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It is known that for any UFD $F,$ if we have a monic polynomial $$f(x) = x^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0 \in F[x], $$ a prime $p$ in $F$, and a positive integer $k\le n-1$ such that the conditions

(1) The elements $a_0, a_1, \dots, a_{k-1}$ are all multiples of $p$

(2) The element $a_0$ is not divisible by $p^2$

(3) The element $a_k$ is not divisible by $p$

are met, then $f$ has an irreducible factor of degree at least $k.$

Is there any analogous result that holds when we work with a general integral domain $F$ (and some prime ideal $P$ of $F$)?

I know that there is a similar result for the specific case of $k=n,$ but i was wondering if there is a similar result in the general case.

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    In general you can show that $f$ can't be written as a product $f_1\cdots f_r$ with $\deg f_i\le k-1$ for all $i$. In order to do this consider $f=f_1\cdots f_r$ mod $\mathfrak p$. – 2017-02-23
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    Related: http://math.stackexchange.com/questions/2154204/in-the-polynomial-ring-over-an-integral-domain-monomials-decompose-into-monomia – 2017-02-23
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    @user26857: That's exactly what I am asking - is that true for arbitrary prime ideals $P,$ and if so, how is it proved? – 2017-02-24
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    And that's exactly what I said: yes, that's true, and I've already told you how to prove it. – 2017-02-24
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    I apologize, but I don't see how the argument follows from your suggestion (had I been able to derive the proof from what you wrote, I would not have had to ask this question in the first place). I'll try to think more about that you wrote however. – 2017-02-24
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    If one works in $R/\mathfrak p$ then $\bar f=x^k\bar g$ with $\bar g\ne 0$ since $\bar a_k\ne 0$ in $R/\mathfrak p$. Since $\bar f=\bar f_1\cdots\bar f_r$, $x$ must divide some $\bar f_i$. Can $x$ also divide some $\bar f_j$ with $j\ne i$? No, otherwise $f_i(0)$ and $f_j(0)$ both belong to $\mathfrak p$. What's wrong with this? Well, then $a_0=f(0)=f_1(0)\cdots f_r(0)\in\mathfrak p^2$. So, $x^k\mid\bar f_i$ and hence $\deg f_i\ge k$. – 2017-02-24
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    Ah okay, thank you very much, that helped. I was confused before because I couldn't see why we would necessarily have $x | \overline{f}_i$ for some $i$, since $(R/P)[x]$ is in general not a UFD. However, since $(R/P)[x]$ is an integral domain, at least one of the $f_i$ will have a root at zero, from which the claim follows. – 2017-02-24
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    Or notice that $x$ is a prime element in $(R/P)[x]$ for the same reason: $R/P$ is an integral domain. – 2017-02-24
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    Ah, you're correct. That's probably a better way to see it. – 2017-02-24

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