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This is a follow-up question to this question. In that question, we learned that if, $T(y) = \mathrm{Res}_{x}(f(x), y - g(x))$ , then $T(g(z))$ is divisible by $f(z)$.

Now, my question is:

If $T(y) = \mathrm{Res}_{x}(f(x), y - g(x)),$ where $f(x), g(x)$ are two monic polynomials in $R[x]$ , where $R = \mathbb{Z}/(p^{k}\mathbb{Z})$, where $p$ is any prime and $k \geq 2$. Suppose further that $T(y) = T_1(y) \cdot T_2(y)$ is a nontrivial factorization of $T(y)$. Then, can we find two non trivial factors $f_1(x), f_2(x)$ such that $f(x) = f_1(x) \cdot f_2(x)$.

Now it is very clear that, if $R$ is a field then, $f_i(x) = \gcd(f(x), T_i(g(x)))$ . But if $R = \mathbb{Z}/(p^{k}\mathbb{Z})$, then this $\gcd$ computation is impossible.

So my question is, can we able to find the factors $f_i(x)$?

Edit 1: I think as $\gcd$ won't work in $R[x] = \mathbb{Z}/(p^{k}\mathbb{Z})[x]$, hence normal calculation in $R$ will not be sufficient. So I was thinking, if we can apply Hensel lifting, that is, by looking into the factorization of $T(y) \bmod{p}$, we can factor $f(x) \bmod{p}$, because then this resulutant works in the field, so we can apply the $\gcd$, then we can lift the factorization of $f(x) \bmod{p}$ to the factorization of $f(x) \bmod{p^k}$.

But there is one more problem. Let us consider a polynomial $f(x) = x^n + f_{n-1}x^{n-1} + \cdots + f_0 \in \mathbb{Z}/(p^k \mathbb{Z})[x]$. Let $-s/r$ be the slope of the lowest line segment of the Newton Polygon of $f(x)$, where $\gcd(s,r) = 1$. That is if $- u/v$ is the slope of any other line segment then $u/v > s/r$, hence $v_p(\alpha) \geq s/r ~\forall \alpha \in Z(f)$ (See the definition for $p$-adic valuation).

Now if we define, define, $\pi := x + \langle x^r - p\rangle$ in the larger ring, $R[x]/\langle x^r - p\rangle$(see this), then we have $\alpha = 0 \bmod{\pi} ~\forall \alpha \in Z(f).$ So basically $f(x) \bmod{\pi} = x^n$.

I also asked this question on MSE.

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  • $\begingroup$ I do not understand even the case of a field. Say, $f(x)=x^2-2$ is irreducible over some field $K$, but for $g(x)=x^2$ the resultant of $x^2-2$ and $y-x^2$ is $(y-2)^2$ that is reducible $\endgroup$ Commented Aug 15 at 9:14
  • $\begingroup$ We are considering two non-trivial factors, that is, two different factors, like not sharing common factors. Actually, in the field case, the statement says T(y) factors into two relatively prime factors, then f(x) factors. But this example is saying that the converse may not hold. But if T(y) is irreducible, then f(x) will also be irreducible. $\endgroup$ Commented Aug 15 at 9:30
  • $\begingroup$ It is not clear to me whether you are asking, "do factors exist?", or, "do we have a way to use the factorization of $T(y)$ to produce a factorization of $f(x)$?" $\endgroup$ Commented Aug 15 at 9:55
  • $\begingroup$ My target is to find a factor of $f(x)$ when the factorisation of $T(y)$ is given. $\endgroup$ Commented Aug 15 at 10:04

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