Monic irreducible polynomials in valued fields

[Mathmos6]

Homework Statement

I am trying to work out a solution to the following problem, where we are working in a field $K$ complete with respect to a discrete valuation, with valuation ring $\mathcal{O}$ and residue field $k$.

Q: Let $f(X)$ be a monic irreducible polynomial in $K[X]$. Show that if $f(0) \in \mathcal{O}$ then $f \in \mathcal{O}[X]$.

I am meant to use the following result I have proved:

Let $f(X) \in \mathcal{O}[X]$ be a polynomial, and suppose $\overline{f}(X) = \phi_1 (X) \phi_2(X)$ where $\phi_1,\,\phi_2 \in k[X]$ are coprime. Show that there exist polynomials $f_1,\,f_2 \in \mathcal{O}[X]$ with $f(X)=f_1(X)f_2(X)$, $\text{deg}(f_1) = \text{deg}(\phi_1)$ and $\overline{f_i} = \phi_i$ for $i=1,\,2$ (where $\overline{\cdot}$ denotes the reduction from $\mathcal{O}$ down into the residue field $k$.)

So, I spoke to the person who wrote the problem sheet who said (briefly) "In this question you should clear denominators and apply Q6." (Q6 being the result I stated above).

I believe I'm meant then to multiply $f$ through by some constant with sufficiently large valuation to get some $g$ which lies in $\mathcal{O}[X]$ (since $\mathcal{O} = \{c \in K: \, v(c) \geq 0\}$), and then I'm not sure where I'm meant to go from there: do I suppose some sort of factorisation and then apply irreducibility to get a contradiction? It also isn't clear to me where the condition on $f(0)$ is applied. I've been confused by this for ages so please, the more help you can give me the better. Many thanks in advance :) ---M

 

[morphism]

Okay, so let's use the hint. Multiply your f by something from K to get a polynomial $g(X) = \sum b_i X^i$ that lives in $\mathcal{O}[X]$ and, moreover, has max{|b_i|}=1. Now consider $\bar{g}(X)$. It will look like
$$b_t X^t + \cdots + b_n X^n \equiv X^t(b_t + \cdots + b_n X^{n-t}) \mod\mathfrak{p}.$$
Here b_t was chosen so that b_1, ..., b_{t-1} are all <1 in abs value. Now use your Q6.

 

[Mathmos6]

Okay, so let's use the hint. Multiply your f by something from K to get a polynomial $g(X) = \sum b_i X^i$ that lives in $\mathcal{O}[X]$ and, moreover, has max{|b_i|}=1. Now consider $\bar{g}(X)$. It will look like
$$b_t X^t + \cdots + b_n X^n \equiv X^t(b_t + \cdots + b_n X^{n-t}) \mod\mathfrak{p}.$$
Here b_t was chosen so that b_1, ..., b_{t-1} are all <1 in abs value. Now use your Q6.

Ok, I think I've got it now, thanks very much for the help!