# 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. – Włodzimierz Holsztyński Feb 9 '16 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. – Włodzimierz Holsztyński Feb 9 '16 at 20:35
• OK, now it makes sense. – Emil Jeřábek Feb 9 '16 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). – Emil Jeřábek Feb 10 '16 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. – Emil Jeřábek Mar 12 '16 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. – Włodzimierz Holsztyński Mar 14 '16 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.