Let R be a non-field UFD, and for x∈R∖{0} letω(x)=ω(upα11...pαnn)=n∑i=1αiFor each i∈N letVi={x∈R | ω(x)=i}∪{0}
Weak Fracture Hypothesis: For any non-field UFD R, there exists i∈N such that Vi is not a group under addition.
Strong Fracture Hypothesis: For any non-field UFD R, for all i∈N+ necessarily Vi is not a group under addition. Equivalently, V1 is not a group under addition.
What would otherwise be the very "strongest" FH (that V0 is also never an additive group) is false, since taking any field F and letting R=F[x], we have V0 is the subset of constant polynomials, which are seen to be a group under addition.
SFH: Assume R violates the SFH with Vi an additive group.
Proposition: The image of Z∖{0} in R is within V0, i.e. z∈Z∖{0} is a unit in R if it is nonzero. Proof: For x∈Vi, we have x+...+x(z times)=zx∈Vi, so that z doesn't append to the prime factorization of x, so is a unit. Since also (−1)2=1, we have −1 is a unit as well, and now all nonzero integers are seen to be units.
Proposition: V0,V1,...,Vi−1 are additive groups. Proof: Assume V0 is not a group so that u1−u2 is divisible by a prime. Now for some prime p we have u1pi−u2pi=(u1−u2)pi∉Vi. Now assume V1 is not a group so that p1−p2∉V1. Observe pi1−p2pi−11=(p1−p2)pi−11∉Vi. The case follows similarly for Vk for any k≤i.
Prime Infinitude: There are an infinite number of distinct primes of R. Proof: Assume P is a complete, finite list of primes in R. Observe x=(∏p∈Pp)−1. Assume x is a unit, in which case x+1=∏p∈Pp is also a unit, a contradiction. So for some q∈P we have q divides x and by construction q divides ∏p∈P so q divides −1, another contradiction.
Proposition: When x∈Va and y∈Vb for a≠b and x≠0≠y, we have x+y∉Va∪Vb. Proof: Assume x+y∈Va. Now (x+y)−x=y∈Va and y=0. The case for Vb holds similarly.
Algebraic Geometry/Rank of V1 and/or Closedness of V0: (First method due to Professor Li) The rank of V1 as a vector space over V0 is greater than 2. Proof: Assume the rank is ≤2, so R≅V0[x,y]/P for some prime ideal P identifying x,y with primes in R. Necessarily P=(f(x,y),p(x)) for some f(x,y)∈V0[x,y], p(x)∈V0[x], each either prime or zero in their respective rings. We see p(x)=0 since each prime in R is transcendental over V0, so simply P=(f(x,y)). Now, V0 must be infinite since V1 must be infinite to satisfy infinitude of primes as above, so choose distinct units u,v. Write Q1=x+u=(α1x+β1y)...(αnx+βny) in R, so Q1−x−u=f(x,y)g(x,y) in V0[x,y]. The left hand side demonstrates a polynomial in V0[x,y] with three homogenous components, so by comparing terms on the right we see the largest homogenous term m(x,y) of f(x,y) divides Q1, so without loss assume α1x+β1y is a factor of m(x,y). But by taking the same approach to Q2=x+v, we see Q2 as a polynomial in V0[x,y] has α1x+β1y as a factor. This is impossible as Q1−Q2=u−v∈V0 (nonzero) while simultaneously α1x+β1y divides Q1−Q2 in V0[x,y] and now too in R.
Theorem: Assume the rank n of V1 as a vector space over V0 is finite, and that V0 is algebraically closed. Then R is not a PID. Proof: By the above n≥3. Consider the surjective ring homomorphism V0[x1,...,xn]→R induced by mapping the variables to the generating primes of V1 over V0. Then by Nullstellensatz, there is a common zero u∈Vn0 of the polynomials generating the kernel of this homomorphism. Consider the nontrivial homomorphism ϕ:R→V0 induced by this zero. Then since ϕ restricts to a linear transformation V1 to V0 over V0, and since n≥3, by rank-nullity the kernel of this transformation is of rank ≥2, i.e. it contains two nonassociate primes, and hence in R the kernel of the full homomorphism cannot be principal.
WFH: Fix some infinite sequence of primes p1,p2,.... For each pair of integers i≤j defineρij:Vi→Vjx↦(j∏k=i+1pk)xThese mappings are seen to satisfy the following properties for i≤j≤kρii=1ρjk∘ρij=ρikAs well, they are group homomorphisms (and are in addition injective), so this directed system satisfies all the properties of a direct limit.
Proposition: For any units u1≠u2 and prime p, we have p+u1 is not within the same fracture as p+u2. Proof: Clearly neither are within V0, so assume p+u1,p+u2∈Vi for positive i. Since Vi is an additive group, we have (p+u1)−(p+u2)=u1−u2∈V0 as well as u1−u2≠0 so that u1−u2∉Vi, a contradiction.
Lemma: V0 is countable. Proof: Fix a prime p and let φ(u) be the index of the fracture p+u is within. By the previous proposition, this mapping is injective.
Lemma: V_1 is countable. Proof: Fix a unit u and let φ(p) be the index of the fracture u+p is within. Assume φ(p_1)=φ(p_2), i.e. p_1+u,p_2+u∈V_i. Since V_i is additive, we have (p_1+u)-(p_2+u)∈V_i while also (p_1+u)-(p_2+u)=p_1-p_2∈V_1. Therefore (p_1+u)-(p_2+u)=0 and p_1=p_2, and the mapping is injective.
Lemma: R is countable. Proof: SinceR=\bigcup_{n∈\mathbb{N}} V_nit remains to show that V_n is countable. We can establish a surjectionφ : V_0 \times \prod_{i=1}^nV_1 → V_n(u,p_1,p_2,...,p_n) \mapsto up_1p_2...p_nSince both V_0 and V_1 are countable, and finite direct products of sets of a given cardinality retain the same cardinality, we have V_n is countable and now R is countable.
Lemma: There are no nontrivial automorphisms of R. Proof: Let φ be a nontrivial automorphism of R. Then since φ is defined by its action on units and primes, either φ doesn't fix some prime p, or φ doesn't fix some unit u in which case either φ doesn't fix any chosen prime p' or φ doesn't fix p=up'; hence in either case we may say φ(p)≠p for some prime p.
Now, write 1+p∈V_k, necessarily k > 1. For general UFDs, automorphisms send units to units and primes to primes, so also φ(1+p)∈V_k. Hence 1+p-φ(1+p)∈V_k. But also 1+p-φ(1+p)=p-φ(p)∈V_1, so since V_1∩V_k=\{0\} we must have p=φ(p), a contradiction.
Module Structure: Note that V_i acts as a vector space over V_0. We may define a bilinear map of V_0-modules \prod_{i=1}^n V_1 → V_n by (p_1,...,p_n) \mapsto p_1...p_n to induce an appropriate homomorphism of V_0-modules \Phi_i : \otimes_{i=1}^n~V_1 → V_n.
Exact Sequences: Since V_0 is a field and V_1 a V_0-module, we can see that V_1 is injective and by an injective homomorphism V_1 → V_n by x \mapsto p_1^{n-1}x we obtain V_n ≅ V_1 \oplus V_n'.
Tensor Algebras: V_1 is a vector space over V_0, and R is a commutative V_0 algebra with a natural inclusion linear transformation V_1 → R, inducing an appropriate extension homomorphism of V_0-algebras Φ: S(V_1) → R. This is nonzero on nonzero 1- and 2-tensor sums, but to this end we claim \ker{Φ} is generated by 3-tensor sums. Proceed by induction on n for an n-tensor sum ∑_{i=1}^nt_i in \text{ker }Φ; for any two simple tensors s_1 and s_2, we have Φ(s_1+s_2)=Φ(s_3) for some simple tensor s_3 due to representation in R. Therefore, for any simple tensors s_1 and s_2 let f(s_1,s_2) be a simple tensor such that s_1+s_2+f(s_1,s_2) is in \text{ker }Φ. Back to the claim at hand, we observe0=Φ(\sum_{i=1}^nt_i)=Φ(t_1+t_2+\sum_{i=3}^nt_i)=Φ(-f(t_1,t_2)+\sum_{i=3}^nt_i)Since -f(t_1,t_2)+\sum_{i=3}^nt_i is an (n-1)-tensor sum in \text{ker }Φ, by induction it is generated by 3-tensor sums in \text{ker }Φ and we have \sum_{i=1}^nt_i=-f(t_1,t_2)+\sum_{i=3}^nt_i+(t_1+t_2+f(t_1,t_2)) to complete the claim.
Furthermore, we can partially predict the behavior of the function f. According to a contradiction of WFH, when s_1 and s_2 are simple tensors in S^n(V_1), then f(s_1,s_2) is in S^n(V_1). Since this function completely defines the 3-tensor sums generating the kernel, we might clearly derive SFH is violated if and only if there is a vector space V over F whose symmetric algebra contains an ideal A such that (i) A is generated by 3-tensor sums (ii) A contains no 1- or 2-tensor sums, and (iii) A for every pair of simple tensors s_1,s_2 there is a simple tensor s_3 such that s_1+s_2+s_3∈A. As well, WFH is violated if and only if there is such V, F, and A such that (iv) when s_1,s_2∈S^n(V) then s_3∈S^n(V).
Rank of V_1: Assume V_1 is generated over V_0 as a vector space by the elements p and q. Then V_0 is algebraically closed. Proof: We first show that the elements p^kq^{n-k} for k∈[0,n] are a basis for V_n over V_0; this is evident because products of the form xp+yq for x,y∈V_0 taken n at a time generate V_n and such products can be expanded to sums in p^kq^{n-k}, and also because they are linearly independent: Assume v_0p^n+v_1qp^{n-1}+...+v_nq^n=0, and further assume v_0≠0; then we have v_0p^n=qr for some r, which is a contradiction by unique factorization. So v_0=0; now divide the sum out by q and apply the independence statement by induction to V_{n-1} to obtain v_i=0 for all i.
Now, observe the polynomial f(x)=x^n+a_{n-1}x^{n-1}+...+a_0 in V_0[x]. Then observe the element p^n+a_{n-1}qp^{n-1}+a_{n-2}q^2p^{n-2}+...+a_0q^n∈V_n. We must be able to write this according to its prime factorization as (p-α_0q)(p-α_1q)...(p-α_nq) for some α_i∈V_0 (the negative signs making the notational argument simpler; note that we may ignore the primes of the form yq for y∈V_0 as the expansion requires the coefficient of p to be nonzero in all multiplicands). We see that in fact α_0,...,α_n behaves exactly as would the solutions β_0,...,β_n in writing f(x)=(x-β_0)...(x-β_n)=(x-α_0)...(x-α_n) so that f(x) splits over V_0.
Open Problems:
A UFD not generated as an additive group by its primes and units would be \mathbb{C}[x], and some UFDs that are would be \mathbb{Z}, \mathbb{Z}[x], \mathbb{Q}[x], and \mathbb{F}_p[x].
If V_0,V_1 generate V_2 as an additive group, then R is generated by V_0,V_1. To see this, assume an ungenerated element x with n=\omega (x) minimal. Then observe the product of the generation of the product of its first n-1 primes, with its last prime, which is a sum of elements in V_1 and V_2, which by the hypothesis can be represented by a sum of elements in V_0 and V_1.
R is not generated as an additive group by V_0 and V_1, and in fact V_2 isn't either. This would presume the existence of nonassociate u_1+p_1,u_2+p_2∈V_2, where necessarily u_1,u_2≠0, so we can write u_1+p_1-(u_1/u_2)(u_2+p_2)=p_1-(u_1/u_2)p_2∈V_1∩V_2=\{0\} but we see u_1+p_1=u_1+(u_1/u_2)p_2 is associate to u_2+p_2 by u_2/u_1.