If $p_i$ is prime then $p_i \Bbb Z \cap p_j \Bbb Z= \emptyset \ \forall i,j \in \Bbb N, i\neq j$.

Question

If $p_i$ is prime then $p_i \Bbb Z \cap p_j \Bbb Z= \emptyset \ \forall i,j \in \Bbb N, i\neq j$. Where $a \Bbb Z=\{x\in \Bbb Z: x=0 \mod a\}$.

I ran into some problem where having this lemma proved would really simplify it, is there any really really simple (to understand) proof of this?

Answer 1

The lemma appears to be false. For example, $p_ip_j \in p_i\mathbb Z \cap p_j\mathbb Z$.

In fact, $p_i\mathbb Z \cap p_j\mathbb Z = p_i p_j \mathbb Z$.

Answer 2

Do you mean $p_i \mathbb{Z} \cap p_j \mathbb{Z} \neq \emptyset$? We have $p_i \mathbb{Z}$ is multiples of $p_i$, $p_j \mathbb{Z}$ is multiples of $p_j$. Both contain $p_ip_j$, so the intersection is NOT empty.