Proof

Assume, $d\ge 0$ is a fixed integer.
Case $d=0$:
- Let $n\in \mathbb Z_0$.
- By definition of $Z_0$ we have $\gcd(n,0)=1$ only for $n=1$, it follows that $\mathbb Z_0=\{1\}$.
- Trivially, $\mathbb Z_0$ is divisor-closed.
Case $d = 1$:
- Let $n\in \mathbb Z_1$.
- Since we have $\gcd(n,1)=1$ for all $n\in \mathbb Z$, it follows $\mathbb Z_1=\mathbb Z$.
- Trivially, $\mathbb Z$ is divisor-closed.
Case $d > 1$:
- There is a unique factorization $|d|=p_1^{e_1}\cdot\ldots\cdot p_k^{e_k}$ with $e_i\ge 0$ for $i=1,\ldots,k$ by the fundamental theorem of arithmetic.
- Let $n\in \mathbb Z_d$ and let $t\mid n$. We have to show that $t\in\mathbb Z_d$.
- If $t=1$, then $\gcd(t,d)=1$. Thus, $t\in \mathbb Z_d$.
- If $t > 1$, then also $n > 1$ and we have two unique factorizations: * $n=q_1^{f_1}\cdot\ldots\cdot q_m^{f_m}$ be with $0\le f_j$ for $j=1,\ldots,m$ and * $t=q_1^{g_1}\cdot\ldots\cdot q_m^{g_m}$ with $0\le g_j\le f_j$ for $j=1,\ldots,m$, since $t\mid n$.
- Because $n\in \mathbb Z_d$, we have $\gcd(n,d)=1$.
- Thus, there is no such index $l$, $l=1,\ldots,\min(m,k)$, for which both exponents $e_l > 0$ and $g_l > 0$.
- Otherwise, $\gcd(q_l^{e_l},q_l^{g_l})=q_l^{\min(e_l,g_l)} > 1 $, because $\min(e_l,g_l)>1$, and we would have found a proper common divisor of $n$ and $d$, contradicting $n\in\mathbb Z_d$.
- For the same reason, $t$ and$d$ are co-prime.
- Thus, $t\in\mathbb Z_d$.
  ∎

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