open import Algebra.Ring.Module.Action
open import Algebra.Group.Subgroup
open import Algebra.Ring.Module
open import Algebra.Group.Ab
open import Algebra.Prelude
open import Algebra.Group
open import Algebra.Ring

open import Cat.Functor.FullSubcategory

open import Data.Power

open import Meta.Bind

module Algebra.Ring.Ideal where

Ideals in rings🔗

An ideal in a ring $R$ is the $\mathbf{Ab}$-enriched analogue of a sieve, when $R$ is considered as an $\mathbf{Ab}$-category with a single object, in that it picks out a sub-$R$-module of $R$, considered as a representable ring, in exactly the same way that a sieve on an object $x : \mathcal{C}$ picks out a subfunctor of $よ(x)$. Since we know that $\operatorname*{\mathbf{B}}R$’s composition is given by $R$’s multiplication, and sieves are subsets closed under precomposition, we instantly deduce that ideals are closed under multiplication.

In the $\mathbf{Ab}$-enriched setting, however, there are some more operations that leaves us in the same $\mathbf{Hom}$-group: addition! More generally, the abelian group operations, i.e. addition, inverse, and the zero morphism. Putting these together we conclude that an ideal in $R$ is a subset of $R$ containing the identity, which is closed under multiplication and addition.

module _ {ℓ} (R : Ring ℓ) where
  private module R = Ring-on (R .snd)

  record is-ideal (𝔞 : ℙ ⌞ R ⌟) : Type (lsuc ℓ) where
    no-eta-equality
    field
      has-rep-subgroup : represents-subgroup R.additive-group 𝔞

      -- Note: these are named after the side the scalar acts on.
      has-*ₗ : ∀ x {y} → y ∈ 𝔞 → (x R.* y) ∈ 𝔞
      has-*ᵣ : ∀ x {y} → y ∈ 𝔞 → (y R.* x) ∈ 𝔞

Note: Since most of the rings over which we want to consider ideals are commutative rings, we will limit ourselves to the definition of two-sided ideals: Those for which we have, for $y \in \mathfrak{a}$ and any element $x : R$, $xy \in \mathfrak{a}$ and $yx \in \mathfrak{a}$.

    open represents-subgroup has-rep-subgroup
      renaming ( has-unit to has-0 ; has-⋆ to has-+ ; has-inv to has-neg )
      public

    ideal→normal : normal-subgroup R.additive-group 𝔞
    ideal→normal .normal-subgroup.has-rep = has-rep-subgroup
    ideal→normal .normal-subgroup.has-conjugate {y = y} x∈𝔞 =
      subst (_∈ 𝔞) (sym (ap (y R.+_) R.+-commutes ∙ R.cancell R.+-invr)) x∈𝔞

    open normal-subgroup ideal→normal hiding (has-rep) public

Since an ideal is a subgroup of $R$’s additive group, its total space inherits a group structure, and since multiplication in $R$ distributes over addition in $R$, the group structure induced on $\mathfrak{a}$ carries a canonical $R$-module structure.

  ideal→module : (𝔞 : ℙ ⌞ R ⌟) → is-ideal 𝔞 → Module R ℓ
  ideal→module 𝔞 x = g .fst , mod where
    open Ring-action
    open is-ideal x
    gr : Group-on _
    gr = rep-subgroup→group-on 𝔞 has-rep-subgroup

    g = from-commutative-group (el! _ , gr) λ x y → Σ-prop-path! R.+-commutes

    mod : Module-on R ⌞ g ⌟
    mod = Action→Module-on R {G = g .snd} λ where
      ._⋆_ r (a , b) → _ , has-*ₗ r b
      .⋆-distribl r x y → Σ-prop-path! R.*-distribl
      .⋆-distribr r s x → Σ-prop-path! R.*-distribr
      .⋆-assoc r s x    → Σ-prop-path! R.*-associative
      .⋆-id x           → Σ-prop-path! R.*-idl

Since a map between modules is a monomorphism when its underlying function is injective, and the first projection from a subset is always injective, we can quickly conclude that the module structure on $\mathfrak{a}$ is a sub-$R$-module of $R$:

  ideal→submodule
    : {𝔞 : ℙ ⌞ R ⌟} (idl : is-ideal 𝔞)
    → ideal→module 𝔞 idl R-Mod.↪ representable-module R
  ideal→submodule {𝔞 = 𝔞} idl = record
    { mor   = total-hom fst (record { linear = λ _ _ _ → refl })
    ; monic = λ {c = c} g h x → Structured-hom-path (R-Mod-structure R) $
      embedding→monic (Subset-proj-embedding λ _ → 𝔞 _ .is-tr) (g .hom) (h .hom) (ap hom x)
    }

Principal ideals🔗

Every element $a : R$ generates an ideal: that of its multiples, which we denote $(a)$. You’ll note that we have to use the $\exists$ quantifier (rather than the $\sigma$ quantifier) to define $(a)$, since in an arbitrary ring, multiplication by $a$ may fail to be injective, so, given $a, b : R$ $b = ca = c'a$, we can’t in general conclude that $c = c'$. Of course, in any ring, multiplication by zero is never injective.

Note that, since our ideals are all two-sided (for simplicity) but our rings may not be commutative (for expressiveness), if we want the ideal generated by an element to be two-sided, then our ring must be commutative.

  principal-ideal
    : (∀ x y → x R.* y ≡ y R.* x)
    → (a : ⌞ R ⌟)
    → is-ideal λ b → elΩ (Σ _ λ c → b ≡ c R.* a)
  principal-ideal comm a = record
    { has-rep-subgroup = record
      { has-unit = pure (_ , sym R.*-zerol)
      ; has-⋆    = λ x y → do
          (xi , p) ← x
          (yi , q) ← y
          pure (xi R.+ yi , ap₂ R._+_ p q ∙ sym R.*-distribr)
      ; has-inv  = λ x → do
          (xi , p) ← x
          pure (R.- xi , ap R.-_ p ∙ R.neg-*-l)
      }
    ; has-*ₗ = λ x y → do
        (yi , q) ← y
        pure (x R.* yi , ap (x R.*_) q ∙ R.*-associative)
    ; has-*ᵣ = λ x y → do
        (yi , q) ← y
        pure ( yi R.* x
             , ap (R._* x) q ·· sym R.*-associative ·· ap (yi R.*_) (comm a x)
             ∙ R.*-associative)
    }