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So assume we are given some $a\in\mathbb{Z}_p^\times$ and we want to figure out if $X^p-a$ has a root in $\mathbb{Q}_p$. We know that such a root must be unique, because given two such roots $\alpha,\beta$, the quotient $\alpha/\beta$ would need to be a non-trivial $p^\textrm{th}$ root of unity and $\mathbb{Q}_p$ does not contain any.

Now we can't apply Hensel, which is the canonical thing to do when looking for roots in $\mathbb{Q}_p$. What other approaches are available?

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Start with the direct sum decomposition: $$\mathbf{Q}_p^*\simeq \langle p\rangle\times\mu_{p-1}\times U_1,$$ where $U_1$ consist of $p$-adic integers congruent to 1 modulo $p$. The first two factors are easy. The last one you need to work on. –  Jyrki Lahtonen Sep 21 '11 at 4:12

3 Answers 3

Let $p$ be an odd prime. Then $\mathbb{Z}_p^{\times} \cong \mathbb{Z}/(p-1) \oplus \mathbb{Z}_p.$ The pro-$p$ sylow of $\mathbb{Z}_p^{\times}$ is $1 + p\mathbb{Z}_p$ and is isomorphic (as a topological group) to $p\mathbb{Z}_p$ via the $p$-adic logarithm. Since $1 + p^2\mathbb{Z}_p$ is a closed subgroup of $1 + p\mathbb{Z}_p$ of index $p$ it must be the case that $1 + p^2\mathbb{Z}_p$ is the pullback of $p^2\mathbb{Z}_p$ under the logarithm. From this it follows

$$ 1 + p^2\mathbb{Z}_p = (1 + p\mathbb{Z}_p)^p.$$

And therefore,

$$(\mathbb{Z}_p^{\times})^p = \mu_{p-1}(1 + p^2\mathbb{Z}_p).$$

Now consider an element $x \in \mathbb{Q}_p^{\times}$ and recall $x = ap^n$ for some $a\in\mathbb{Z}_p^{\times}$ and $n\in\mathbb{Z}.$ Then $x$ has a $p$-th root in $\mathbb{Q}_p$ if and only if $a$ has a root in $\mathbb{Z}_p^{\times}$ and $p|n.$

Hence,

$$(\mathbb{Q}_p^{\times})^p = \mathbb{Z}_p^{\times}\langle p^p \rangle = \mu_{p-1}(1 + p^2\mathbb{Z}_p)\langle p^p \rangle.$$

Now try to answer the question for $p =2.$

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Since $a^p \equiv a\ ({\rm mod}\ p)$, we can replace $a$ by $a/a^p$ and just consider the case $a \equiv 1 ({\rm mod}\ p)$. Now we can still use Hensel's Lemma, just not in the trivial form. We need the general form that if we can find an approximation with $|f(a_0)| < |f'(a_0)|^2$, then $f$ has a root. See the example at Wikipedia in which the case $p=3$ is worked out. In general, if $c \in \mathbb{Z}_p$ and $c \equiv 1 ({\rm mod}\ p)$ then $c$ has a $p$th root in $\mathbb{Z}_p$ iff it has a $p$th root mod $p^2$, except when $p=2$, where you have to check mod 8.

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Let's take $\alpha\equiv 1\mod p^2$. Then $\alpha = 1 + p^2\beta\in\mathbb{Z}_p$, where $\beta\in\mathbb{Z}_p$. What we want to do is make Hensel's lemma work for us, after changing the equation a bit. We have $f(x) = x^p - \alpha = x^p - (1 + p^2\beta)$. We see that if $x$ is to be a solution, it must satisfy $x\equiv 1\mod p$, so $x = 1 + py$ for some $y\in\mathbb{Z}_p$. Now we have a new polynomial equation: $$ f(y) = (1 + py)^p - (1 + p^2\beta) = \sum_{i = 0}^p \begin{pmatrix} p \\ i\end{pmatrix}(py)^i - (1 + p^2\beta), $$ which reduces to $$ f(y) = \sum_{i = 1}^p \begin{pmatrix} p \\ i\end{pmatrix}(py)^i - p^2\beta. $$ So long as $p\neq 2$, we can set this equal to zero and cancel a $p^2$ from each term, and get $$ 0 = p^2 y + \begin{pmatrix} p \\ 2\end{pmatrix}(py)^2 + \ldots + (py)^p - p^2\beta = y + \begin{pmatrix} p \\ 2\end{pmatrix} y^2 + \ldots + p^{p-2}y^p - \beta. $$ Mod $p$, we can solve this equation: $$ y + \begin{pmatrix} p \\ 2\end{pmatrix} y^2 + \ldots + p^{p-2}y^p - \beta \equiv y - \beta \equiv y - \beta_0\mod p, $$ where $\beta = \beta_0 + \beta_1 p + \beta_2 p^2 + \ldots$ by $y = \beta_0$. Mod $p$, our derivative is always nonzero: $$ \frac{d}{dy}\left[y - \beta_0\right] \equiv 1\mod p, $$ so we can use Hensel's lemma and lift our modified solution mod $p$ to a solution in $\mathbb{Q}_p$. Therefore, if $\alpha\in 1 + p^2\mathbb{Z}_p$ and $p\neq 2$, there exists a $p$th root of $\alpha$ in $\mathbb{Q}_p$.

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