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!