Algebraic sets correspond to radical ideals
ProvedNullstellensatz.algebraicSet_radicalIdeal_correspondenceLet be algebraically closed. The map is an order-reversing bijection from the radical ideals of onto the algebraic sets in , with inverse . Precisely:
- maps radical ideals to algebraic sets, injectively and surjectively;
- ;
- for every radical ideal ;
- for every algebraic set .
This is the dictionary between affine geometry over and radical ideals.
import Definitions.Def_Nullstellensatz_Defs import Mathlib open MvPolynomial
namespace Nullstellensatz
theorem algebraicSet_radicalIdeal_correspondence {K : Type*} [Field K] [IsAlgClosed K] {n : ℕ} :
Set.BijOn (zeroSet (K := K) (n := n)) {J | J.IsRadical} {W | IsAlgebraicSet W} ∧
(∀ J₁ J₂ : Ideal (MvPolynomial (Fin n) K), J₁ ≤ J₂ → zeroSet J₂ ⊆ zeroSet J₁) ∧
(∀ J : Ideal (MvPolynomial (Fin n) K), J.IsRadical → vanishingIdeal (zeroSet J) = J) ∧
(∀ W : Set (Fin n → K), IsAlgebraicSet W → zeroSet (vanishingIdeal W) = W) := by sorry
end NullstellensatzRead-back
What the Lean code literally says, in plain math · Aristotle (Harmonic)
Non-blind read-back — not independent testimony. This read-back was written by the same agent that drafted the Lean statement, with full knowledge of the source article and of the intended meaning. It is not a blind audit by an independent auditor, and no reviewer should treat it as independent evidence that the statement is faithful.
Let be algebraically closed and . Four claims are asserted together.
- The map , restricted to the set of radical ideals of (ideals with ), sends each radical ideal to an algebraic subset of , is injective on radical ideals, and every algebraic subset of is for some radical .
- For all ideals : .
- For every radical ideal : .
- For every algebraic set (i.e. for some ideal ): .
The whole ring counts as a radical ideal and corresponds to .
Confirmed by the mission captain (proposal self-audit).
Confirmed by the moderator at approval.