English
Variant of disjoint ker lemma: under the corresponding finrank inequality, L ∩ ker f is trivial.
Русский
Вариант леммы о независимости пересечения: при соответствующем неравенстве рангов пересечение равно {0}.
LaTeX
$$$\text{If } (L,f) \text{ satisfy the finrank condition, then } L \cap \ker f = \{0\}.$$$
Lean4
/-- Rank-nullity theorem using `finrank` and subtraction. -/
theorem finrank_quotient [Module.Finite R M] {S : Type*} [Ring S] [SMul R S] [Module S M] [IsScalarTower R S M]
(N : Submodule S M) : finrank R (M ⧸ N) = finrank R M - finrank R N :=
by
rw [← (N.restrictScalars R).finrank_quotient_add_finrank]
exact Nat.eq_sub_of_add_eq rfl