P. 337 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 33601-33700   *Has distinct variable group(s)

Type	Label	Description
Statement
21.3.10.52  Vector Space Dimension
Syntax	cldim 33601	Extend class notation with the dimension of a vector space.
class dim
Definition	df-dim 33602	Define the dimension of a vector space as the cardinality of its bases. Note that by lvecdim 21074, all bases are equinumerous. (Contributed by Thierry Arnoux, 6-May-2023.)
⊢ dim = (𝑓 ∈ V ↦ ∪ (♯ “ (LBasis‘𝑓)))
Theorem	dimval 33603	The dimension of a vector space 𝐹 is the cardinality of one of its bases. (Contributed by Thierry Arnoux, 6-May-2023.)
⊢ 𝐽 = (LBasis‘𝐹)    ⇒   ⊢ ((𝐹 ∈ LVec ∧ 𝑆 ∈ 𝐽) → (dim‘𝐹) = (♯‘𝑆))
Theorem	dimvalfi 33604	The dimension of a vector space 𝐹 is the cardinality of one of its bases. This version of dimval 33603 does not depend on the axiom of choice, but it is limited to the case where the base 𝑆 is finite. (Contributed by Thierry Arnoux, 24-May-2023.)
⊢ 𝐽 = (LBasis‘𝐹)    ⇒   ⊢ ((𝐹 ∈ LVec ∧ 𝑆 ∈ 𝐽 ∧ 𝑆 ∈ Fin) → (dim‘𝐹) = (♯‘𝑆))
Theorem	dimcl 33605	Closure of the vector space dimension. (Contributed by Thierry Arnoux, 18-May-2023.)
⊢ (𝑉 ∈ LVec → (dim‘𝑉) ∈ ℕ0*)
Theorem	lmimdim 33606	Module isomorphisms preserve vector space dimensions. (Contributed by Thierry Arnoux, 25-Feb-2025.)
⊢ (𝜑 → 𝐹 ∈ (𝑆 LMIso 𝑇))    &   ⊢ (𝜑 → 𝑆 ∈ LVec)    ⇒   ⊢ (𝜑 → (dim‘𝑆) = (dim‘𝑇))
Theorem	lmicdim 33607	Module isomorphisms preserve vector space dimensions. (Contributed by Thierry Arnoux, 25-Mar-2025.)
⊢ (𝜑 → 𝑆 ≃𝑚 𝑇)    &   ⊢ (𝜑 → 𝑆 ∈ LVec)    ⇒   ⊢ (𝜑 → (dim‘𝑆) = (dim‘𝑇))
Theorem	lvecdim0i 33608	A vector space of dimension zero is reduced to its identity element. (Contributed by Thierry Arnoux, 31-Jul-2023.)
⊢ 0 = (0g‘𝑉)    ⇒   ⊢ ((𝑉 ∈ LVec ∧ (dim‘𝑉) = 0) → (Base‘𝑉) = { 0 })
Theorem	lvecdim0 33609	A vector space of dimension zero is reduced to its identity element, biconditional version. (Contributed by Thierry Arnoux, 31-Jul-2023.)
⊢ 0 = (0g‘𝑉)    ⇒   ⊢ (𝑉 ∈ LVec → ((dim‘𝑉) = 0 ↔ (Base‘𝑉) = { 0 }))
Theorem	lssdimle 33610	The dimension of a linear subspace is less than or equal to the dimension of the parent vector space. This is corollary 5.4 of [Lang] p. 141. (Contributed by Thierry Arnoux, 20-May-2023.)
⊢ 𝑋 = (𝑊 ↾s 𝑈)    ⇒   ⊢ ((𝑊 ∈ LVec ∧ 𝑈 ∈ (LSubSp‘𝑊)) → (dim‘𝑋) ≤ (dim‘𝑊))
Theorem	dimpropd 33611*	If two structures have the same components (properties), they have the same dimension. (Contributed by Thierry Arnoux, 18-May-2023.)
⊢ (𝜑 → 𝐵 = (Base‘𝐾))    &   ⊢ (𝜑 → 𝐵 = (Base‘𝐿))    &   ⊢ (𝜑 → 𝐵 ⊆ 𝑊)    &   ⊢ ((𝜑 ∧ (𝑥 ∈ 𝑊 ∧ 𝑦 ∈ 𝑊)) → (𝑥(+g‘𝐾)𝑦) = (𝑥(+g‘𝐿)𝑦))    &   ⊢ ((𝜑 ∧ (𝑥 ∈ 𝑃 ∧ 𝑦 ∈ 𝐵)) → (𝑥( ·𝑠 ‘𝐾)𝑦) ∈ 𝑊)    &   ⊢ ((𝜑 ∧ (𝑥 ∈ 𝑃 ∧ 𝑦 ∈ 𝐵)) → (𝑥( ·𝑠 ‘𝐾)𝑦) = (𝑥( ·𝑠 ‘𝐿)𝑦))    &   ⊢ 𝐹 = (Scalar‘𝐾)    &   ⊢ 𝐺 = (Scalar‘𝐿)    &   ⊢ (𝜑 → 𝑃 = (Base‘𝐹))    &   ⊢ (𝜑 → 𝑃 = (Base‘𝐺))    &   ⊢ ((𝜑 ∧ (𝑥 ∈ 𝑃 ∧ 𝑦 ∈ 𝑃)) → (𝑥(+g‘𝐹)𝑦) = (𝑥(+g‘𝐺)𝑦))    &   ⊢ (𝜑 → 𝐾 ∈ LVec)    &   ⊢ (𝜑 → 𝐿 ∈ LVec)    ⇒   ⊢ (𝜑 → (dim‘𝐾) = (dim‘𝐿))
Theorem	rlmdim 33612	The left vector space induced by a ring over itself has dimension 1. (Contributed by Thierry Arnoux, 5-Aug-2023.) Generalize to division rings. (Revised by SN, 22-Mar-2025.)
⊢ 𝑉 = (ringLMod‘𝐹)    ⇒   ⊢ (𝐹 ∈ DivRing → (dim‘𝑉) = 1)
Theorem	rgmoddimOLD 33613	Obsolete version of rlmdim 33612 as of 21-Mar-2025. (Contributed by Thierry Arnoux, 5-Aug-2023.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ 𝑉 = (ringLMod‘𝐹)    ⇒   ⊢ (𝐹 ∈ Field → (dim‘𝑉) = 1)
Theorem	frlmdim 33614	Dimension of a free left module. (Contributed by Thierry Arnoux, 20-May-2023.)
⊢ 𝐹 = (𝑅 freeLMod 𝐼)    ⇒   ⊢ ((𝑅 ∈ DivRing ∧ 𝐼 ∈ 𝑉) → (dim‘𝐹) = (♯‘𝐼))
Theorem	tnglvec 33615	Augmenting a structure with a norm conserves left vector spaces. (Contributed by Thierry Arnoux, 20-May-2023.)
⊢ 𝑇 = (𝐺 toNrmGrp 𝑁)    ⇒   ⊢ (𝑁 ∈ 𝑉 → (𝐺 ∈ LVec ↔ 𝑇 ∈ LVec))
Theorem	tngdim 33616	Dimension of a left vector space augmented with a norm. (Contributed by Thierry Arnoux, 20-May-2023.)
⊢ 𝑇 = (𝐺 toNrmGrp 𝑁)    ⇒   ⊢ ((𝐺 ∈ LVec ∧ 𝑁 ∈ 𝑉) → (dim‘𝐺) = (dim‘𝑇))
Theorem	rrxdim 33617	Dimension of the generalized Euclidean space. (Contributed by Thierry Arnoux, 20-May-2023.)
⊢ 𝐻 = (ℝ^‘𝐼)    ⇒   ⊢ (𝐼 ∈ 𝑉 → (dim‘𝐻) = (♯‘𝐼))
Theorem	matdim 33618	Dimension of the space of square matrices. (Contributed by Thierry Arnoux, 20-May-2023.)
⊢ 𝐴 = (𝐼 Mat 𝑅)    &   ⊢ 𝑁 = (♯‘𝐼)    ⇒   ⊢ ((𝐼 ∈ Fin ∧ 𝑅 ∈ DivRing) → (dim‘𝐴) = (𝑁 · 𝑁))
Theorem	lbslsat 33619	A nonzero vector 𝑋 is a basis of a line spanned by the singleton 𝑋. Spans of nonzero singletons are sometimes called "atoms", see df-lsatoms 38976 and for example lsatlspsn 38993. (Contributed by Thierry Arnoux, 20-May-2023.)
⊢ 𝑉 = (Base‘𝑊)    &   ⊢ 𝑁 = (LSpan‘𝑊)    &   ⊢ 0 = (0g‘𝑊)    &   ⊢ 𝑌 = (𝑊 ↾s (𝑁‘{𝑋}))    ⇒   ⊢ ((𝑊 ∈ LVec ∧ 𝑋 ∈ 𝑉 ∧ 𝑋 ≠ 0 ) → {𝑋} ∈ (LBasis‘𝑌))
Theorem	lsatdim 33620	A line, spanned by a nonzero singleton, has dimension 1. (Contributed by Thierry Arnoux, 20-May-2023.)
⊢ 𝑉 = (Base‘𝑊)    &   ⊢ 𝑁 = (LSpan‘𝑊)    &   ⊢ 0 = (0g‘𝑊)    &   ⊢ 𝑌 = (𝑊 ↾s (𝑁‘{𝑋}))    ⇒   ⊢ ((𝑊 ∈ LVec ∧ 𝑋 ∈ 𝑉 ∧ 𝑋 ≠ 0 ) → (dim‘𝑌) = 1)
Theorem	drngdimgt0 33621	The dimension of a vector space that is also a division ring is greater than zero. (Contributed by Thierry Arnoux, 29-Jul-2023.)
⊢ ((𝐹 ∈ LVec ∧ 𝐹 ∈ DivRing) → 0 < (dim‘𝐹))
Theorem	lmhmlvec2 33622	A homomorphism of left vector spaces has a left vector space as codomain. (Contributed by Thierry Arnoux, 7-May-2023.)
⊢ ((𝑉 ∈ LVec ∧ 𝐹 ∈ (𝑉 LMHom 𝑈)) → 𝑈 ∈ LVec)
Theorem	kerlmhm 33623	The kernel of a vector space homomorphism is a vector space itself. (Contributed by Thierry Arnoux, 7-May-2023.)
⊢ 0 = (0g‘𝑈)    &   ⊢ 𝐾 = (𝑉 ↾s (◡𝐹 “ { 0 }))    ⇒   ⊢ ((𝑉 ∈ LVec ∧ 𝐹 ∈ (𝑉 LMHom 𝑈)) → 𝐾 ∈ LVec)
Theorem	imlmhm 33624	The image of a vector space homomorphism is a vector space itself. (Contributed by Thierry Arnoux, 7-May-2023.)
⊢ 𝐼 = (𝑈 ↾s ran 𝐹)    ⇒   ⊢ ((𝑉 ∈ LVec ∧ 𝐹 ∈ (𝑉 LMHom 𝑈)) → 𝐼 ∈ LVec)
Theorem	ply1degltdimlem 33625*	Lemma for ply1degltdim 33626. (Contributed by Thierry Arnoux, 20-Feb-2025.)
⊢ 𝑃 = (Poly1‘𝑅)    &   ⊢ 𝐷 = (deg1‘𝑅)    &   ⊢ 𝑆 = (◡𝐷 “ (-∞[,)𝑁))    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    &   ⊢ (𝜑 → 𝑅 ∈ DivRing)    &   ⊢ 𝐸 = (𝑃 ↾s 𝑆)    &   ⊢ 𝐹 = (𝑛 ∈ (0..^𝑁) ↦ (𝑛(.g‘(mulGrp‘𝑃))(var1‘𝑅)))    ⇒   ⊢ (𝜑 → ran 𝐹 ∈ (LBasis‘𝐸))
Theorem	ply1degltdim 33626	The space 𝑆 of the univariate polynomials of degree less than 𝑁 has dimension 𝑁. (Contributed by Thierry Arnoux, 20-Feb-2025.)
⊢ 𝑃 = (Poly1‘𝑅)    &   ⊢ 𝐷 = (deg1‘𝑅)    &   ⊢ 𝑆 = (◡𝐷 “ (-∞[,)𝑁))    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    &   ⊢ (𝜑 → 𝑅 ∈ DivRing)    &   ⊢ 𝐸 = (𝑃 ↾s 𝑆)    ⇒   ⊢ (𝜑 → (dim‘𝐸) = 𝑁)
Theorem	lindsunlem 33627	Lemma for lindsun 33628. (Contributed by Thierry Arnoux, 9-May-2023.)
⊢ 𝑁 = (LSpan‘𝑊)    &   ⊢ 0 = (0g‘𝑊)    &   ⊢ (𝜑 → 𝑊 ∈ LVec)    &   ⊢ (𝜑 → 𝑈 ∈ (LIndS‘𝑊))    &   ⊢ (𝜑 → 𝑉 ∈ (LIndS‘𝑊))    &   ⊢ (𝜑 → ((𝑁‘𝑈) ∩ (𝑁‘𝑉)) = { 0 })    &   ⊢ 𝑂 = (0g‘(Scalar‘𝑊))    &   ⊢ 𝐹 = (Base‘(Scalar‘𝑊))    &   ⊢ (𝜑 → 𝐶 ∈ 𝑈)    &   ⊢ (𝜑 → 𝐾 ∈ (𝐹 ∖ {𝑂}))    &   ⊢ (𝜑 → (𝐾( ·𝑠 ‘𝑊)𝐶) ∈ (𝑁‘((𝑈 ∪ 𝑉) ∖ {𝐶})))    ⇒   ⊢ (𝜑 → ⊥)
Theorem	lindsun 33628	Condition for the union of two independent sets to be an independent set. (Contributed by Thierry Arnoux, 9-May-2023.)
⊢ 𝑁 = (LSpan‘𝑊)    &   ⊢ 0 = (0g‘𝑊)    &   ⊢ (𝜑 → 𝑊 ∈ LVec)    &   ⊢ (𝜑 → 𝑈 ∈ (LIndS‘𝑊))    &   ⊢ (𝜑 → 𝑉 ∈ (LIndS‘𝑊))    &   ⊢ (𝜑 → ((𝑁‘𝑈) ∩ (𝑁‘𝑉)) = { 0 })    ⇒   ⊢ (𝜑 → (𝑈 ∪ 𝑉) ∈ (LIndS‘𝑊))
Theorem	lbsdiflsp0 33629	The linear spans of two disjunct independent sets only have a trivial intersection. This can be seen as the opposite direction of lindsun 33628. (Contributed by Thierry Arnoux, 17-May-2023.)
⊢ 𝐽 = (LBasis‘𝑊)    &   ⊢ 𝑁 = (LSpan‘𝑊)    &   ⊢ 0 = (0g‘𝑊)    ⇒   ⊢ ((𝑊 ∈ LVec ∧ 𝐵 ∈ 𝐽 ∧ 𝑉 ⊆ 𝐵) → ((𝑁‘(𝐵 ∖ 𝑉)) ∩ (𝑁‘𝑉)) = { 0 })
Theorem	dimkerim 33630	Given a linear map 𝐹 between vector spaces 𝑉 and 𝑈, the dimension of the vector space 𝑉 is the sum of the dimension of 𝐹 's kernel and the dimension of 𝐹's image. Second part of theorem 5.3 in [Lang] p. 141 This can also be described as the Rank-nullity theorem, (dim‘𝐼) being the rank of 𝐹 (the dimension of its image), and (dim‘𝐾) its nullity (the dimension of its kernel). (Contributed by Thierry Arnoux, 17-May-2023.)
⊢ 0 = (0g‘𝑈)    &   ⊢ 𝐾 = (𝑉 ↾s (◡𝐹 “ { 0 }))    &   ⊢ 𝐼 = (𝑈 ↾s ran 𝐹)    ⇒   ⊢ ((𝑉 ∈ LVec ∧ 𝐹 ∈ (𝑉 LMHom 𝑈)) → (dim‘𝑉) = ((dim‘𝐾) +𝑒 (dim‘𝐼)))
Theorem	qusdimsum 33631	Let 𝑊 be a vector space, and let 𝑋 be a subspace. Then the dimension of 𝑊 is the sum of the dimension of 𝑋 and the dimension of the quotient space of 𝑋. First part of theorem 5.3 in [Lang] p. 141. (Contributed by Thierry Arnoux, 20-May-2023.)
⊢ 𝑋 = (𝑊 ↾s 𝑈)    &   ⊢ 𝑌 = (𝑊 /s (𝑊 ~QG 𝑈))    ⇒   ⊢ ((𝑊 ∈ LVec ∧ 𝑈 ∈ (LSubSp‘𝑊)) → (dim‘𝑊) = ((dim‘𝑋) +𝑒 (dim‘𝑌)))
Theorem	fedgmullem1 33632*	Lemma for fedgmul 33634. (Contributed by Thierry Arnoux, 20-Jul-2023.)
⊢ 𝐴 = ((subringAlg ‘𝐸)‘𝑉)    &   ⊢ 𝐵 = ((subringAlg ‘𝐸)‘𝑈)    &   ⊢ 𝐶 = ((subringAlg ‘𝐹)‘𝑉)    &   ⊢ 𝐹 = (𝐸 ↾s 𝑈)    &   ⊢ 𝐾 = (𝐸 ↾s 𝑉)    &   ⊢ (𝜑 → 𝐸 ∈ DivRing)    &   ⊢ (𝜑 → 𝐹 ∈ DivRing)    &   ⊢ (𝜑 → 𝐾 ∈ DivRing)    &   ⊢ (𝜑 → 𝑈 ∈ (SubRing‘𝐸))    &   ⊢ (𝜑 → 𝑉 ∈ (SubRing‘𝐹))    &   ⊢ 𝐷 = (𝑗 ∈ 𝑌, 𝑖 ∈ 𝑋 ↦ (𝑖(.r‘𝐸)𝑗))    &   ⊢ 𝐻 = (𝑗 ∈ 𝑌, 𝑖 ∈ 𝑋 ↦ ((𝐺‘𝑗)‘𝑖))    &   ⊢ (𝜑 → 𝑋 ∈ (LBasis‘𝐶))    &   ⊢ (𝜑 → 𝑌 ∈ (LBasis‘𝐵))    &   ⊢ (𝜑 → 𝑍 ∈ (Base‘𝐴))    &   ⊢ (𝜑 → 𝐿:𝑌⟶(Base‘(Scalar‘𝐵)))    &   ⊢ (𝜑 → 𝐿 finSupp (0g‘(Scalar‘𝐵)))    &   ⊢ (𝜑 → 𝑍 = (𝐵 Σg (𝑗 ∈ 𝑌 ↦ ((𝐿‘𝑗)( ·𝑠 ‘𝐵)𝑗))))    &   ⊢ (𝜑 → 𝐺:𝑌⟶((Base‘(Scalar‘𝐶)) ↑m 𝑋))    &   ⊢ ((𝜑 ∧ 𝑗 ∈ 𝑌) → (𝐺‘𝑗) finSupp (0g‘(Scalar‘𝐶)))    &   ⊢ ((𝜑 ∧ 𝑗 ∈ 𝑌) → (𝐿‘𝑗) = (𝐶 Σg (𝑖 ∈ 𝑋 ↦ (((𝐺‘𝑗)‘𝑖)( ·𝑠 ‘𝐶)𝑖))))    ⇒   ⊢ (𝜑 → (𝐻 finSupp (0g‘(Scalar‘𝐴)) ∧ 𝑍 = (𝐴 Σg (𝐻 ∘f ( ·𝑠 ‘𝐴)𝐷))))
Theorem	fedgmullem2 33633*	Lemma for fedgmul 33634. (Contributed by Thierry Arnoux, 20-Jul-2023.)
⊢ 𝐴 = ((subringAlg ‘𝐸)‘𝑉)    &   ⊢ 𝐵 = ((subringAlg ‘𝐸)‘𝑈)    &   ⊢ 𝐶 = ((subringAlg ‘𝐹)‘𝑉)    &   ⊢ 𝐹 = (𝐸 ↾s 𝑈)    &   ⊢ 𝐾 = (𝐸 ↾s 𝑉)    &   ⊢ (𝜑 → 𝐸 ∈ DivRing)    &   ⊢ (𝜑 → 𝐹 ∈ DivRing)    &   ⊢ (𝜑 → 𝐾 ∈ DivRing)    &   ⊢ (𝜑 → 𝑈 ∈ (SubRing‘𝐸))    &   ⊢ (𝜑 → 𝑉 ∈ (SubRing‘𝐹))    &   ⊢ 𝐷 = (𝑗 ∈ 𝑌, 𝑖 ∈ 𝑋 ↦ (𝑖(.r‘𝐸)𝑗))    &   ⊢ 𝐻 = (𝑗 ∈ 𝑌, 𝑖 ∈ 𝑋 ↦ ((𝐺‘𝑗)‘𝑖))    &   ⊢ (𝜑 → 𝑋 ∈ (LBasis‘𝐶))    &   ⊢ (𝜑 → 𝑌 ∈ (LBasis‘𝐵))    &   ⊢ (𝜑 → 𝑊 ∈ (Base‘((Scalar‘𝐴) freeLMod (𝑌 × 𝑋))))    &   ⊢ (𝜑 → (𝐴 Σg (𝑊 ∘f ( ·𝑠 ‘𝐴)𝐷)) = (0g‘𝐴))    ⇒   ⊢ (𝜑 → 𝑊 = ((𝑌 × 𝑋) × {(0g‘(Scalar‘𝐴))}))
Theorem	fedgmul 33634	The multiplicativity formula for degrees of field extensions. Given 𝐸 a field extension of 𝐹, itself a field extension of 𝐾, we have [𝐸:𝐾] = [𝐸:𝐹][𝐹:𝐾]. Proposition 1.2 of [Lang], p. 224. Here (dim‘𝐴) is the degree of the extension 𝐸 of 𝐾, (dim‘𝐵) is the degree of the extension 𝐸 of 𝐹, and (dim‘𝐶) is the degree of the extension 𝐹 of 𝐾. This proof is valid for infinite dimensions, and is actually valid for division ring extensions, not just field extensions. (Contributed by Thierry Arnoux, 25-Jul-2023.)
⊢ 𝐴 = ((subringAlg ‘𝐸)‘𝑉)    &   ⊢ 𝐵 = ((subringAlg ‘𝐸)‘𝑈)    &   ⊢ 𝐶 = ((subringAlg ‘𝐹)‘𝑉)    &   ⊢ 𝐹 = (𝐸 ↾s 𝑈)    &   ⊢ 𝐾 = (𝐸 ↾s 𝑉)    &   ⊢ (𝜑 → 𝐸 ∈ DivRing)    &   ⊢ (𝜑 → 𝐹 ∈ DivRing)    &   ⊢ (𝜑 → 𝐾 ∈ DivRing)    &   ⊢ (𝜑 → 𝑈 ∈ (SubRing‘𝐸))    &   ⊢ (𝜑 → 𝑉 ∈ (SubRing‘𝐹))    ⇒   ⊢ (𝜑 → (dim‘𝐴) = ((dim‘𝐵) ·e (dim‘𝐶)))
Theorem	dimlssid 33635	If the dimension of a linear subspace 𝐿 is the dimension of the whole vector space 𝐸, then 𝐿 is the whole space. (Contributed by Thierry Arnoux, 3-Aug-2025.)
⊢ 𝐵 = (Base‘𝐸)    &   ⊢ (𝜑 → 𝐸 ∈ LVec)    &   ⊢ (𝜑 → (dim‘𝐸) ∈ ℕ0)    &   ⊢ (𝜑 → 𝐿 ∈ (LSubSp‘𝐸))    &   ⊢ (𝜑 → (dim‘(𝐸 ↾s 𝐿)) = (dim‘𝐸))    ⇒   ⊢ (𝜑 → 𝐿 = 𝐵)
Theorem	lvecendof1f1o 33636	If an endomorphism 𝑈 of a vector space 𝐸 of finite dimension is injective, then it is bijective. Item (b) of Corollary of Proposition 9 in [BourbakiAlg1] p. 298 . (Contributed by Thierry Arnoux, 3-Aug-2025.)
⊢ 𝐵 = (Base‘𝐸)    &   ⊢ (𝜑 → 𝐸 ∈ LVec)    &   ⊢ (𝜑 → (dim‘𝐸) ∈ ℕ0)    &   ⊢ (𝜑 → 𝑈 ∈ (𝐸 LMHom 𝐸))    &   ⊢ (𝜑 → 𝑈:𝐵–1-1→𝐵)    ⇒   ⊢ (𝜑 → 𝑈:𝐵–1-1-onto→𝐵)
Theorem	lactlmhm 33637*	In an associative algebra 𝐴, left-multiplication by a fixed element of the algebra is a module homomorphism, analogous to ringlghm 20228. (Contributed by Thierry Arnoux, 3-Aug-2025.)
⊢ 𝐵 = (Base‘𝐴)    &   ⊢ · = (.r‘𝐴)    &   ⊢ 𝐹 = (𝑥 ∈ 𝐵 ↦ (𝐶 · 𝑥))    &   ⊢ (𝜑 → 𝐴 ∈ AssAlg)    &   ⊢ (𝜑 → 𝐶 ∈ 𝐵)    ⇒   ⊢ (𝜑 → 𝐹 ∈ (𝐴 LMHom 𝐴))
Theorem	assalactf1o 33638*	In an associative algebra 𝐴, left-multiplication by a fixed element of the algebra is bijective. See also lactlmhm 33637. (Contributed by Thierry Arnoux, 3-Aug-2025.)
⊢ 𝐵 = (Base‘𝐴)    &   ⊢ · = (.r‘𝐴)    &   ⊢ 𝐹 = (𝑥 ∈ 𝐵 ↦ (𝐶 · 𝑥))    &   ⊢ (𝜑 → 𝐴 ∈ AssAlg)    &   ⊢ 𝐸 = (RLReg‘𝐴)    &   ⊢ 𝐾 = (Scalar‘𝐴)    &   ⊢ (𝜑 → 𝐾 ∈ DivRing)    &   ⊢ (𝜑 → (dim‘𝐴) ∈ ℕ0)    &   ⊢ (𝜑 → 𝐶 ∈ 𝐸)    ⇒   ⊢ (𝜑 → 𝐹:𝐵–1-1-onto→𝐵)
Theorem	assarrginv 33639	If an element 𝑋 of an associative algebra 𝐴 over a division ring 𝐾 is regular, then it is a unit. Proposition 2. in Chapter 5. of [BourbakiAlg2] p. 113. (Contributed by Thierry Arnoux, 3-Aug-2025.)
⊢ 𝐸 = (RLReg‘𝐴)    &   ⊢ 𝑈 = (Unit‘𝐴)    &   ⊢ 𝐾 = (Scalar‘𝐴)    &   ⊢ (𝜑 → 𝐴 ∈ AssAlg)    &   ⊢ (𝜑 → 𝐾 ∈ DivRing)    &   ⊢ (𝜑 → (dim‘𝐴) ∈ ℕ0)    &   ⊢ (𝜑 → 𝑋 ∈ 𝐸)    ⇒   ⊢ (𝜑 → 𝑋 ∈ 𝑈)
Theorem	assafld 33640	If an algebra 𝐴 of finite degree over a division ring 𝐾 is an integral domain, then it is a field. Corollary of Proposition 2. in Chapter 5. of [BourbakiAlg2] p. 113. (Contributed by Thierry Arnoux, 3-Aug-2025.)
⊢ 𝐾 = (Scalar‘𝐴)    &   ⊢ (𝜑 → 𝐴 ∈ AssAlg)    &   ⊢ (𝜑 → 𝐴 ∈ IDomn)    &   ⊢ (𝜑 → 𝐾 ∈ DivRing)    &   ⊢ (𝜑 → (dim‘𝐴) ∈ ℕ0)    ⇒   ⊢ (𝜑 → 𝐴 ∈ Field)
21.3.11  Field Extensions
Syntax	cfldext 33641	Syntax for the field extension relation.
class /FldExt
Syntax	cfinext 33642	Syntax for the finite field extension relation.
class /FinExt
Syntax	cextdg 33643	Syntax for the field extension degree operation.
class [:]
Definition	df-fldext 33644*	Definition of the field extension relation. (Contributed by Thierry Arnoux, 29-Jul-2023.)
⊢ /FldExt = {⟨𝑒, 𝑓⟩ ∣ ((𝑒 ∈ Field ∧ 𝑓 ∈ Field) ∧ (𝑓 = (𝑒 ↾s (Base‘𝑓)) ∧ (Base‘𝑓) ∈ (SubRing‘𝑒)))}
Definition	df-extdg 33645*	Definition of the field extension degree operation. (Contributed by Thierry Arnoux, 29-Jul-2023.)
⊢ [:] = (𝑒 ∈ V, 𝑓 ∈ (/FldExt “ {𝑒}) ↦ (dim‘((subringAlg ‘𝑒)‘(Base‘𝑓))))
Definition	df-finext 33646*	Definition of the finite field extension relation. (Contributed by Thierry Arnoux, 29-Jul-2023.)
⊢ /FinExt = {⟨𝑒, 𝑓⟩ ∣ (𝑒/FldExt𝑓 ∧ (𝑒[:]𝑓) ∈ ℕ0)}
Theorem	relfldext 33647	The field extension is a relation. (Contributed by Thierry Arnoux, 29-Jul-2023.)
⊢ Rel /FldExt
Theorem	brfldext 33648	The field extension relation explicited. (Contributed by Thierry Arnoux, 29-Jul-2023.)
⊢ ((𝐸 ∈ Field ∧ 𝐹 ∈ Field) → (𝐸/FldExt𝐹 ↔ (𝐹 = (𝐸 ↾s (Base‘𝐹)) ∧ (Base‘𝐹) ∈ (SubRing‘𝐸))))
Theorem	ccfldextrr 33649	The field of the complex numbers is an extension of the field of the real numbers. (Contributed by Thierry Arnoux, 20-Jul-2023.)
⊢ ℂfld/FldExtℝfld
Theorem	fldextfld1 33650	A field extension is only defined if the extension is a field. (Contributed by Thierry Arnoux, 29-Jul-2023.)
⊢ (𝐸/FldExt𝐹 → 𝐸 ∈ Field)
Theorem	fldextfld2 33651	A field extension is only defined if the subfield is a field. (Contributed by Thierry Arnoux, 29-Jul-2023.)
⊢ (𝐸/FldExt𝐹 → 𝐹 ∈ Field)
Theorem	fldextsubrg 33652	Field extension implies a subring relation. (Contributed by Thierry Arnoux, 29-Jul-2023.)
⊢ 𝑈 = (Base‘𝐹)    ⇒   ⊢ (𝐸/FldExt𝐹 → 𝑈 ∈ (SubRing‘𝐸))
Theorem	sdrgfldext 33653	A field 𝐸 and any sub-division-ring 𝐹 of 𝐸 form a field extension. (Contributed by Thierry Arnoux, 26-Oct-2025.)
⊢ 𝐵 = (Base‘𝐸)    &   ⊢ (𝜑 → 𝐸 ∈ Field)    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐸))    ⇒   ⊢ (𝜑 → 𝐸/FldExt(𝐸 ↾s 𝐹))
Theorem	fldextress 33654	Field extension implies a structure restriction relation. (Contributed by Thierry Arnoux, 29-Jul-2023.)
⊢ (𝐸/FldExt𝐹 → 𝐹 = (𝐸 ↾s (Base‘𝐹)))
Theorem	brfinext 33655	The finite field extension relation explicited. (Contributed by Thierry Arnoux, 29-Jul-2023.)
⊢ (𝐸/FldExt𝐹 → (𝐸/FinExt𝐹 ↔ (𝐸[:]𝐹) ∈ ℕ0))
Theorem	extdgval 33656	Value of the field extension degree operation. (Contributed by Thierry Arnoux, 29-Jul-2023.)
⊢ (𝐸/FldExt𝐹 → (𝐸[:]𝐹) = (dim‘((subringAlg ‘𝐸)‘(Base‘𝐹))))
Theorem	fldextsdrg 33657	Deduce sub-division-ring from field extension. (Contributed by Thierry Arnoux, 26-Oct-2025.)
⊢ 𝐵 = (Base‘𝐹)    &   ⊢ (𝜑 → 𝐸/FldExt𝐹)    ⇒   ⊢ (𝜑 → 𝐵 ∈ (SubDRing‘𝐸))
Theorem	fldextsralvec 33658	The subring algebra associated with a field extension is a vector space. (Contributed by Thierry Arnoux, 3-Aug-2023.)
⊢ (𝐸/FldExt𝐹 → ((subringAlg ‘𝐸)‘(Base‘𝐹)) ∈ LVec)
Theorem	extdgcl 33659	Closure of the field extension degree operation. (Contributed by Thierry Arnoux, 29-Jul-2023.)
⊢ (𝐸/FldExt𝐹 → (𝐸[:]𝐹) ∈ ℕ0*)
Theorem	extdggt0 33660	Degrees of field extension are greater than zero. (Contributed by Thierry Arnoux, 30-Jul-2023.)
⊢ (𝐸/FldExt𝐹 → 0 < (𝐸[:]𝐹))
Theorem	fldexttr 33661	Field extension is a transitive relation. (Contributed by Thierry Arnoux, 29-Jul-2023.)
⊢ ((𝐸/FldExt𝐹 ∧ 𝐹/FldExt𝐾) → 𝐸/FldExt𝐾)
Theorem	fldextid 33662	The field extension relation is reflexive. (Contributed by Thierry Arnoux, 30-Jul-2023.)
⊢ (𝐹 ∈ Field → 𝐹/FldExt𝐹)
Theorem	extdgid 33663	A trivial field extension has degree one. (Contributed by Thierry Arnoux, 4-Aug-2023.)
⊢ (𝐸 ∈ Field → (𝐸[:]𝐸) = 1)
Theorem	fldsdrgfldext 33664	A sub-division-ring of a field forms a field extension. (Contributed by Thierry Arnoux, 19-Oct-2025.)
⊢ 𝐺 = (𝐹 ↾s 𝐴)    &   ⊢ (𝜑 → 𝐹 ∈ Field)    &   ⊢ (𝜑 → 𝐴 ∈ (SubDRing‘𝐹))    ⇒   ⊢ (𝜑 → 𝐹/FldExt𝐺)
Theorem	fldsdrgfldext2 33665	A sub-sub-division-ring of a field forms a field extension. (Contributed by Thierry Arnoux, 19-Oct-2025.)
⊢ 𝐺 = (𝐹 ↾s 𝐴)    &   ⊢ (𝜑 → 𝐹 ∈ Field)    &   ⊢ (𝜑 → 𝐴 ∈ (SubDRing‘𝐹))    &   ⊢ (𝜑 → 𝐵 ∈ (SubDRing‘𝐺))    &   ⊢ 𝐻 = (𝐹 ↾s 𝐵)    ⇒   ⊢ (𝜑 → 𝐺/FldExt𝐻)
Theorem	extdgmul 33666	The multiplicativity formula for degrees of field extensions. Given 𝐸 a field extension of 𝐹, itself a field extension of 𝐾, the degree of the extension 𝐸/FldExt𝐾 is the product of the degrees of the extensions 𝐸/FldExt𝐹 and 𝐹/FldExt𝐾. Proposition 1.2 of [Lang], p. 224. (Contributed by Thierry Arnoux, 30-Jul-2023.)
⊢ ((𝐸/FldExt𝐹 ∧ 𝐹/FldExt𝐾) → (𝐸[:]𝐾) = ((𝐸[:]𝐹) ·e (𝐹[:]𝐾)))
Theorem	finexttrb 33667	The extension 𝐸 of 𝐾 is finite if and only if 𝐸 is finite over 𝐹 and 𝐹 is finite over 𝐾. Corollary 1.3 of [Lang] , p. 225. (Contributed by Thierry Arnoux, 30-Jul-2023.)
⊢ ((𝐸/FldExt𝐹 ∧ 𝐹/FldExt𝐾) → (𝐸/FinExt𝐾 ↔ (𝐸/FinExt𝐹 ∧ 𝐹/FinExt𝐾)))
Theorem	extdg1id 33668	If the degree of the extension 𝐸/FldExt𝐹 is 1, then 𝐸 and 𝐹 are identical. (Contributed by Thierry Arnoux, 6-Aug-2023.)
⊢ ((𝐸/FldExt𝐹 ∧ (𝐸[:]𝐹) = 1) → 𝐸 = 𝐹)
Theorem	extdg1b 33669	The degree of the extension 𝐸/FldExt𝐹 is 1 iff 𝐸 and 𝐹 are the same structure. (Contributed by Thierry Arnoux, 6-Aug-2023.)
⊢ (𝐸/FldExt𝐹 → ((𝐸[:]𝐹) = 1 ↔ 𝐸 = 𝐹))
Theorem	fldgenfldext 33670	A subfield 𝐹 extended with a set 𝐴 forms a field extension. (Contributed by Thierry Arnoux, 22-Jun-2025.)
⊢ 𝐵 = (Base‘𝐸)    &   ⊢ 𝐾 = (𝐸 ↾s 𝐹)    &   ⊢ 𝐿 = (𝐸 ↾s (𝐸 fldGen (𝐹 ∪ 𝐴)))    &   ⊢ (𝜑 → 𝐸 ∈ Field)    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐸))    &   ⊢ (𝜑 → 𝐴 ⊆ 𝐵)    ⇒   ⊢ (𝜑 → 𝐿/FldExt𝐾)
Theorem	fldextchr 33671	The characteristic of a subfield is the same as the characteristic of the larger field. (Contributed by Thierry Arnoux, 20-Aug-2023.)
⊢ (𝐸/FldExt𝐹 → (chr‘𝐹) = (chr‘𝐸))
Theorem	evls1fldgencl 33672	Closure of the subring polynomial evaluation in the field extention. (Contributed by Thierry Arnoux, 2-Apr-2025.)
⊢ 𝐵 = (Base‘𝐸)    &   ⊢ 𝑂 = (𝐸 evalSub1 𝐹)    &   ⊢ 𝑃 = (Poly1‘(𝐸 ↾s 𝐹))    &   ⊢ 𝑈 = (Base‘𝑃)    &   ⊢ (𝜑 → 𝐸 ∈ Field)    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐸))    &   ⊢ (𝜑 → 𝐴 ∈ 𝐵)    &   ⊢ (𝜑 → 𝐺 ∈ 𝑈)    ⇒   ⊢ (𝜑 → ((𝑂‘𝐺)‘𝐴) ∈ (𝐸 fldGen (𝐹 ∪ {𝐴})))
Theorem	ccfldsrarelvec 33673	The subring algebra of the complex numbers over the real numbers is a left vector space. (Contributed by Thierry Arnoux, 20-Aug-2023.)
⊢ ((subringAlg ‘ℂfld)‘ℝ) ∈ LVec
Theorem	ccfldextdgrr 33674	The degree of the field extension of the complex numbers over the real numbers is 2. (Suggested by GL, 4-Aug-2023.) (Contributed by Thierry Arnoux, 20-Aug-2023.)
⊢ (ℂfld[:]ℝfld) = 2
Theorem	fldextrspunlsplem 33675*	Lemma for fldextrspunlsp 33676: First direction. Part of the proof of Proposition 5, Chapter 5, of [BourbakiAlg2] p. 116. (Contributed by Thierry Arnoux, 13-Oct-2025.)
⊢ 𝐾 = (𝐿 ↾s 𝐹)    &   ⊢ 𝐼 = (𝐿 ↾s 𝐺)    &   ⊢ 𝐽 = (𝐿 ↾s 𝐻)    &   ⊢ (𝜑 → 𝐿 ∈ Field)    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐼))    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐽))    &   ⊢ (𝜑 → 𝐺 ∈ (SubDRing‘𝐿))    &   ⊢ (𝜑 → 𝐻 ∈ (SubDRing‘𝐿))    &   ⊢ 𝑁 = (RingSpan‘𝐿)    &   ⊢ 𝐶 = (𝑁‘(𝐺 ∪ 𝐻))    &   ⊢ 𝐸 = (𝐿 ↾s 𝐶)    &   ⊢ (𝜑 → 𝐵 ∈ (LBasis‘((subringAlg ‘𝐽)‘𝐹)))    &   ⊢ (𝜑 → 𝐵 ∈ Fin)    &   ⊢ (𝜑 → 𝑃:𝐻⟶𝐺)    &   ⊢ (𝜑 → 𝑃 finSupp (0g‘𝐿))    &   ⊢ (𝜑 → 𝑋 = (𝐿 Σg (𝑓 ∈ 𝐻 ↦ ((𝑃‘𝑓)(.r‘𝐿)𝑓))))    ⇒   ⊢ (𝜑 → ∃𝑎 ∈ (𝐺 ↑m 𝐵)(𝑎 finSupp (0g‘𝐿) ∧ 𝑋 = (𝐿 Σg (𝑏 ∈ 𝐵 ↦ ((𝑎‘𝑏)(.r‘𝐿)𝑏)))))
Theorem	fldextrspunlsp 33676	Lemma for fldextrspunfld 33678. The subring generated by the union of two field extensions 𝐺 and 𝐻 is the vector sub- 𝐺 space generated by a basis 𝐵 of 𝐻. Part of the proof of Proposition 5, Chapter 5, of [BourbakiAlg2] p. 116. (Contributed by Thierry Arnoux, 13-Oct-2025.)
⊢ 𝐾 = (𝐿 ↾s 𝐹)    &   ⊢ 𝐼 = (𝐿 ↾s 𝐺)    &   ⊢ 𝐽 = (𝐿 ↾s 𝐻)    &   ⊢ (𝜑 → 𝐿 ∈ Field)    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐼))    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐽))    &   ⊢ (𝜑 → 𝐺 ∈ (SubDRing‘𝐿))    &   ⊢ (𝜑 → 𝐻 ∈ (SubDRing‘𝐿))    &   ⊢ 𝑁 = (RingSpan‘𝐿)    &   ⊢ 𝐶 = (𝑁‘(𝐺 ∪ 𝐻))    &   ⊢ 𝐸 = (𝐿 ↾s 𝐶)    &   ⊢ (𝜑 → 𝐵 ∈ (LBasis‘((subringAlg ‘𝐽)‘𝐹)))    &   ⊢ (𝜑 → 𝐵 ∈ Fin)    ⇒   ⊢ (𝜑 → 𝐶 = ((LSpan‘((subringAlg ‘𝐿)‘𝐺))‘𝐵))
Theorem	fldextrspunlem1 33677	Lemma for fldextrspunfld 33678. Part of the proof of Proposition 5, Chapter 5, of [BourbakiAlg2] p. 116. (Contributed by Thierry Arnoux, 13-Oct-2025.)
⊢ 𝐾 = (𝐿 ↾s 𝐹)    &   ⊢ 𝐼 = (𝐿 ↾s 𝐺)    &   ⊢ 𝐽 = (𝐿 ↾s 𝐻)    &   ⊢ (𝜑 → 𝐿 ∈ Field)    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐼))    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐽))    &   ⊢ (𝜑 → 𝐺 ∈ (SubDRing‘𝐿))    &   ⊢ (𝜑 → 𝐻 ∈ (SubDRing‘𝐿))    &   ⊢ (𝜑 → (𝐽[:]𝐾) ∈ ℕ0)    &   ⊢ 𝑁 = (RingSpan‘𝐿)    &   ⊢ 𝐶 = (𝑁‘(𝐺 ∪ 𝐻))    &   ⊢ 𝐸 = (𝐿 ↾s 𝐶)    ⇒   ⊢ (𝜑 → (dim‘((subringAlg ‘𝐸)‘𝐺)) ≤ (𝐽[:]𝐾))
Theorem	fldextrspunfld 33678	The ring generated by the union of two field extensions is a field. Part of the proof of Proposition 5, Chapter 5, of [BourbakiAlg2] p. 116. (Contributed by Thierry Arnoux, 13-Oct-2025.)
⊢ 𝐾 = (𝐿 ↾s 𝐹)    &   ⊢ 𝐼 = (𝐿 ↾s 𝐺)    &   ⊢ 𝐽 = (𝐿 ↾s 𝐻)    &   ⊢ (𝜑 → 𝐿 ∈ Field)    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐼))    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐽))    &   ⊢ (𝜑 → 𝐺 ∈ (SubDRing‘𝐿))    &   ⊢ (𝜑 → 𝐻 ∈ (SubDRing‘𝐿))    &   ⊢ (𝜑 → (𝐽[:]𝐾) ∈ ℕ0)    &   ⊢ 𝑁 = (RingSpan‘𝐿)    &   ⊢ 𝐶 = (𝑁‘(𝐺 ∪ 𝐻))    &   ⊢ 𝐸 = (𝐿 ↾s 𝐶)    ⇒   ⊢ (𝜑 → 𝐸 ∈ Field)
Theorem	fldextrspunlem2 33679	Part of the proof of Proposition 5, Chapter 5, of [BourbakiAlg2] p. 116. (Contributed by Thierry Arnoux, 13-Oct-2025.)
⊢ 𝐾 = (𝐿 ↾s 𝐹)    &   ⊢ 𝐼 = (𝐿 ↾s 𝐺)    &   ⊢ 𝐽 = (𝐿 ↾s 𝐻)    &   ⊢ (𝜑 → 𝐿 ∈ Field)    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐼))    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐽))    &   ⊢ (𝜑 → 𝐺 ∈ (SubDRing‘𝐿))    &   ⊢ (𝜑 → 𝐻 ∈ (SubDRing‘𝐿))    &   ⊢ (𝜑 → (𝐽[:]𝐾) ∈ ℕ0)    &   ⊢ 𝑁 = (RingSpan‘𝐿)    &   ⊢ 𝐶 = (𝑁‘(𝐺 ∪ 𝐻))    &   ⊢ 𝐸 = (𝐿 ↾s 𝐶)    ⇒   ⊢ (𝜑 → 𝐶 = (𝐿 fldGen (𝐺 ∪ 𝐻)))
Theorem	fldextrspundgle 33680	Inequality involving the degree of two different field extensions 𝐼 and 𝐽 of a same field 𝐹. Part of the proof of Proposition 5, Chapter 5, of [BourbakiAlg2] p. 116. (Contributed by Thierry Arnoux, 13-Oct-2025.)
⊢ 𝐾 = (𝐿 ↾s 𝐹)    &   ⊢ 𝐼 = (𝐿 ↾s 𝐺)    &   ⊢ 𝐽 = (𝐿 ↾s 𝐻)    &   ⊢ (𝜑 → 𝐿 ∈ Field)    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐼))    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐽))    &   ⊢ (𝜑 → 𝐺 ∈ (SubDRing‘𝐿))    &   ⊢ (𝜑 → 𝐻 ∈ (SubDRing‘𝐿))    &   ⊢ (𝜑 → (𝐽[:]𝐾) ∈ ℕ0)    &   ⊢ 𝐸 = (𝐿 ↾s (𝐿 fldGen (𝐺 ∪ 𝐻)))    ⇒   ⊢ (𝜑 → (𝐸[:]𝐼) ≤ (𝐽[:]𝐾))
Theorem	fldextrspundglemul 33681	Given two field extensions 𝐼 / 𝐾 and 𝐽 / 𝐾 of the same field 𝐾, 𝐽 / 𝐾 being finite, and the composiste field 𝐸 = 𝐼𝐽, the degree of the extension of the composite field 𝐸 / 𝐾 is at most the product of the field extension degrees of 𝐼 / 𝐾 and 𝐽 / 𝐾. (Contributed by Thierry Arnoux, 19-Oct-2025.)
⊢ 𝐾 = (𝐿 ↾s 𝐹)    &   ⊢ 𝐼 = (𝐿 ↾s 𝐺)    &   ⊢ 𝐽 = (𝐿 ↾s 𝐻)    &   ⊢ (𝜑 → 𝐿 ∈ Field)    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐼))    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐽))    &   ⊢ (𝜑 → 𝐺 ∈ (SubDRing‘𝐿))    &   ⊢ (𝜑 → 𝐻 ∈ (SubDRing‘𝐿))    &   ⊢ (𝜑 → (𝐽[:]𝐾) ∈ ℕ0)    &   ⊢ 𝐸 = (𝐿 ↾s (𝐿 fldGen (𝐺 ∪ 𝐻)))    ⇒   ⊢ (𝜑 → (𝐸[:]𝐾) ≤ ((𝐼[:]𝐾) ·e (𝐽[:]𝐾)))
Theorem	fldextrspundgdvdslem 33682	Lemma for fldextrspundgdvds 33683. (Contributed by Thierry Arnoux, 19-Oct-2025.)
⊢ 𝐾 = (𝐿 ↾s 𝐹)    &   ⊢ 𝐼 = (𝐿 ↾s 𝐺)    &   ⊢ 𝐽 = (𝐿 ↾s 𝐻)    &   ⊢ (𝜑 → 𝐿 ∈ Field)    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐼))    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐽))    &   ⊢ (𝜑 → 𝐺 ∈ (SubDRing‘𝐿))    &   ⊢ (𝜑 → 𝐻 ∈ (SubDRing‘𝐿))    &   ⊢ (𝜑 → (𝐽[:]𝐾) ∈ ℕ0)    &   ⊢ 𝐸 = (𝐿 ↾s (𝐿 fldGen (𝐺 ∪ 𝐻)))    &   ⊢ (𝜑 → (𝐼[:]𝐾) ∈ ℕ)    ⇒   ⊢ (𝜑 → (𝐸[:]𝐼) ∈ ℕ0)
Theorem	fldextrspundgdvds 33683	Given two finite extensions 𝐼 / 𝐾 and 𝐽 / 𝐾 of the same field 𝐾, the degree of the extension 𝐼 / 𝐾 divides the degree of the extension 𝐸 / 𝐾, 𝐸 being the composite field 𝐼𝐽. (Contributed by Thierry Arnoux, 19-Oct-2025.)
⊢ 𝐾 = (𝐿 ↾s 𝐹)    &   ⊢ 𝐼 = (𝐿 ↾s 𝐺)    &   ⊢ 𝐽 = (𝐿 ↾s 𝐻)    &   ⊢ (𝜑 → 𝐿 ∈ Field)    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐼))    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐽))    &   ⊢ (𝜑 → 𝐺 ∈ (SubDRing‘𝐿))    &   ⊢ (𝜑 → 𝐻 ∈ (SubDRing‘𝐿))    &   ⊢ (𝜑 → (𝐽[:]𝐾) ∈ ℕ0)    &   ⊢ 𝐸 = (𝐿 ↾s (𝐿 fldGen (𝐺 ∪ 𝐻)))    &   ⊢ (𝜑 → (𝐼[:]𝐾) ∈ ℕ)    ⇒   ⊢ (𝜑 → (𝐼[:]𝐾) ∥ (𝐸[:]𝐾))
Theorem	fldext2rspun 33684*	Given two field extensions 𝐼 / 𝐾 and 𝐽 / 𝐾, 𝐼 / 𝐾 being a quadratic extension, and the degree of 𝐽 / 𝐾 being a power of 2, the degree of the extension 𝐸 / 𝐾 is a power of 2 , 𝐸 being the composite field 𝐼𝐽. (Contributed by Thierry Arnoux, 19-Oct-2025.)
⊢ 𝐾 = (𝐿 ↾s 𝐹)    &   ⊢ 𝐼 = (𝐿 ↾s 𝐺)    &   ⊢ 𝐽 = (𝐿 ↾s 𝐻)    &   ⊢ (𝜑 → 𝐿 ∈ Field)    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐼))    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐽))    &   ⊢ (𝜑 → 𝐺 ∈ (SubDRing‘𝐿))    &   ⊢ (𝜑 → 𝐻 ∈ (SubDRing‘𝐿))    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    &   ⊢ (𝜑 → (𝐼[:]𝐾) = 2)    &   ⊢ (𝜑 → (𝐽[:]𝐾) = (2↑𝑁))    &   ⊢ 𝐸 = (𝐿 ↾s (𝐿 fldGen (𝐺 ∪ 𝐻)))    ⇒   ⊢ (𝜑 → ∃𝑛 ∈ ℕ0 (𝐸[:]𝐾) = (2↑𝑛))
21.3.11.1  Algebraic numbers
Syntax	cirng 33685	Integral subring of a ring.
class IntgRing
Definition	df-irng 33686*	Define the subring of elements of a ring 𝑟 integral over a subset 𝑠. (Contributed by Mario Carneiro, 2-Dec-2014.) (Revised by Thierry Arnoux, 28-Jan-2025.)
⊢ IntgRing = (𝑟 ∈ V, 𝑠 ∈ V ↦ ∪ 𝑓 ∈ (Monic1p‘(𝑟 ↾s 𝑠))(◡((𝑟 evalSub1 𝑠)‘𝑓) “ {(0g‘𝑟)}))
Theorem	irngval 33687*	The elements of a field 𝑅 integral over a subset 𝑆. In the case of a subfield, those are the algebraic numbers over the field 𝑆 within the field 𝑅. That is, the numbers 𝑋 which are roots of monic polynomials 𝑃(𝑋) with coefficients in 𝑆. (Contributed by Thierry Arnoux, 28-Jan-2025.)
⊢ 𝑂 = (𝑅 evalSub1 𝑆)    &   ⊢ 𝑈 = (𝑅 ↾s 𝑆)    &   ⊢ 𝐵 = (Base‘𝑅)    &   ⊢ 0 = (0g‘𝑅)    &   ⊢ (𝜑 → 𝑅 ∈ Ring)    &   ⊢ (𝜑 → 𝑆 ⊆ 𝐵)    ⇒   ⊢ (𝜑 → (𝑅 IntgRing 𝑆) = ∪ 𝑓 ∈ (Monic1p‘𝑈)(◡(𝑂‘𝑓) “ { 0 }))
Theorem	elirng 33688*	Property for an element 𝑋 of a field 𝑅 to be integral over a subring 𝑆. (Contributed by Thierry Arnoux, 28-Jan-2025.)
⊢ 𝑂 = (𝑅 evalSub1 𝑆)    &   ⊢ 𝑈 = (𝑅 ↾s 𝑆)    &   ⊢ 𝐵 = (Base‘𝑅)    &   ⊢ 0 = (0g‘𝑅)    &   ⊢ (𝜑 → 𝑅 ∈ CRing)    &   ⊢ (𝜑 → 𝑆 ∈ (SubRing‘𝑅))    ⇒   ⊢ (𝜑 → (𝑋 ∈ (𝑅 IntgRing 𝑆) ↔ (𝑋 ∈ 𝐵 ∧ ∃𝑓 ∈ (Monic1p‘𝑈)((𝑂‘𝑓)‘𝑋) = 0 )))
Theorem	irngss 33689	All elements of a subring 𝑆 are integral over 𝑆. This is only true in the case of a nonzero ring, since there are no integral elements in a zero ring (see 0ringirng 33691). (Contributed by Thierry Arnoux, 28-Jan-2025.)
⊢ 𝑂 = (𝑅 evalSub1 𝑆)    &   ⊢ 𝑈 = (𝑅 ↾s 𝑆)    &   ⊢ 𝐵 = (Base‘𝑅)    &   ⊢ 0 = (0g‘𝑅)    &   ⊢ (𝜑 → 𝑅 ∈ CRing)    &   ⊢ (𝜑 → 𝑆 ∈ (SubRing‘𝑅))    &   ⊢ (𝜑 → 𝑅 ∈ NzRing)    ⇒   ⊢ (𝜑 → 𝑆 ⊆ (𝑅 IntgRing 𝑆))
Theorem	irngssv 33690	An integral element is an element of the base set. (Contributed by Thierry Arnoux, 28-Jan-2025.)
⊢ 𝑂 = (𝑅 evalSub1 𝑆)    &   ⊢ 𝑈 = (𝑅 ↾s 𝑆)    &   ⊢ 𝐵 = (Base‘𝑅)    &   ⊢ 0 = (0g‘𝑅)    &   ⊢ (𝜑 → 𝑅 ∈ CRing)    &   ⊢ (𝜑 → 𝑆 ∈ (SubRing‘𝑅))    ⇒   ⊢ (𝜑 → (𝑅 IntgRing 𝑆) ⊆ 𝐵)
Theorem	0ringirng 33691	A zero ring 𝑅 has no integral elements. (Contributed by Thierry Arnoux, 5-Feb-2025.)
⊢ 𝑂 = (𝑅 evalSub1 𝑆)    &   ⊢ 𝑈 = (𝑅 ↾s 𝑆)    &   ⊢ 𝐵 = (Base‘𝑅)    &   ⊢ 0 = (0g‘𝑅)    &   ⊢ (𝜑 → 𝑅 ∈ CRing)    &   ⊢ (𝜑 → 𝑆 ∈ (SubRing‘𝑅))    &   ⊢ (𝜑 → ¬ 𝑅 ∈ NzRing)    ⇒   ⊢ (𝜑 → (𝑅 IntgRing 𝑆) = ∅)
Theorem	irngnzply1lem 33692	In the case of a field 𝐸, a root 𝑋 of some nonzero polynomial 𝑃 with coefficients in a subfield 𝐹 is integral over 𝐹. (Contributed by Thierry Arnoux, 5-Feb-2025.)
⊢ 𝑂 = (𝐸 evalSub1 𝐹)    &   ⊢ 𝑍 = (0g‘(Poly1‘𝐸))    &   ⊢ 0 = (0g‘𝐸)    &   ⊢ (𝜑 → 𝐸 ∈ Field)    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐸))    &   ⊢ 𝐵 = (Base‘𝐸)    &   ⊢ (𝜑 → 𝑃 ∈ dom 𝑂)    &   ⊢ (𝜑 → 𝑃 ≠ 𝑍)    &   ⊢ (𝜑 → ((𝑂‘𝑃)‘𝑋) = 0 )    &   ⊢ (𝜑 → 𝑋 ∈ 𝐵)    ⇒   ⊢ (𝜑 → 𝑋 ∈ (𝐸 IntgRing 𝐹))
Theorem	irngnzply1 33693*	In the case of a field 𝐸, the roots of nonzero polynomials 𝑝 with coefficients in a subfield 𝐹 are exactly the integral elements over 𝐹. Roots of nonzero polynomials are called algebraic numbers, so this shows that in the case of a field, elements integral over 𝐹 are exactly the algebraic numbers. In this formula, dom 𝑂 represents the polynomials, and 𝑍 the zero polynomial. (Contributed by Thierry Arnoux, 5-Feb-2025.)
⊢ 𝑂 = (𝐸 evalSub1 𝐹)    &   ⊢ 𝑍 = (0g‘(Poly1‘𝐸))    &   ⊢ 0 = (0g‘𝐸)    &   ⊢ (𝜑 → 𝐸 ∈ Field)    &   ⊢ (𝜑 → 𝐹 ∈ (SubDRing‘𝐸))    ⇒   ⊢ (𝜑 → (𝐸 IntgRing 𝐹) = ∪ 𝑝 ∈ (dom 𝑂 ∖ {𝑍})(◡(𝑂‘𝑝) “ { 0 }))
21.3.11.2  Algebraic extensions
Syntax	calgext 33694	Syntax for the algebraic field extension relation.
class /AlgExt
Definition	df-algext 33695*	Definition of the algebraic extension relation. (Contributed by Thierry Arnoux, 29-Jul-2023.)
⊢ /AlgExt = {⟨𝑒, 𝑓⟩ ∣ (𝑒/FldExt𝑓 ∧ (𝑒 IntgRing 𝑓) = (Base‘𝑒))}
21.3.11.3  Minimal polynomials
Syntax	cminply 33696	Extend class notation with the minimal polynomial builder function.
class minPoly
Definition	df-minply 33697*	Define the minimal polynomial builder function. (Contributed by Thierry Arnoux, 19-Jan-2025.)
⊢ minPoly = (𝑒 ∈ V, 𝑓 ∈ V ↦ (𝑥 ∈ (Base‘𝑒) ↦ ((idlGen1p‘(𝑒 ↾s 𝑓))‘{𝑝 ∈ dom (𝑒 evalSub1 𝑓) ∣ (((𝑒 evalSub1 𝑓)‘𝑝)‘𝑥) = (0g‘𝑒)})))
Theorem	ply1annidllem 33698*	Write the set 𝑄 of polynomials annihilating an element 𝐴 as the kernel of the ring homomorphism 𝐹 mapping polynomials 𝑝 to their subring evaluation at a given point 𝐴. (Contributed by Thierry Arnoux, 9-Feb-2025.)
⊢ 𝑂 = (𝑅 evalSub1 𝑆)    &   ⊢ 𝑃 = (Poly1‘(𝑅 ↾s 𝑆))    &   ⊢ 𝐵 = (Base‘𝑅)    &   ⊢ (𝜑 → 𝑅 ∈ CRing)    &   ⊢ (𝜑 → 𝑆 ∈ (SubRing‘𝑅))    &   ⊢ (𝜑 → 𝐴 ∈ 𝐵)    &   ⊢ 0 = (0g‘𝑅)    &   ⊢ 𝑄 = {𝑞 ∈ dom 𝑂 ∣ ((𝑂‘𝑞)‘𝐴) = 0 }    &   ⊢ 𝐹 = (𝑝 ∈ (Base‘𝑃) ↦ ((𝑂‘𝑝)‘𝐴))    ⇒   ⊢ (𝜑 → 𝑄 = (◡𝐹 “ { 0 }))
Theorem	ply1annidl 33699*	The set 𝑄 of polynomials annihilating an element 𝐴 forms an ideal. (Contributed by Thierry Arnoux, 9-Feb-2025.)
⊢ 𝑂 = (𝑅 evalSub1 𝑆)    &   ⊢ 𝑃 = (Poly1‘(𝑅 ↾s 𝑆))    &   ⊢ 𝐵 = (Base‘𝑅)    &   ⊢ (𝜑 → 𝑅 ∈ CRing)    &   ⊢ (𝜑 → 𝑆 ∈ (SubRing‘𝑅))    &   ⊢ (𝜑 → 𝐴 ∈ 𝐵)    &   ⊢ 0 = (0g‘𝑅)    &   ⊢ 𝑄 = {𝑞 ∈ dom 𝑂 ∣ ((𝑂‘𝑞)‘𝐴) = 0 }    ⇒   ⊢ (𝜑 → 𝑄 ∈ (LIdeal‘𝑃))
Theorem	ply1annnr 33700*	The set 𝑄 of polynomials annihilating an element 𝐴 is not the whole polynomial ring. (Contributed by Thierry Arnoux, 22-Mar-2025.)
⊢ 𝑂 = (𝑅 evalSub1 𝑆)    &   ⊢ 𝑃 = (Poly1‘(𝑅 ↾s 𝑆))    &   ⊢ 𝐵 = (Base‘𝑅)    &   ⊢ (𝜑 → 𝑅 ∈ CRing)    &   ⊢ (𝜑 → 𝑆 ∈ (SubRing‘𝑅))    &   ⊢ (𝜑 → 𝐴 ∈ 𝐵)    &   ⊢ 0 = (0g‘𝑅)    &   ⊢ 𝑄 = {𝑞 ∈ dom 𝑂 ∣ ((𝑂‘𝑞)‘𝐴) = 0 }    &   ⊢ 𝑈 = (Base‘𝑃)    &   ⊢ (𝜑 → 𝑅 ∈ NzRing)    ⇒   ⊢ (𝜑 → 𝑄 ≠ 𝑈)