑅 ) ∘ ( 𝐷 × { 0 } ) ) )
45 eqid ( ℤRHom ‘ 𝑅 ) = ( ℤRHom ‘ 𝑅 )
46 45 zrhrhm ( 𝑅 ∈ Ring → ( ℤRHom ‘ 𝑅 ) ∈ ( ℤring RingHom 𝑅 ) )
47 zringbas ℤ = ( Base ‘ ℤring )
48 eqid ( Base ‘ 𝑅 ) = ( Base ‘ 𝑅 )
49 47 48 rhmf ( ( ℤRHom ‘ 𝑅 ) ∈ ( ℤring RingHom 𝑅 ) → ( ℤRHom ‘ 𝑅 ) : ℤ ⟶ ( Base ‘ 𝑅 ) )
50 3 46 49 3syl ( 𝜑 → ( ℤRHom ‘ 𝑅 ) : ℤ ⟶ ( Base ‘ 𝑅 ) )
51 50 ffnd ( 𝜑 → ( ℤRHom ‘ 𝑅 ) Fn ℤ )
52 0zd ( 𝜑 → 0 ∈ ℤ )
53 fcoconst ( ( ( ℤRHom ‘ 𝑅 ) Fn ℤ ∧ 0 ∈ ℤ ) → ( ( ℤRHom ‘ 𝑅 ) ∘ ( 𝐷 × { 0 } ) ) = ( 𝐷 × { ( ( ℤRHom ‘ 𝑅 ) ‘ 0 ) } ) )
54 51 52 53 syl2anc ( 𝜑 → ( ( ℤRHom ‘ 𝑅 ) ∘ ( 𝐷 × { 0 } ) ) = ( 𝐷 × { ( ( ℤRHom ‘ 𝑅 ) ‘ 0 ) } ) )
55 eqid ( 0g𝑅 ) = ( 0g𝑅 )
56 45 55 zrh0 ( 𝑅 ∈ Ring → ( ( ℤRHom ‘ 𝑅 ) ‘ 0 ) = ( 0g𝑅 ) )
57 3 56 syl ( 𝜑 → ( ( ℤRHom ‘ 𝑅 ) ‘ 0 ) = ( 0g𝑅 ) )
58 57 sneqd ( 𝜑 → { ( ( ℤRHom ‘ 𝑅 ) ‘ 0 ) } = { ( 0g𝑅 ) } )
59 58 xpeq2d ( 𝜑 → ( 𝐷 × { ( ( ℤRHom ‘ 𝑅 ) ‘ 0 ) } ) = ( 𝐷 × { ( 0g𝑅 ) } ) )
60 44 54 59 3eqtrd ( 𝜑 → ( ( ℤRHom ‘ 𝑅 ) ∘ ( ( 𝟭 ‘ 𝐷 ) ‘ ( ( 𝟭 ‘ 𝐼 ) “ { 𝑐 ∈ 𝒫 𝐼 ∣ ( ♯ ‘ 𝑐 ) = 𝐾 } ) ) ) = ( 𝐷 × { ( 0g𝑅 ) } ) )
61 1 2 3 17 esplyfval ( 𝜑 → ( ( 𝐼 eSymPoly 𝑅 ) ‘ 𝐾 ) = ( ( ℤRHom ‘ 𝑅 ) ∘ ( ( 𝟭 ‘ 𝐷 ) ‘ ( ( 𝟭 ‘ 𝐼 ) “ { 𝑐 ∈ 𝒫 𝐼 ∣ ( ♯ ‘ 𝑐 ) = 𝐾 } ) ) ) )
62 eqid ( 𝐼 mPoly 𝑅 ) = ( 𝐼 mPoly 𝑅 )
63 1 psrbasfsupp 𝐷 = { ∈ ( ℕ0m 𝐼 ) ∣ ( “ ℕ ) ∈ Fin }
64 3 ringgrpd ( 𝜑𝑅 ∈ Grp )
65 62 63 55 5 2 64 mpl0 ( 𝜑𝑍 = ( 𝐷 × { ( 0g𝑅 ) } ) )
66 60 61 65 3eqtr4d ( 𝜑 → ( ( 𝐼 eSymPoly 𝑅 ) ‘ 𝐾 ) = 𝑍 )