Why must a zero of $f \in \mathbb{Z}_p[X_1, \dots, X_m] ~ (\text{mod } p^n)$ be simple in order to lift to $\mathbb{Z}_p$? In chapter II, section 2.2, of J-P. Serre's A Course in Arithmetic, we have the following theorem:

Theorem 1: Let $f \in \mathbb{Z}_p[X_1, \dots, X_m]$, $x = (x_i) \in \left( \mathbb{Z}_p \right)^m$, $n,k \in \mathbb{Z}$ and $j$ an integer such that $0 \leq j \leq m$. Suppose that $0 < 2k < n$ and that 
  $$ f(x) \equiv 0 ~(\text{mod } p^n) \quad \text{and} \quad \nu_p \left( \frac{\partial f}{\partial X_j} \right) = k$$
  Then there exists a zero $y$ of $f$ in $\left(\mathbb{Z}_p\right)^m$ which is congruent to $x$ modulo $p^{n-k}$.

Immediately after the proof, the following corollary is given:

Corollary 1. Every simple zero of the reduction modulo $p$ of a polynomial $f$ lifts to a zero of $f$ with coefficients in $\mathbb{Z}_p$.

Here a simple zero of $f$ is a zero $x$ such that at least one of the partial derivatives $\partial f / \partial X_j$ is non-zero at $x$. 
I do not understand why the corollary requires the condition that the zero be simple. It feels like I am missing something quite obvious. In any case, this leads to: 
Question: Why is this simple-zero hypothesis necessary in Corollary 1 above?
 A: A simple explanation (no pun intended) is that there are counter examples : just consider $f(x) \in \mathbb Z[x]$ defined by $f(x) = p$. All its roots mod $p$ do not lift mod $p^2$. 
A more fundamental explanation which explains there the argument breaks down is this : just consider the one variable case $f \in \mathbb Z[X]$for a second, where you write 
$$
f(X) = \sum_{i=0}^n a_i (X-x_0)^i 
$$
and the assumption that $x_0$ is a simple zero amounts to $a_0 \equiv 0 \pmod p$, $a_1 \not\equiv 0 \pmod p$. Trying to find a lift is trying to find $b_0 \equiv a_0 \pmod p$ such that 
$$
a_0 + a_1(b_0 - a_0) + a_2 (b_0 - a_0)^2 + \cdots \equiv 0 \pmod{p^2}.
$$
I put dots because you see that $(b_0-a_0)^i = (kp)^i \equiv 0 \pmod{p^2}$ for $i \ge 2$. Writing $b_0 - a_0 = kp$ for some $k \in \{0,\cdots,p-1\}$, you see that if $a_1 \equiv 0 \pmod p$, then the sum above equals the integer $a_0$ modulo $p^2$, hence you have no "choice" of lifting ; either the integer $a_0$ was already a zero modulo $p^2$ (in which case any value of $b_0 \equiv a_0 \pmod p$ does the trick) or it wasn't and you have no options for a solution modulo $p^2$.
Hope that helps,
