The following might be super easy but I am failing to have a complete proof. Consider the ring $\mathbb{Z}_{p^k}$ and identify $\mathbb{Z}_p$ with $\{0, \cdots , p-1\} \subset \mathbb{Z}_{p^k}$. If we have linearly independent vectors in $\mathbb{Z}_p^n$, with the above identification, they can be viewed as vectors in $\mathbb{Z}_{p^k}^n$. Are they still linearly independent over $\mathbb{Z}_{p^k}$?
Or more generally, if $A$ is a matrix over $\mathbb{Z}_p$, can we say that $$\text{rank }_{_{\mathbb{Z}_{p}}} (A) = \text{rank }_{_{\mathbb{Z}_{p^k}}}(A)?$$
Yes, independence is preserved.
Suppose that $v_1, v_2, \dots, v_m$ are linearly independent in $\mathbb Z_p^n$, and let $\lambda_1, \dots, \lambda_m \in \mathbb Z_{p^k}$ be such that $$\lambda_1 v_1 + \lambda_2 v_2 + \dots + \lambda_m v_m \equiv 0 \in \mathbb Z_{p^k}^n.$$ Setting $\lambda_i' = \lambda_i \bmod p$, and taking the equation above modulo $p$, we get $$\lambda_1' v_1 + \lambda_2' v_2 + \dots + \lambda_m' v_m \equiv 0 \in \mathbb Z_p^n,$$ so we must have $\lambda_i' = 0$ for all $i$. Therefore $p \mid \lambda_i$ for all $i$, and we can rewrite our original equation as $$\frac{\lambda_1}{p} v_1 + \frac{\lambda_2}{p} v_2 + \dots + \frac{\lambda_m}{p} v_m \equiv 0 \in \mathbb Z_{p^{k-1}}^n.$$ Repeating this argument, we can conclude for ever-increasing $j$ that $p^j \mid \lambda_i$ for all $i$ and that $$\frac{\lambda_1}{p^j} v_1 + \frac{\lambda_2}{p^j} v_2 + \dots + \frac{\lambda_m}{p^j} v_m \equiv 0 \in \mathbb Z_{p^{k-j}}^n$$ until we finally conclude it for $j=k-1$. At that point, we have a linear relation between $v_1, \dots, v_n$ in $\mathbb Z_p$, so we can conclude that all coefficients are $0$. This implies $\lambda_1 = \dots = \lambda_m = 0$, concluding the standard independence argument.