Showing reducibility of a polynomial in a Discrete Valuation Ring Let $R$ be a complete discrete valuation ring with uniformiser $\pi$. I would like to show that a polynomial $f$ in $R[X]$ is reducible. Does it suffice to show that $f$ is reducible in $\frac{R}{\pi^i}[X]$ for all $i\in\mathbb{Z}_{\geq{1}}$?

Thoughts: I think it does because $R$ complete means it is the inverse limit of the  $\frac{R}{\pi^i}$'s. Moreover each factorisation of $f$ in $\frac{R}{\pi^i}[X]$ passes down to $j\leq i$. But I kind of need to pass down from $i$ equal to infinity to make this work. I feel like I've seen something like this before somewhere else as well where it was made to work but I can't remember the argument... 

The example I'm interested in is when $R$ is a $\mathbb{Z}_p$ and $\pi=p$ where $p$ is a rational prime.
If $R/(\pi)$ is a finite field, then $R$ is compact for the $\pi$-adic topology, which gives an easy proof of this.

Denote $R_n[X]$ the set of polynomials of degree at most $n$ with coefficients in $R$. You can give it the product $\pi$-adic topology (two polynomials are close when all their coefficients are close) and it is also compact.

Your hypothesis says that you have factorisations $f = P_kQ_k \pmod {\pi^k}$, that is, you have an infinite sequence $(P_k,Q_k)$ such that $\lim (P_k Q_k) = f$ and $1 \le \deg P_k, \deg Q_k &lt; \deg f$, and so $(P_k, Q_k)$ are in the compact space $R_{\deg f-1}[X]^2$.

Therefore the sequence $(P_k,Q_k)$ has one (maybe several) limit point $(P,Q)$. And since the multiplication map is continuous, you must have $PQ = f$, and $\deg P, \deg Q &lt; \deg f$ so this is a nontrivial factorisation.



Concretely, you can just look at your sequence $(P_k,Q_k)$, first select an infinite subsequence where all the $(P_k,Q_k) \pmod \pi$ agree to some $(p_1,q_1) \in (R/\pi)[X]^2$, then select an infinite subsequence where all the $(P_k,Q_k) \pmod {\pi^2}$ agree to some $(p_2,q_2) \in (R/\pi^2) [X]^2$, and so on.

The family $(p_l,q_l)$ you get is compatible, and defines an element $(p,q)$ in the projective limit $R[X]^2$
Here it is. There is a well known result (exercise ?) saying that $\mathbf Q_{p}(w)$ , where $w$ is a primitive p-th root of 1, is equal to $\mathbf Q_{p}(\sqrt[p-1]-p)$, so the polynomial $X^{p-1} + p$ is irreducible over $\mathbf Z_{p}$ , but is reducible modulo all powers of p .

Edit. This counter-example is not correct, see Mathmo's remark.