Let $u,v\in\mathbb{Z},$ and let $f=X^4+uX^2+v.$ Let $p$ be a prime number, and let $r\geq 1.$

In a paper I'm reading, one can find the following result.

**Proposition.** If $f$ is reducible modulo $p^r,$ then it is the product of two monic polynomials of $\mathbb{Z}/p^r[X]$ of degree $2$.

The authors pretend that the proof of this result is similar to the case where we consider polynomials in $\mathbb{Z}[X]$.

For the case of polynomials in $\mathbb{Z}[X]$, this is easy: since $\mathbb{Z}$ is an integral domain, the sum of the degrees of two factors of $f$ is $4$, and each of the leading coefficients is invertible, so $f$ has either a monic linear factor or a monic quadratic factor. In the first case, $f$ has a root $m$, but $-m$ is also a root, and we are done if $m\neq 0$ (if $m=0$ the result is obviously true).

Unfortunately, weird things may happen in $\mathbb{Z}/p^r[X].$ The degree of a product is not the sum of the degrees, so the proof above does not work anymore. Worse: for example, if $u=v=0,p=2,r=2$, then for any polynomial $g\in \mathbb{Z}/4[X]$, we have $f=X^4=(X^2+2g)^2,$ so $f$ may have divisors of arbitrary degree.

However, note there is still a decomposition as in the statement of the proposition, since $f=X^2\cdot X^2.$

I'm trying to find a proof of the result, but for the moment, I'm struggling. If $f$ is a product of two coprime polynomials modulo $p$, then by Hensel's lemma, the decomposition may be lifted to $\mathbb{Z}/p^r[X],$ but honestly if I can avoid the use of Hensel's lemma, I would be happier.

If $f$ is the square of an irreducible polynomial modulo $p$, I don't know...

**Question.** Am I missing something obvious? Is the proposition true (I believe so...)? Does somebody have some arguments to fill the gap, or a complete proof?