Is it correct to say that prime numbers don't exist on $\mathbb{R}$ and $\mathbb{Q}$?

Question

A prime number is defined as: "A non invertible and non zero numer $p$ of a ring $A$ is called a prime number if any time it divides a product of two numbers, it also divides one of the factors". In $\mathbb{R}$ and $\mathbb{Q}$ any number is divisible by any number, except for $0$. So take $7$ and $3\cdot 49$ for example. On $\mathbb{Z}$ and $\mathbb{N}$ we have that $7|3\cdot49=3\cdot7\cdot7\land7|7\land7\nmid3$, so $7$ is a prime number on $\mathbb{Z}$ and $\mathbb{N}$. But on $\mathbb{R}$ and $\mathbb{Q}$, $7$ divides both $3$ and $7$ (recall that in a ring $A$ a number $a$ is said do divide a number $b$ if there's a number $c\in A$ s.t. $b=a\cdot c$). So, having said that, can I say that prime numbers don't exist on $\mathbb{R}$ and $\mathbb{Q}$?

Answer 1

It is true that $\mathbb R$ and $\mathbb Q$ do not contain any prime elements, but the reasoning you have provided does not seem correct to me. If $R$ is a commutative ring, we say that $r\in R$ is a prime element if the following three conditions hold:

1. $r\neq0.$
2. $r$ is not a unit.
3. For all $a,b\in R$, if $r$ divides $ab$, then $r$ divides $a$ or $r$ divides $b.$

In the case of $\mathbb R$, the element $7$ satisfies both (1) and (3). However, $7$ is a unit, so it does not satisfy (2). The same comments apply to $\mathbb Q$.

More generally, if $K$ is a field, then $K$ does not contain any prime elements. This is because, if $x\in K$, then either $x=0$ or $x\neq0$. If $x=0$ then, by definition, $x$ is not prime. If $x\neq0$, then by the definition of a field, $x$ is a unit, so again $x$ is not prime.