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.


$\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.).


$\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\}\ $


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 $\ Q_{_{>0}}$.


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\ $?

  • $\begingroup$ 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. $\endgroup$ – Włodzimierz Holsztyński Feb 9 '16 at 19:46
  • $\begingroup$ @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. $\endgroup$ – Włodzimierz Holsztyński Feb 9 '16 at 20:35
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    $\begingroup$ OK, now it makes sense. $\endgroup$ – Emil Jeřábek Feb 9 '16 at 20:39
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    $\begingroup$ 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). $\endgroup$ – Emil Jeřábek Feb 10 '16 at 10:43
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    $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ Emil, thank you for your answer. And let's still hope for more. $\endgroup$ – 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).


(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 $\ S$.


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

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.


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