[proofplan] We show that $\mathbb{Z}_p$ is a principal ideal domain with unique prime element $p$ (up to units). The proof uses the ideal classification from the theorem [Ideals of $\mathbb{Z}_p$ and Quotients](/theorems/???): every non-zero ideal of $\mathbb{Z}_p$ is of the form $p^n \mathbb{Z}_p$. From this classification, the PID property is immediate (every ideal is principal, generated by $p^n$). We then verify that $p$ is prime and that it is the unique irreducible element up to units, by examining the unit group $\mathbb{Z}_p^\times = \{x \in \mathbb{Z}_p : |x|_p = 1\}$. [/proofplan] [step:Establish that $\mathbb{Z}_p$ is an integral domain] Since $\mathbb{Z}_p$ is a subring of the field $\mathbb{Q}_p$, it has no zero divisors: if $x, y \in \mathbb{Z}_p$ with $xy = 0$, then $x = 0$ or $y = 0$ (since $\mathbb{Q}_p$ is a field). Therefore $\mathbb{Z}_p$ is an integral domain. [/step] [step:Deduce the PID property from the ideal classification] By the theorem [Ideals of $\mathbb{Z}_p$ and Quotients](/theorems/???), every non-zero ideal of $\mathbb{Z}_p$ has the form $p^n \mathbb{Z}_p$ for some $n \geq 0$. Each such ideal is principal, generated by $p^n$. The zero ideal $(0)$ is also principal (generated by $0$). Therefore every ideal of $\mathbb{Z}_p$ is principal. Since $\mathbb{Z}_p$ is an integral domain and every ideal is principal, $\mathbb{Z}_p$ is a principal ideal domain. [/step] [step:Show that $p$ is a prime element of $\mathbb{Z}_p$] An element $\pi \in \mathbb{Z}_p$ is prime if $\pi \neq 0$, $\pi \notin \mathbb{Z}_p^\times$, and whenever $\pi \mid ab$ in $\mathbb{Z}_p$, either $\pi \mid a$ or $\pi \mid b$. We verify these conditions for $\pi = p$. First, $p \neq 0$. Second, $|p|_p = p^{-1} < 1$, so $p \notin \mathbb{Z}_p^\times$ (since $\mathbb{Z}_p^\times = \{x \in \mathbb{Z}_p : |x|_p = 1\}$ by the theorem [Properties of the Valuation Ring](/theorems/???)). For the divisibility condition: suppose $p \mid ab$, meaning $ab \in p\mathbb{Z}_p$, equivalently $|ab|_p \leq p^{-1}$. Since $|ab|_p = |a|_p |b|_p$ and the values of $|\cdot|_p$ on $\mathbb{Z}_p \setminus \{0\}$ are $\{p^{-k} : k \geq 0\}$, we need $|a|_p |b|_p \leq p^{-1}$. Since $|a|_p, |b|_p \in \{p^{-k} : k \geq 0\} \cup \{0\}$ and each is at most $1$, at least one factor must satisfy $|\cdot|_p \leq p^{-1}$. (If both were $1$, the product would be $1 > p^{-1}$.) Therefore $|a|_p \leq p^{-1}$ or $|b|_p \leq p^{-1}$, giving $a \in p\mathbb{Z}_p$ or $b \in p\mathbb{Z}_p$. Hence $p \mid a$ or $p \mid b$. [guided] We check that $p$ satisfies the definition of a prime element. The non-zero and non-unit conditions are immediate from $|p|_p = p^{-1} \neq 0$ and $|p|_p \neq 1$. For the divisibility property, suppose $p \mid ab$, i.e., $ab \in p\mathbb{Z}_p$. This means $v_p(ab) \geq 1$ (or $ab = 0$). If $ab = 0$, then $a = 0$ or $b = 0$ (since $\mathbb{Z}_p$ is an integral domain), and in either case $p$ divides $a$ or $b$. If $ab \neq 0$, then $v_p(ab) = v_p(a) + v_p(b) \geq 1$, so at least one of $v_p(a)$ or $v_p(b)$ is $\geq 1$. This means $a \in p\mathbb{Z}_p$ or $b \in p\mathbb{Z}_p$, i.e., $p \mid a$ or $p \mid b$. The argument works because the $p$-adic valuation is additive on products and takes values in $\mathbb{Z}_{\geq 0}$ on $\mathbb{Z}_p$. This is the fundamental reason that $p$ behaves as a prime: the valuation detects divisibility by $p$. [/guided] [/step] [step:Prove uniqueness: $p$ is the only prime element up to units] Let $\pi \in \mathbb{Z}_p$ be any prime element. Since $\pi \neq 0$ and $\pi \notin \mathbb{Z}_p^\times$, we have $v_p(\pi) \geq 1$. The ideal $\pi \mathbb{Z}_p$ is a non-zero proper ideal of $\mathbb{Z}_p$, so by the ideal classification, $\pi \mathbb{Z}_p = p^n \mathbb{Z}_p$ for some $n \geq 1$. We claim $n = 1$. Since $\pi$ is prime, the ideal $\pi \mathbb{Z}_p$ is a prime ideal. The ideal $p^n \mathbb{Z}_p$ is prime if and only if whenever $ab \in p^n \mathbb{Z}_p$, either $a \in p^n \mathbb{Z}_p$ or $b \in p^n \mathbb{Z}_p$. If $n \geq 2$, take $a = b = p$: then $ab = p^2 \in p^n \mathbb{Z}_p$ (since $n \leq 2$ would require $n = 2$ and $p^2 \in p^2\mathbb{Z}_p$, which holds). But $p \in p^2 \mathbb{Z}_p$ would require $v_p(p) \geq 2$, i.e., $1 \geq 2$, a contradiction. More generally, for $n \geq 2$, take $a = b = p^{\lceil n/2 \rceil}$: then $v_p(ab) = 2\lceil n/2 \rceil \geq n$, so $ab \in p^n \mathbb{Z}_p$, but $v_p(a) = \lceil n/2 \rceil < n$ for $n \geq 2$, so $a \notin p^n \mathbb{Z}_p$ and similarly $b \notin p^n \mathbb{Z}_p$. This contradicts primality of $p^n \mathbb{Z}_p$. Therefore $n = 1$, giving $\pi \mathbb{Z}_p = p \mathbb{Z}_p$. This means $\pi = p \cdot u$ for some $u \in \mathbb{Z}_p^\times$ (since $\pi \in p\mathbb{Z}_p$ gives $\pi = p u'$ with $u' \in \mathbb{Z}_p$, and $p = \pi w$ with $w \in \mathbb{Z}_p$, so $p = p u' w$, giving $u' w = 1$, hence $u' \in \mathbb{Z}_p^\times$). Thus $\pi$ is an associate of $p$. [guided] We want to show that any prime element $\pi$ of $\mathbb{Z}_p$ must be a unit multiple of $p$. The strategy uses two facts: the ideal classification tells us $\pi \mathbb{Z}_p = p^n \mathbb{Z}_p$ for some $n \geq 1$, and the primality of $\pi$ forces $n = 1$. Since $\pi$ is prime, $\pi \mathbb{Z}_p$ is a prime ideal. By the ideal classification, $\pi \mathbb{Z}_p = p^n \mathbb{Z}_p$ for some unique $n \geq 1$ (it cannot be $n = 0$ because $p^0 \mathbb{Z}_p = \mathbb{Z}_p$, but $\pi$ is not a unit). Why must $n = 1$? If $n \geq 2$, we can exhibit a failure of the prime ideal property. Consider $a = b = p$. Then $ab = p^2$, and $v_p(p^2) = 2 \geq n$ when $n = 2$, so $ab \in p^n \mathbb{Z}_p$. But $v_p(p) = 1 < 2 = n$, so $p \notin p^n \mathbb{Z}_p$. Neither factor lies in $p^n \mathbb{Z}_p$, contradicting the primality of $p^n \mathbb{Z}_p$. For $n > 2$, a similar argument works with $a = p^{\lceil n/2 \rceil}$ and $b = p^{\lfloor n/2 \rfloor}$: the product has valuation $n$ but each factor has valuation $< n$. So $n = 1$, meaning $\pi \mathbb{Z}_p = p \mathbb{Z}_p$. From $\pi \in p\mathbb{Z}_p$, write $\pi = pu$ with $u \in \mathbb{Z}_p$. From $p \in \pi \mathbb{Z}_p$, write $p = \pi w$ with $w \in \mathbb{Z}_p$. Then $p = puw$, so $uw = 1$, and $u \in \mathbb{Z}_p^\times$. Therefore $\pi = pu$ is a unit multiple of $p$. [/guided] [/step]