Theorems · Theorem · commutative algebra
Submodule.finiteQuotientOfFreeOfRankEq
∀ {M : Type u_3} [inst : AddCommGroup M] [Module.Free ℤ M] [Module.Finite ℤ M] (N : Submodule ℤ M),
Module.finrank ℤ ↥N = Module.finrank ℤ M → Finite (M ⧸ N)A submodule of full rank of a free finite ℤ-module has a finite quotient.
It can't be an instance because of the side condition Module.finrank ℤ N = Module.finrank ℤ M.
- Cited by
- 3 results in Mathlib
- Foundations
- Depth 131 from the axioms · uses propext, Classical.choice, Quot.sound
Around this declaration
Dashed lines are statement dependencies; solid lines are citations in proofs.
Cites18
Mathlib declarations this one mentions in its statement or cites explicitly in its proof. Plumbing is filtered out.
- AddCommGroupstatement and proof · cited by 12,871
- Submodulestatement and proof · cited by 7,192
- Finitestatement · cited by 3,029
- HasQuotient.Quotientstatement and proof · cited by 2,301
- Module.finrankstatement and proof · cited by 1,770
- Module.Basisproof · cited by 1,477
- AddEquivproof · cited by 1,087
- Module.Finitestatement and proof · cited by 1,032
- ZModproof · cited by 1,024
- Module.Freestatement and proof · cited by 597
- AddEquiv.symmproof · cited by 530
- Module.Free.ChooseBasisIndexproof · cited by 133
Cited by3
Results whose statement or proof uses this declaration.
- Ideal.finiteQuotientOfFreeOfNeBotproof · cited by 3
- Submodule.finiteQuotient_iffproof · cited by 1
- Ideal.exists_prime_and_absNorm_eq_powproof · cited by 1