A homogeneous prime that is homogeneously prime but not prime #
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In src/ring_theory/graded_algebra/radical.lean,  we assumed that the underline grading is indexed
by a linear_ordered_cancel_add_comm_monoid to prove that a homogeneous ideal is prime if and only
if it is homogeneously prime. This file is aimed to show that even if this assumption isn't strictly
necessary, the assumption of "being cancellative" is. We construct a counterexample where the
underlying indexing set is a linear_ordered_add_comm_monoid but is not cancellative and the
statement is false.
We achieve this by considering the ring R=ℤ/4ℤ. We first give the two element set ι = {0, 1} a
structure of linear ordered additive commutative monoid by setting 0 + 0 = 0 and _ + _ = 1 and
0 < 1. Then we use ι to grade R² by setting {(a, a) | a ∈ R} to have grade 0; and
{(0, b) | b ∈ R} to have grade 1. Then the ideal I = span {(0, 2)} ⊆ ℤ/4ℤ is homogeneous and not
prime. But it is homogeneously prime, i.e. if (a, b), (c, d) are two homogeneous elements then
(a, b) * (c, d) ∈ I implies either (a, b) ∈ I or (c, d) ∈ I.
Tags #
homogeneous, prime
Equations
- counterexample.counterexample_not_prime_but_homogeneous_prime.two.linear_ordered_add_comm_monoid = {le := linear_order.le with_zero.linear_order, lt := linear_order.lt with_zero.linear_order, le_refl := counterexample.counterexample_not_prime_but_homogeneous_prime.two.linear_ordered_add_comm_monoid._proof_1, le_trans := counterexample.counterexample_not_prime_but_homogeneous_prime.two.linear_ordered_add_comm_monoid._proof_2, lt_iff_le_not_le := counterexample.counterexample_not_prime_but_homogeneous_prime.two.linear_ordered_add_comm_monoid._proof_3, le_antisymm := counterexample.counterexample_not_prime_but_homogeneous_prime.two.linear_ordered_add_comm_monoid._proof_4, le_total := counterexample.counterexample_not_prime_but_homogeneous_prime.two.linear_ordered_add_comm_monoid._proof_5, decidable_le := linear_order.decidable_le with_zero.linear_order, decidable_eq := linear_order.decidable_eq with_zero.linear_order, decidable_lt := linear_order.decidable_lt with_zero.linear_order, max := linear_order.max with_zero.linear_order, max_def := counterexample.counterexample_not_prime_but_homogeneous_prime.two.linear_ordered_add_comm_monoid._proof_6, min := linear_order.min with_zero.linear_order, min_def := counterexample.counterexample_not_prime_but_homogeneous_prime.two.linear_ordered_add_comm_monoid._proof_7, add := add_comm_monoid.add with_zero.add_comm_monoid, add_assoc := counterexample.counterexample_not_prime_but_homogeneous_prime.two.linear_ordered_add_comm_monoid._proof_8, zero := add_comm_monoid.zero with_zero.add_comm_monoid, zero_add := counterexample.counterexample_not_prime_but_homogeneous_prime.two.linear_ordered_add_comm_monoid._proof_9, add_zero := counterexample.counterexample_not_prime_but_homogeneous_prime.two.linear_ordered_add_comm_monoid._proof_10, nsmul := add_comm_monoid.nsmul with_zero.add_comm_monoid, nsmul_zero' := counterexample.counterexample_not_prime_but_homogeneous_prime.two.linear_ordered_add_comm_monoid._proof_11, nsmul_succ' := counterexample.counterexample_not_prime_but_homogeneous_prime.two.linear_ordered_add_comm_monoid._proof_12, add_comm := counterexample.counterexample_not_prime_but_homogeneous_prime.two.linear_ordered_add_comm_monoid._proof_13, add_le_add_left := counterexample.counterexample_not_prime_but_homogeneous_prime.two.linear_ordered_add_comm_monoid._proof_14}
 
The grade 0 part of R² is {(a, a) | a ∈ R}.
The grade 1 part of R² is {(0, b) | b ∈ R}.
Given the above grading (see submodule_z and submodule_o),
we turn R² into a graded ring.
Equations
- counterexample.counterexample_not_prime_but_homogeneous_prime.grading R 1 = counterexample.counterexample_not_prime_but_homogeneous_prime.submodule_o R
 - counterexample.counterexample_not_prime_but_homogeneous_prime.grading R 0 = counterexample.counterexample_not_prime_but_homogeneous_prime.submodule_z R
 
Instances for counterexample.counterexample_not_prime_but_homogeneous_prime.grading
        
        
    R² ≅ {(a, a) | a ∈ R} ⨁ {(0, b) | b ∈ R} by (x, y) ↦ (x, x) + (0, y - x).
Equations
- counterexample.counterexample_not_prime_but_homogeneous_prime.grading.decompose = {to_fun := λ (zz : zmod 4 × zmod 4), ⇑(direct_sum.of (λ (i : counterexample.counterexample_not_prime_but_homogeneous_prime.two), ↥(counterexample.counterexample_not_prime_but_homogeneous_prime.grading (zmod 4) i)) 0) ⟨(zz.fst, zz.fst), _⟩ + ⇑(direct_sum.of (λ (i : counterexample.counterexample_not_prime_but_homogeneous_prime.two), ↥(counterexample.counterexample_not_prime_but_homogeneous_prime.grading (zmod 4) i)) 1) ⟨(0, zz.snd - zz.fst), _⟩, map_zero' := counterexample.counterexample_not_prime_but_homogeneous_prime.grading.decompose._proof_6, map_add' := counterexample.counterexample_not_prime_but_homogeneous_prime.grading.decompose._proof_7}
 
The counterexample is the ideal I = span {(2, 2)}.