Let $R$ be a non-field UFD, and for $x∈R \setminus \{0\}$ let$$\omega(x)=\omega(up_1^{\alpha_1}...p_n^{\alpha_n})=\sum_{i=1}^n \alpha_i$$For each $i∈\mathbb{N}$ let$$V_i = \{x∈R~|~\omega(x)=i\}∪\{0\}$$
Weak Fracture Hypothesis: For any non-field UFD $R$, there exists $i∈\mathbb{N}$ such that $V_i$ is not a group under addition.
Strong Fracture Hypothesis: For any non-field UFD $R$, for all $i∈\mathbb{N}^+$ necessarily $V_i$ is not a group under addition. Equivalently, $V_1$ is not a group under addition.
What would otherwise be the very "strongest" FH (that $V_0$ is also never an additive group) is false, since taking any field $F$ and letting $R=F[x]$, we have $V_0$ is the subset of constant polynomials, which are seen to be a group under addition.
SFH: Assume $R$ violates the SFH with $V_i$ an additive group.
Proposition: The image of $\mathbb{Z} \setminus \{0\}$ in $R$ is within $V_0$, i.e. $z∈\mathbb{Z} \setminus \{0\}$ is a unit in $R$ if it is nonzero. Proof: For $x∈V_i$, we have $x+...+x \text{(z times)} = zx∈V_i$, 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: $V_0,V_1,...,V_{i-1}$ are additive groups. Proof: Assume $V_0$ is not a group so that $u_1-u_2$ is divisible by a prime. Now for some prime $p$ we have $u_1p^i-u_2p^i=(u_1-u_2)p^i∉V_i$. Now assume $V_1$ is not a group so that $p_1-p_2∉V_1$. Observe $p_1^i-p_2p_1^{i-1}=(p_1-p_2)p_1^{i-1}∉V_i$. The case follows similarly for $V_k$ 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 $x=(\prod_{p∈P} p) - 1$. Assume $x$ is a unit, in which case $x+1=\prod_{p∈P} p$ is also a unit, a contradiction. So for some $q∈P$ we have $q$ divides $x$ and by construction $q$ divides $\prod_{p∈P}$ so $q$ divides $-1$, another contradiction.
Proposition: When $x∈V_a$ and $y∈V_b$ for $a≠b$ and $x≠0≠y$, we have $x+y∉V_a \cup V_b$. Proof: Assume $x+y∈V_a$. Now $(x+y)-x=y∈V_a$ and $y=0$. The case for $V_b$ holds similarly.
Algebraic Geometry/Rank of $V_1$ and/or Closedness of $V_0$: (First method due to Professor Li) The rank of $V_1$ as a vector space over $V_0$ is greater than $2$. Proof: Assume the rank is $\leq 2$, so $R \cong V_0[x,y]/P$ for some prime ideal $P$ identifying $x,y$ with primes in $R$. Necessarily $P=(f(x,y),p(x))$ for some $f(x,y) \in V_0[x,y]$, $p(x) \in V_0[x]$, each either prime or zero in their respective rings. We see $p(x)=0$ since each prime in $R$ is transcendental over $V_0$, so simply $P=(f(x,y))$. Now, $V_0$ must be infinite since $V_1$ must be infinite to satisfy infinitude of primes as above, so choose distinct units $u,v$. Write $Q_1=x+u=(\alpha_1x+\beta_1y)...(\alpha_nx+\beta_ny)$ in $R$, so $Q_1-x-u=f(x,y)g(x,y)$ in $V_0[x,y]$. The left hand side demonstrates a polynomial in $V_0[x,y]$ 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 $Q_1$, so without loss assume $\alpha_1x+\beta_1y$ is a factor of $m(x,y)$. But by taking the same approach to $Q_2=x+v$, we see $Q_2$ as a polynomial in $V_0[x,y]$ has $\alpha_1x+\beta_1y$ as a factor. This is impossible as $Q_1-Q_2 = u-v \in V_0$ (nonzero) while simultaneously $\alpha_1x+\beta_1y$ divides $Q_1-Q_2$ in $V_0[x,y]$ and now too in $R$.
Theorem: Assume the rank $n$ of $V_1$ as a vector space over $V_0$ is finite, and that $V_0$ is algebraically closed. Then $R$ is not a PID. Proof: By the above $n \geq 3$. Consider the surjective ring homomorphism $V_0[x_1,...,x_n] → R$ induced by mapping the variables to the generating primes of $V_1$ over $V_0$. Then by Nullstellensatz, there is a common zero $u \in V_0^n$ of the polynomials generating the kernel of this homomorphism. Consider the nontrivial homomorphism $\phi : R → V_0$ induced by this zero. Then since $\phi$ restricts to a linear transformation $V_1$ to $V_0$ over $V_0$, and since $n \geq 3$, by rank-nullity the kernel of this transformation is of rank $\geq 2$, 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 $p_1,p_2,...$. For each pair of integers $i≤j$ define$$\rho_{ij} : V_i → V_j$$$$x \mapsto (\prod_{k=i+1}^jp_k)x$$These mappings are seen to satisfy the following properties for $i≤j≤k$$$\rho_{ii}=1$$$$\rho_{jk} \circ \rho_{ij}=\rho_{ik}$$As 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 $u_1≠u_2$ and prime $p$, we have $p+u_1$ is not within the same fracture as $p+u_2$. Proof: Clearly neither are within $V_0$, so assume $p+u_1,p+u_2∈V_i$ for positive $i$. Since $V_i$ is an additive group, we have $(p+u_1)-(p+u_2)=u_1-u_2∈V_0$ as well as $u_1-u_2≠0$ so that $u_1-u_2∉V_i$, a contradiction.
Lemma: $V_0$ 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: Since$$R=\bigcup_{n∈\mathbb{N}} V_n$$it remains to show that $V_n$ is countable. We can establish a surjection$$φ : V_0 \times \prod_{i=1}^nV_1 → V_n$$$$(u,p_1,p_2,...,p_n) \mapsto up_1p_2...p_n$$Since 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 observe$$0=Φ(\sum_{i=1}^nt_i)=Φ(t_1+t_2+\sum_{i=3}^nt_i)=Φ(-f(t_1,t_2)+\sum_{i=3}^nt_i)$$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 $u_2/u_1$.
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