# Primary structures in $\mathbb Q$

I'll formulate a topic restricted here to the positive rational numbers $\ \mathbb Q_{_{>0}},\$, then will pose a question (Q2) plus some related, to which I don't know the answers nor reference. The idea is to understand how general is the notion of the set of all prime numbers: do we or don't we have other similar objects which I call primary structures.

STANDARD TERMINOLOGY

$\mathbb N := \{1\ 2\ \ldots\}\$ -- the set of natural numbers;
$\mathbb Z_+:=\mathbb N\cup\{0\}\$ -- the set on nonnegative integers;
$\mathbb Z\$ -- the ring of (rational) integers;
$\mathbb Q\$ -- the field of rational numbers;
$\mathbb Q_{_{>0}} := \{x\in \mathbb Q : x > 0\}\$ -- the set of positive rational numbers.
$\mathbb P:= (2\ 3\ 5\ 7\ 11\ \ldots)\$ -- the sequence of all primes (i.e. $P_1:=2,\$ and $\ P_5:= 11,\$ etc.).

SPECIAL TERMINOLOGY

$\Omega := \{f\in \mathbb Z^\mathbb N:\sum|f|<\infty\}\$ -- integer sequences with finitely many non-zero values; $\Lambda := \{f\in \mathbb Z_+^\mathbb N:\sum f<\infty\} = \mathbb Z_+^\mathbb N\cap \Omega;$
$x^f := \prod_{n\in\mathbb N} x_n^{f(n)}\$ for every sequence $x:=(x_1\ x_2\ \ldots)\$ of positive rationals, and for $\ f\in \Omega$;
$x^* := \{x^f:f\in\Lambda\}\$ -- the multiplicative monoid generated by terms of sequence $x$.
$\mathbb Q(x) := \{x^f:f\in\Omega\}\$

DEFINITIONS

Let $\ S:=(S_1\ S_2\ \ldots)\in \mathbb Q_{_{>0}}\!^\mathbb N$.

D1. $\$ Sequence $S\$ has the unique decomposition property (u.d.p. for short) $\Leftarrow:\Rightarrow$

$$\forall_{f\ g\in\Omega}\ \left(S^f = S^g\,\ \Rightarrow\,\ f=g\right)$$

Note 1. $\$ Replacing $\ \Omega\$ by $\ \Lambda\$ would not affect the above definition.
Note 2. $\ \forall_{x\ y\in S^*}\ \left(x\cdot y=1\,\ \Rightarrow\,\ x=y=1\right)$

There are plenty of sequences with the u.d.p. However one extra condition will narrow the choice drastically:

D2. $\$ A u.d.p. sequence $\ S\$ is called a primary structure $\ \Leftarrow:\Rightarrow\ S^*\$ is an additive semigroup in $\ Q_{_{>0}}$.

QUESTIONS

Let $\ S\$ be an arbitrary primary structure. Is it true that:

Q1: $\ \forall_{n\in\mathbb N}\ S_n > 1\ ?$

Q2: Is $\ S\$ a permutation of $\ \mathbb P$?

If NO to Q2 (just in case :-), we may still wonder about:

Q3: If every prime appears in $\ S\$ is it true that only primes appear in $\ S\$ (i.e. that $\ S\$ is a permutation of $\ \mathbb P$)?

Q4. $\ \mathbb Q(S) = \mathbb Q\$?

• I could add examples of u.d.p. sequences for which it is not obvious that they are not additive monoids (i.e. that they are not primary structures). The QUESTION is already a bit long but I could do it. Feb 9, 2016 at 19:46
• @EmilJeřábek, thank you for pointing to my typos. I've fixed them by now. It was not just $\ S\$ but the induced multiplicative monoid $\ S^*;\$ and it was also supposed to be an additive semigroup -- not an additive monoid. Sorry for the typos. Feb 9, 2016 at 20:35
• OK, now it makes sense. Feb 9, 2016 at 20:39
• Q3 and Q4 are easy: Q3 follows from the fact that every positive rational is a ratio of products of primes, and Q4 from the fact that by D2, $S^*$ includes $\mathbb N$. Which actually implies a generalization of Q3: any primary structure is maximal (it is not properly included in another udp). Feb 10, 2016 at 10:43
• OK. I was hoping for someone (possibly me) to throw some light on the real question, but as this didn't happen, I've expanded the comment to an answer. Mar 12, 2016 at 10:14

Since there were so far no takers for the main question, let me state for the record that Q3 and Q4 are true, as already mentioned in the comments.

If $S$ is a primary structure, then $S^*$ is an additive semigroup containing $1$, whence $\mathbb N\subseteq S^*$, and a fortiori $\mathbb Q(S)=\mathbb Q$. Thus Q4 is true.

Notice that the properties of being a u.d.p. and being a primary structure are invariant under permutations, and any u.d.p. is an injective sequence. Thus, there is no loss in treating u.d.p.s and primary structures as sets rather that sequences (where a set of rationals is defined to be a u.d.p. if some/every its injective enumeration is a u.d.p., and likewise for p.s.).

With this in mind, every p.s. is a maximal u.d.p.: that is, there are no p.s. $S$ and u.d.p. $S'$ such that $S\subsetneq S'$. Indeed, any $a\in S'\smallsetminus S$ is also in $\mathbb Q(S)$ by the argument above, which gives a nontrivial multiplicative relation among elements of $S'$.

In particular, this implies a positive answer to Q3, as the set of primes is itself a p.s.

• Emil, thank you for your answer. And let's still hope for more. Mar 14, 2016 at 1:55

The example below illustrates the questions without solving any of them.

EXAMPLE $\$ Let $\ \mathbb P=(P_1\ P_2\ \ldots)\$ be the increasing sequence of all primes. Let

(i) $\ S_1 := 2$;
(ii) $\ \forall_{n>1}\ S_n := \frac{P_n}{P_{n-1}}$.

Then sequence $\ S:=(S_1\ S_2\ \ldots)\$ has the following properties:

(1) $\ S\$ has the u.d.p.;
(2) $\ \mathbb Q(S) = \mathbb Q$;
(3) $\ S^* \supseteq\mathbb N$;
(4) $\ S\$ is not a primary structure (i.e. $\ S^*\$ is not closed under addition).

PROOF

(1) Let's apply the $\ \Lambda\$ version of the definition of u.d.p. Let $\ f\ g\in\Lambda\$ be such that $\ S^f=S^g.\$ If $\ f\$ is the zero sequence then $\ S^f=1,\$ and it's clear that also $\ g\$ is a zero sequence. Otherwise $\ f\$ is a nonzero sequence. Then there exists a unique largest prime $\ P_n\$ which appears in the standard prime decomposition of the rational number $\ S^f\$ (the same primes in the similar way appears in $\ S^g\,$ since $\ S^f=S^g).\$ Then $\ f(n)>0\$ and $\ g(n)>0,\$ hence

$$S^f=S_n\cdot S^{f'}\qquad and \qquad S^g=S_n\cdot S^{g'}$$

where $\ f'(k)=f(k)\$ and $\ g'(k)=g(k)\$ for every $\ k\ne n,\$ and $\ f(n)-f'(n) = g(n)-g'(n) = 1.\$ Thus $\ S^{f'} = S^{g'}\$ which by induction on $\ \sum f\$ implies that $\ f'=g'.\$ This induction proves that the u.d.p. holds for $\ S$.

(2&3)

$$\forall_{n\in\mathbb N}\ P_n = \prod_{k=1}^n\,S_k$$

Thus $\ \mathbb P\subseteq S^*.\$ It follows that $\ \mathbb N\subseteq S^*,\$ and $\ \mathbb Q\subseteq \mathbb Q(S)$.

(4) Note that $\ P(3)=5\$ and $\ P_8=19.\$ Thus $$S_2 + S_3\ = \frac{19}{2\cdot 3}\ = \ S_1^{-1}\cdot \prod_{k=3}^8 S_k\ \notin\ S^*$$

because $\ S_2+S_3\ = S^f\$ where $\ f(1) = -1,\$ hence $\ f\notin \Lambda,\$ (apply the $\Omega$ version of the definition of the u.d.p.). But, of course, $\ S_2\ S_3\in S^*.\$ Thus $\ S^*\$ is not closed under addition.

END of proof

Well, just a measly example.