Prime decomposition over the real numbers?

Question

Over the natural numbers, each number can be uniquely decomposed as a product of prime numbers. Can we generalize this idea to the reals? Here is my attempt:

Theorem Can there exist a subset $P\subseteq \mathbb R$ of the reals such that

1. For every $0\neq x\in\mathbb R$ exist $F_x := \{(x_1,p_1),\dots,(x_n,p_n)\}\subseteq P\times \mathbb N $ with $x=\prod\limits_{i=1}^n x_i^{p_i}$
2. For every $x\neq \pm 1$ this decomposition is unique, aside from a representation of $\pm 1$ (i.e. For 2 decompositions $F_x=\{(x_1,p_1),\dots,(x_n,p_n)\}$ and $F_x'=\{(x_1',p_1'),\dots,(x_m',p_m')\}$ there exist $F_1:=\{(z_1,q_1),\dots,(z_a,q_a)\}\subseteq X$ and $F_1':=\{(z_1',q_1'),\dots,(z_b',q_b')\}\subset X'$ with $\prod\limits_{i=1}^p z_i^{q_i}=\pm 1$ and $\prod\limits_{i=1}^q z_i'^{q_i'}=\pm 1$ such that $F_x\setminus F_1 = F_x'\setminus F_1'$

This is essentially Integer factorization generalized on the reals with an additional condition to allow uniqueness. Obviously this set is ambiguous, but the decomposition for each number, given such a set $P$, is unique.

My attemps/results until now First of all, I am pretty sure this theorem is true. I am aware of the fact that it is very likely to involve the Axiom of Choice.

• If $P$ exists, then it is uncountable.
  Why? If $P$ were countable, the set of finite subsets of $P$ is countable. Per definition of $P$, that is equinumerous to $\mathbb R$ via $\sigma:\{(x_1,p_1),\dots,(x_n,p_n) \} \mapsto \prod\limits_{i=1}^nx_i^{p_i}$.

• $\sigma:\mathbb (Q\times N)^*\to \mathbb R\setminus \{0\}$ ($(Q\times \mathbb N)^*$ is the set of all finite subsets on $Q\times \mathbb N$, $Q\subseteq\mathbb R$) is surjective and multiplicative, i.e. $A\cap B=\varnothing \Rightarrow \sigma(A\dot\cup B)=\sigma(A)\sigma(B)$

• $X \sim Y :\Leftrightarrow X$ only differs by a representation of 1 from $Y$ is a equivalence relation.
  Why? reflexivity and symmetry are trivial. For transitivity, $X\sim Y\Leftrightarrow X\setminus F_1 = Y\setminus F_1' \Leftrightarrow \sigma (X) = \sigma(X\setminus F_1\; \dot\cup\; F_1)=\sigma(X\setminus F_1)\cdot 1 = \sigma(Y\setminus F_1')\cdot 1=\sigma(Y)$

• Over the rationals, such a (countable) set exists: $P_{\mathbb Q}\bigcup\limits_{p\text{ prime}}\{p,\frac{1}{p}\}$

• Iteratively adding numbers, which arent expressible by the numbers in the set, to $P_\mathbb Q$ works fine, yet will never yield the set we are searching for.

I am having trouble with finishing the last argument, i.e. doing the step from countably to uncountably infinite elements. So how can you prove this theorem, or is there a counterexample?

Answer 1

You are trying to prove that the reals are a UFD, up to units. That is true, but not in the way you would like. Since $\mathbb{R}$ is a field, every nonzero element is a unit (not just $\pm 1$). Hence, you need to expand the exception to include everything, which alas trivializes the result.

As desired, here's a further explanation. Suppose you had such a unique factorization $x=x_1^{p_1}\cdots x_n^{p_n}$, with $p_i\in\mathbb{N}$. But then you'd also need one for $\frac{1}{x}$, and by uniqueness you would need $\frac{1}{x}=x_1^{-p_1}\cdots x_n^{-p_n}$; but now the exponents are no longer positive.