Metamath Proof Explorer


Theorem rtelextdg2

Description: If an element X is a solution of a quadratic equation, then it is either in the base field, or the degree of its field extension is exactly 2. (Contributed by Thierry Arnoux, 22-Jun-2025)

Ref Expression
Hypotheses rtelextdg2.1 𝐾 = ( 𝐸s 𝐹 )
rtelextdg2.2 𝐿 = ( 𝐸s ( 𝐸 fldGen ( 𝐹 ∪ { 𝑋 } ) ) )
rtelextdg2.3 0 = ( 0g𝐸 )
rtelextdg2.4 𝑃 = ( Poly1𝐾 )
rtelextdg2.5 𝑉 = ( Base ‘ 𝐸 )
rtelextdg2.6 · = ( .r𝐸 )
rtelextdg2.7 + = ( +g𝐸 )
rtelextdg2.8 = ( .g ‘ ( mulGrp ‘ 𝐸 ) )
rtelextdg2.9 ( 𝜑𝐸 ∈ Field )
rtelextdg2.10 ( 𝜑𝐹 ∈ ( SubDRing ‘ 𝐸 ) )
rtelextdg2.11 ( 𝜑𝑋𝑉 )
rtelextdg2.12 ( 𝜑𝐴𝐹 )
rtelextdg2.13 ( 𝜑𝐵𝐹 )
rtelextdg2.14 ( 𝜑 → ( ( 2 𝑋 ) + ( ( 𝐴 · 𝑋 ) + 𝐵 ) ) = 0 )
Assertion rtelextdg2 ( 𝜑 → ( 𝑋𝐹 ∨ ( 𝐿 [:] 𝐾 ) = 2 ) )

Proof

Step Hyp Ref Expression
1 rtelextdg2.1 𝐾 = ( 𝐸s 𝐹 )
2 rtelextdg2.2 𝐿 = ( 𝐸s ( 𝐸 fldGen ( 𝐹 ∪ { 𝑋 } ) ) )
3 rtelextdg2.3 0 = ( 0g𝐸 )
4 rtelextdg2.4 𝑃 = ( Poly1𝐾 )
5 rtelextdg2.5 𝑉 = ( Base ‘ 𝐸 )
6 rtelextdg2.6 · = ( .r𝐸 )
7 rtelextdg2.7 + = ( +g𝐸 )
8 rtelextdg2.8 = ( .g ‘ ( mulGrp ‘ 𝐸 ) )
9 rtelextdg2.9 ( 𝜑𝐸 ∈ Field )
10 rtelextdg2.10 ( 𝜑𝐹 ∈ ( SubDRing ‘ 𝐸 ) )
11 rtelextdg2.11 ( 𝜑𝑋𝑉 )
12 rtelextdg2.12 ( 𝜑𝐴𝐹 )
13 rtelextdg2.13 ( 𝜑𝐵𝐹 )
14 rtelextdg2.14 ( 𝜑 → ( ( 2 𝑋 ) + ( ( 𝐴 · 𝑋 ) + 𝐵 ) ) = 0 )
15 9 flddrngd ( 𝜑𝐸 ∈ DivRing )
16 5 sdrgss ( 𝐹 ∈ ( SubDRing ‘ 𝐸 ) → 𝐹𝑉 )
17 10 16 syl ( 𝜑𝐹𝑉 )
18 11 snssd ( 𝜑 → { 𝑋 } ⊆ 𝑉 )
19 17 18 unssd ( 𝜑 → ( 𝐹 ∪ { 𝑋 } ) ⊆ 𝑉 )
20 5 15 19 fldgenssid ( 𝜑 → ( 𝐹 ∪ { 𝑋 } ) ⊆ ( 𝐸 fldGen ( 𝐹 ∪ { 𝑋 } ) ) )
21 ssun2 { 𝑋 } ⊆ ( 𝐹 ∪ { 𝑋 } )
22 snidg ( 𝑋𝑉𝑋 ∈ { 𝑋 } )
23 11 22 syl ( 𝜑𝑋 ∈ { 𝑋 } )
24 21 23 sselid ( 𝜑𝑋 ∈ ( 𝐹 ∪ { 𝑋 } ) )
25 20 24 sseldd ( 𝜑𝑋 ∈ ( 𝐸 fldGen ( 𝐹 ∪ { 𝑋 } ) ) )
26 25 adantr ( ( 𝜑 ∧ ( 𝐿 [:] 𝐾 ) = 1 ) → 𝑋 ∈ ( 𝐸 fldGen ( 𝐹 ∪ { 𝑋 } ) ) )
27 5 1 2 9 10 18 fldgenfldext ( 𝜑𝐿 /FldExt 𝐾 )
28 extdg1id ( ( 𝐿 /FldExt 𝐾 ∧ ( 𝐿 [:] 𝐾 ) = 1 ) → 𝐿 = 𝐾 )
29 27 28 sylan ( ( 𝜑 ∧ ( 𝐿 [:] 𝐾 ) = 1 ) → 𝐿 = 𝐾 )
30 29 fveq2d ( ( 𝜑 ∧ ( 𝐿 [:] 𝐾 ) = 1 ) → ( Base ‘ 𝐿 ) = ( Base ‘ 𝐾 ) )
31 5 15 19 fldgenssv ( 𝜑 → ( 𝐸 fldGen ( 𝐹 ∪ { 𝑋 } ) ) ⊆ 𝑉 )
32 2 5 ressbas2 ( ( 𝐸 fldGen ( 𝐹 ∪ { 𝑋 } ) ) ⊆ 𝑉 → ( 𝐸 fldGen ( 𝐹 ∪ { 𝑋 } ) ) = ( Base ‘ 𝐿 ) )
33 31 32 syl ( 𝜑 → ( 𝐸 fldGen ( 𝐹 ∪ { 𝑋 } ) ) = ( Base ‘ 𝐿 ) )
34 33 adantr ( ( 𝜑 ∧ ( 𝐿 [:] 𝐾 ) = 1 ) → ( 𝐸 fldGen ( 𝐹 ∪ { 𝑋 } ) ) = ( Base ‘ 𝐿 ) )
35 1 5 ressbas2 ( 𝐹𝑉𝐹 = ( Base ‘ 𝐾 ) )
36 17 35 syl ( 𝜑𝐹 = ( Base ‘ 𝐾 ) )
37 36 adantr ( ( 𝜑 ∧ ( 𝐿 [:] 𝐾 ) = 1 ) → 𝐹 = ( Base ‘ 𝐾 ) )
38 30 34 37 3eqtr4d ( ( 𝜑 ∧ ( 𝐿 [:] 𝐾 ) = 1 ) → ( 𝐸 fldGen ( 𝐹 ∪ { 𝑋 } ) ) = 𝐹 )
39 26 38 eleqtrd ( ( 𝜑 ∧ ( 𝐿 [:] 𝐾 ) = 1 ) → 𝑋𝐹 )
40 simpr ( ( 𝜑 ∧ ( 𝐿 [:] 𝐾 ) = 2 ) → ( 𝐿 [:] 𝐾 ) = 2 )
41 1zzd ( 𝜑 → 1 ∈ ℤ )
42 2z 2 ∈ ℤ
43 42 a1i ( 𝜑 → 2 ∈ ℤ )
44 extdgcl ( 𝐿 /FldExt 𝐾 → ( 𝐿 [:] 𝐾 ) ∈ ℕ0* )
45 27 44 syl ( 𝜑 → ( 𝐿 [:] 𝐾 ) ∈ ℕ0* )
46 2nn0 2 ∈ ℕ0
47 46 a1i ( 𝜑 → 2 ∈ ℕ0 )
48 eqid ( var1𝐾 ) = ( var1𝐾 )
49 eqid ( +g𝑃 ) = ( +g𝑃 )
50 eqid ( .r𝑃 ) = ( .r𝑃 )
51 eqid ( .g ‘ ( mulGrp ‘ 𝑃 ) ) = ( .g ‘ ( mulGrp ‘ 𝑃 ) )
52 eqid ( algSc ‘ 𝑃 ) = ( algSc ‘ 𝑃 )
53 eqid ( ( 2 ( .g ‘ ( mulGrp ‘ 𝑃 ) ) ( var1𝐾 ) ) ( +g𝑃 ) ( ( ( ( algSc ‘ 𝑃 ) ‘ 𝐴 ) ( .r𝑃 ) ( var1𝐾 ) ) ( +g𝑃 ) ( ( algSc ‘ 𝑃 ) ‘ 𝐵 ) ) ) = ( ( 2 ( .g ‘ ( mulGrp ‘ 𝑃 ) ) ( var1𝐾 ) ) ( +g𝑃 ) ( ( ( ( algSc ‘ 𝑃 ) ‘ 𝐴 ) ( .r𝑃 ) ( var1𝐾 ) ) ( +g𝑃 ) ( ( algSc ‘ 𝑃 ) ‘ 𝐵 ) ) )
54 1 2 3 4 5 6 7 8 9 10 11 12 13 14 48 49 50 51 52 53 rtelextdg2lem ( 𝜑 → ( 𝐿 [:] 𝐾 ) ≤ 2 )
55 xnn0lenn0nn0 ( ( ( 𝐿 [:] 𝐾 ) ∈ ℕ0* ∧ 2 ∈ ℕ0 ∧ ( 𝐿 [:] 𝐾 ) ≤ 2 ) → ( 𝐿 [:] 𝐾 ) ∈ ℕ0 )
56 45 47 54 55 syl3anc ( 𝜑 → ( 𝐿 [:] 𝐾 ) ∈ ℕ0 )
57 56 nn0zd ( 𝜑 → ( 𝐿 [:] 𝐾 ) ∈ ℤ )
58 extdggt0 ( 𝐿 /FldExt 𝐾 → 0 < ( 𝐿 [:] 𝐾 ) )
59 27 58 syl ( 𝜑 → 0 < ( 𝐿 [:] 𝐾 ) )
60 zgt0ge1 ( ( 𝐿 [:] 𝐾 ) ∈ ℤ → ( 0 < ( 𝐿 [:] 𝐾 ) ↔ 1 ≤ ( 𝐿 [:] 𝐾 ) ) )
61 60 biimpa ( ( ( 𝐿 [:] 𝐾 ) ∈ ℤ ∧ 0 < ( 𝐿 [:] 𝐾 ) ) → 1 ≤ ( 𝐿 [:] 𝐾 ) )
62 57 59 61 syl2anc ( 𝜑 → 1 ≤ ( 𝐿 [:] 𝐾 ) )
63 41 43 57 62 54 elfzd ( 𝜑 → ( 𝐿 [:] 𝐾 ) ∈ ( 1 ... 2 ) )
64 fz12pr ( 1 ... 2 ) = { 1 , 2 }
65 63 64 eleqtrdi ( 𝜑 → ( 𝐿 [:] 𝐾 ) ∈ { 1 , 2 } )
66 elpri ( ( 𝐿 [:] 𝐾 ) ∈ { 1 , 2 } → ( ( 𝐿 [:] 𝐾 ) = 1 ∨ ( 𝐿 [:] 𝐾 ) = 2 ) )
67 65 66 syl ( 𝜑 → ( ( 𝐿 [:] 𝐾 ) = 1 ∨ ( 𝐿 [:] 𝐾 ) = 2 ) )
68 39 40 67 orim12da ( 𝜑 → ( 𝑋𝐹 ∨ ( 𝐿 [:] 𝐾 ) = 2 ) )