Intersection of ideals of $\mathbb{Z}$

Question

Let $R=\mathbb{Z}$, and let $p$ a fixed prime. We condiser $$\bigcap_{k\ge 1} (p^k).$$ I know that this intersection is empty for the Foundamental Theorem, but How can I properly formalize the proof?

Thanks

Answer 1

Follows what I would consider a more-or-less formal proof of the assertion than $\bigcap_{k \ge 1} (p^k) = \{0\}$, where $p \in \Bbb P$ is a prime:

If

$\displaystyle \bigcap_{k\ge 1} (p^k) \ne \{0\}, \tag 1$

then

$\exists 0 \ne z \in \Bbb Z, \; z \in \displaystyle \bigcap_{k \ge 1} (p^k), \tag 2$

and since $\bigcap_{k \ge 1} (p^k)$ is an ideal,

$z \in \displaystyle \bigcap_{k \ge 1} (p^k) \Longleftrightarrow -z \in \displaystyle \bigcap_{k \ge 1} (p^k); \tag 3$

thus we may assume

$z > 0; \tag 4$

now by (2) we have

$\forall k \ge 1, \; z \in (p^k); \tag 5$

then

$\forall k \ge 1 \; \exists 0 < m_k \in \Bbb Z, z = m_k p^k; \tag 6$

we observe that if $z \ne 0$ then this forces $z > 1$; however, then this contradicts the fundamental theorem of arithmetic, which asserts that the prime factorization of every positive integer is unique. We thus conclude that (1) is false, i.e. that

$\displaystyle \bigcap_{k\ge 1} (p^k) = \{0\}. \tag 7$