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Monic irreducible polynomials in valued fields

  1. Jan 29, 2012 #1
    1. The problem statement, all variables and given/known data
    I am trying to work out a solution to the following problem, where we are working in a field [itex]K[/itex] complete with respect to a discrete valuation, with valuation ring [itex]\mathcal{O}[/itex] and residue field [itex]k[/itex].

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

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

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

    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 [itex]f[/itex] through by some constant with sufficiently large valuation to get some [itex]g[/itex] which lies in [itex]\mathcal{O}[X][/itex] (since [itex]\mathcal{O} = \{c \in K: \, v(c) \geq 0\}[/itex]), 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 [itex]f(0)[/itex] 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
  2. jcsd
  3. Jan 29, 2012 #2


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    Okay, so let's use the hint. Multiply your f by something from K to get a polynomial [itex]g(X) = \sum b_i X^i[/itex] that lives in [itex]\mathcal{O}[X][/itex] and, moreover, has max{|b_i|}=1. Now consider [itex]\bar{g}(X)[/itex]. It will look like
    [tex]b_t X^t + \cdots + b_n X^n \equiv X^t(b_t + \cdots + b_n X^{n-t}) \mod\mathfrak{p}.[/tex]
    Here b_t was chosen so that b_1, ..., b_{t-1} are all <1 in abs value. Now use your Q6.
  4. Jan 30, 2012 #3
    Ok, I think I've got it now, thanks very much for the help!
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