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Let $5$ divide $p-1$. Therefore, we have $$1+x+x^2+x^3+x^4=(x-\alpha)(x-{\alpha}^2)(x-\alpha^3)(x-\alpha^4)=f_1f_2f_3f_4$$ over $F_p,$ where $\alpha$ is an element of order $5$ in ${F_p}^\times.$ We can easily prove that $\left\langle x-\alpha^i\right\rangle$ and $\left\langle x-\alpha^j\right\rangle$ are co-prime in $F_p[x]$ for $1 \leq i,j \leq 4.$

Hensel lemma implies \begin{align} 1+x+x^2+x^3+x^4&=g_1g_2g_3g_4 \end{align} in $\mathbb{Z}_{p^n}[x]$ where $g_i's$ are co-prime and $\bar{g}_i=f_i.$ Here $\bar{g}_i$ means $g_i$ modulo $p$.

My question is that can we write $g_i=(x-\beta_i)$ for some $\beta_i \in {\mathbb{Z}}_{p^n}?$, i.e. can a simple root $\alpha_i$ be extended to $\beta_i$ such that $\bar{\beta}_i=\alpha_i$ ? If yes, then how to prove it?

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    $\begingroup$ You can use Hensel's lemma, already mentioned in your title. What do you understand "Hensel's lemma" to mean? I am not sure why you are doubting we can write $g_i = x - \beta_i$ when you know the $g_i$'s exist. $\endgroup$
    – KConrad
    Commented Feb 21 at 18:54
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    $\begingroup$ The distinct roots mod $p$ each lift to a unique root in the $p$-adic integers. In $\mathbf Z_p[x]$, which is a UFD, the polynomial is a product of four monic linear factors. Now reduce that mod $p^n$. $\endgroup$
    – KConrad
    Commented Feb 22 at 9:39
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    $\begingroup$ There is no need for any argument by contradiction. If a monic polynomial in $\mathbf Z[x]$ with degree $4$ has $4$ roots $r_i$ in $\mathbf Z$ then the polynomial is $\prod_{i=1}^4 (x-r_i)$. The same reasoning holds in $\mathbf Z_p[x]$. $\endgroup$
    – KConrad
    Commented Feb 22 at 14:51
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    $\begingroup$ I don't know why you say it is not true mod $p^n$. When $A$ is any commutative ring and $f(x) \in A[x]$ satisfies $f(a) = 0$ for some $a \in A$, we can write $f(x) = (x-a)q(x)$ where $\deg q = \deg f - 1$. If also $f(b) = 0$ and $b-a$ is a unit in $A$, then $0 = f(b) = (b-a)q(b)$, so $q(b) = 0$, so $q(x) = (x-b)r(x)$. Then $f(x) = (x-a)(x-b)r(x)$. You can keep repeating this as long as you have roots in $A$ whose differences are units in $A$. When $A = \mathbf Z/(p^n)$ where $a$ and $b$ are roots with $a\not\equiv b\bmod p$, $a-b \bmod p^n$ is a unit. $\endgroup$
    – KConrad
    Commented Feb 22 at 20:04
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    $\begingroup$ You seem to be using just one version of Hensel's lemma, involving a coprime polynomial factorization. There are many related results called Hensel's lemma, with no one result necessarily implying all the rest. Look up a version of Hensel's lemma specifically about lifting roots of polynomials, not about coprime polynomial factorizations. $\endgroup$
    – KConrad
    Commented Feb 22 at 20:06

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