Metamath Proof Explorer


Theorem mhpsclcl

Description: A scalar (or constant) polynomial has degree 0. Compare deg1scl . In other contexts, there may be an exception for the zero polynomial, but under df-mhp the zero polynomial can be any degree (see mhp0cl ) so there is no exception. (Contributed by SN, 25-May-2024)

Ref Expression
Hypotheses mhpsclcl.h H = I mHomP R
mhpsclcl.p P = I mPoly R
mhpsclcl.a A = algSc P
mhpsclcl.k K = Base R
mhpsclcl.i φ I V
mhpsclcl.r φ R Ring
mhpsclcl.c φ C K
Assertion mhpsclcl φ A C H 0

Proof

Step Hyp Ref Expression
1 mhpsclcl.h H = I mHomP R
2 mhpsclcl.p P = I mPoly R
3 mhpsclcl.a A = algSc P
4 mhpsclcl.k K = Base R
5 mhpsclcl.i φ I V
6 mhpsclcl.r φ R Ring
7 mhpsclcl.c φ C K
8 eqid h 0 I | h -1 Fin = h 0 I | h -1 Fin
9 eqid 0 R = 0 R
10 5 adantr φ d h 0 I | h -1 Fin I V
11 6 adantr φ d h 0 I | h -1 Fin R Ring
12 7 adantr φ d h 0 I | h -1 Fin C K
13 2 8 9 4 3 10 11 12 mplascl φ d h 0 I | h -1 Fin A C = y h 0 I | h -1 Fin if y = I × 0 C 0 R
14 eqeq1 y = d y = I × 0 d = I × 0
15 14 ifbid y = d if y = I × 0 C 0 R = if d = I × 0 C 0 R
16 15 adantl φ d h 0 I | h -1 Fin y = d if y = I × 0 C 0 R = if d = I × 0 C 0 R
17 simpr φ d h 0 I | h -1 Fin d h 0 I | h -1 Fin
18 fvexd φ 0 R V
19 7 18 ifexd φ if d = I × 0 C 0 R V
20 19 adantr φ d h 0 I | h -1 Fin if d = I × 0 C 0 R V
21 13 16 17 20 fvmptd φ d h 0 I | h -1 Fin A C d = if d = I × 0 C 0 R
22 21 neeq1d φ d h 0 I | h -1 Fin A C d 0 R if d = I × 0 C 0 R 0 R
23 iffalse ¬ d = I × 0 if d = I × 0 C 0 R = 0 R
24 23 necon1ai if d = I × 0 C 0 R 0 R d = I × 0
25 fconstmpt I × 0 = k I 0
26 25 oveq2i fld 𝑠 0 I × 0 = fld 𝑠 0 k I 0
27 nn0subm 0 SubMnd fld
28 eqid fld 𝑠 0 = fld 𝑠 0
29 28 submmnd 0 SubMnd fld fld 𝑠 0 Mnd
30 27 29 ax-mp fld 𝑠 0 Mnd
31 cnfld0 0 = 0 fld
32 28 31 subm0 0 SubMnd fld 0 = 0 fld 𝑠 0
33 27 32 ax-mp 0 = 0 fld 𝑠 0
34 33 gsumz fld 𝑠 0 Mnd I V fld 𝑠 0 k I 0 = 0
35 30 10 34 sylancr φ d h 0 I | h -1 Fin fld 𝑠 0 k I 0 = 0
36 26 35 eqtrid φ d h 0 I | h -1 Fin fld 𝑠 0 I × 0 = 0
37 oveq2 d = I × 0 fld 𝑠 0 d = fld 𝑠 0 I × 0
38 37 eqeq1d d = I × 0 fld 𝑠 0 d = 0 fld 𝑠 0 I × 0 = 0
39 36 38 syl5ibrcom φ d h 0 I | h -1 Fin d = I × 0 fld 𝑠 0 d = 0
40 24 39 syl5 φ d h 0 I | h -1 Fin if d = I × 0 C 0 R 0 R fld 𝑠 0 d = 0
41 22 40 sylbid φ d h 0 I | h -1 Fin A C d 0 R fld 𝑠 0 d = 0
42 41 ralrimiva φ d h 0 I | h -1 Fin A C d 0 R fld 𝑠 0 d = 0
43 eqid Base P = Base P
44 0nn0 0 0
45 44 a1i φ 0 0
46 2 43 4 3 5 6 mplasclf φ A : K Base P
47 46 7 ffvelrnd φ A C Base P
48 1 2 43 9 8 5 6 45 47 ismhp3 φ A C H 0 d h 0 I | h -1 Fin A C d 0 R fld 𝑠 0 d = 0
49 42 48 mpbird φ A C H 0