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I know that the polynomial $f(x) = x^p -x - \frac{1}{p} \in \mathbb{Q}_{p}[x]$ is irreducible. So, let $\alpha$ be a root of $f(x)$, and $K = \mathbb{Q}_p(\alpha)$. Let $O_K$ be the valuation ring of $K$ with respect to the unique extension of the p-adic valuation to $K$. Let $\mathfrak{p}_K$ be the unique maximal ideal of $O_K$. We know that $O_K$ is also the integral closure of $\mathbb{Z}_p$ in $K$.

I am trying to show that there exists $\beta_i \in O_K$ for $i =0,...,p-1$ such that $\alpha + \beta_i$ is a root of $f(x)$ and $\beta_i \equiv i$ (mod $\mathfrak{p}_k$).

My guess is the following:

I know from Hensel's lemma that the polynomial $x^p - x$ splits in $\mathbb{Z}_p$ with roots $\beta_i \in \mathbb{Z}_p$ for $i =0,...,p-1$ such that $\beta_i \equiv i$ (mod $p\mathbb{Z}_p$). So clearly, $\beta_i \equiv i$ (mod $\mathfrak{p}_k$). What remains to show is that $\alpha + \beta_i$ are the roots that I seek.

I think I am on the right track, but I don't know how to show that $\alpha + \beta_i$ are roots of $f(x)$. Any help or hint would be appreciated.

Of course, my guess could be totally wrong. In that case I have no idea how to come up with the desired $\beta_i 's$.

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The polynomial $x^p-x$ takes sums to sums only in characteristic $p$, so I don't think that $\alpha+\beta_i$ will be your roots?? –  Jyrki Lahtonen Apr 23 '12 at 6:04
    
Do you think there is some other natural choice of $\beta_i$ then? I can't think of any other way to come up with the $\beta_i 's$. –  Rankeya Apr 23 '12 at 6:06
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1 Answer 1

up vote 5 down vote accepted

Let $g(x) = f(x + \alpha)$. I claim that the coefficients of $g$ lie in $\mathcal{O}_K$, and not just $K$; to see this, write out $g$ and expand the first term using the binomial theorem; note that the valuation of $\alpha$, while negative, is greater than $-1$ (the valuation is normalized from $\mathbb{Z}_p$).

Since we have checked this, it now makes sense to reduce $g$ mod the maximal ideal of $\mathcal{O}_K$ so that we can apply Hensel's lemma, but after reducing, the polynomial $g$ is just $x^p - x$, so over $\mathcal{O}_K$ it has roots congruent to all the residue classes mod $p$. This is equivalent to your claim since $\beta$ is a root of $g$ if and only if $\beta + \alpha$ is a root of $f$.

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This is really nice. Thanks. –  Rankeya Apr 23 '12 at 6:39
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