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In Gouvea, proof of the corollary uses Hensel lemma that, in turn, uses derivative. …

The equation $x_j^p - x_j = y_j$ has a unique solution $x_j \in \mathbb Z_p$ congruent to $i$ mod $p$ for each $y_j \in p \mathbb Z_p$ by Hensel's lemma. … It is continuous, and it is easy to see the inverse provided by Hensel's lemma is continuous as well. Thus it satisfies your desiderata (assuming a topological isomorphism was meant). …

If $g(z)$ were a polynomial then you'd know $g(z)$ has a zero in $\mathfrak m_K$ by Hensel's lemma. … By Hensel's lemma for power series, the conditions $|g(0)| < 1$ and $|g'(0)| = 1$ implies there is some $x_0 \in \mathfrak m_K$ such that $g(x_0) = 0$, so $f(x_0) = z_0$. …

I had always thought it was true when $S=\{\text{Bad primes of }E/K, \text{ infinite places of }K\}$ until now because rational points on reduction mod $v$ lifts to rational points on $K_v$ thanks to Hensel … lemma. …

I imagine one might approach this by some kind of "Hensel lifting", whereby we have solutions $x_i^2 - dy_i^2 \equiv \alpha \pmod{\mathfrak{p}_F^i}$ for $i=0,1,2,\ldots$, and construct $(x_{i+1},y_{i+1 … I don't know if such lifting is feasible, since the polynomial doesn't necessarily have a root, so the hypotheses of Hensel's lemma cannot be satisfied in general. …

.$ Hensel lemma implies \begin{align} 1+x+x^2+x^3+x^4&=g_1g_2g_3g_4 \end{align} in $\mathbb{Z}_{p^n}[x]$ where $g_i's$ are co-prime and $\bar{g}_i=f_i.$ Here $\bar{g}_i$ means $g_i$ modulo $p$. …

Adjoining the $n$th root of this generator gives an unramified extension by Hensel's lemma. …

^{\infty} \sum_{m=1}^{n+1} \text{sinc}(\pi (n^2 + n - mz)) \end{equation} Why I am interested in this specific series: One can prove (this hinges on a combination of the Chinese Remainder Theorem and Hensel's … Lemma) that when $z$ is a positive integer, \begin{equation} 2^{\omega(z)}=\sum_{n=0}^{\infty} \sum_{m=1}^{n+1} \text{sinc}(\pi (n^2 + n - mz)) \end{equation} where $\omega(z)$ gives the number of …

A Hensel's-lemma type approximation, using surjectivity of $\operatorname{tr}_{k_F/k_E}$ (where $k$ denotes residue fields), shows that $N_{E/F} : 1 + \mathfrak p_E \to 1 + \mathfrak p_F$ is surjective …

Have you tried using Hensel's lemma to show $|T - \alpha|_P < 1$ for some $\alpha$ in $\mathbf F_{q^d}$ and then show $T-\alpha$ is a uniformizer in the completion? …

I believe that Hensel's lemma will also put these in bijection with the idempotents over $E_n$ and over the spherical group algebra over the $K(h)$-local sphere; in both cases $\pi_0$ is a group algebra …

Is it true that $Z_{V,p}(p^{-s})$ is the Hasse Weil zeta function for the lift of a positive characteristic variety in the way of Hensel's lemma and Newton's method in positive characteristic? …

$p$” methods in mathematics (starting with the Hensel lemma!) so would love to know if any of them apply here. …

indexed) version of Rayner's construction is some sort of "completion" $\widehat R$ of a higher-rank valuation ring, so every element of $1+\mathfrak m_{\widehat R}$ should be a square if you believe in Hensel's … lemma/Newton's method (this is Theorem 1 in Rayner). …

lemma (the one about lifting relatively prime factorizations, not just about lifting a simple root) we'd get a factorization of $F(x)$ in $\mathbf Z_p[x]$ into nonconstant monic factors, which contradicts … The polynomial $t^{p^f} - t$ splits completely over the field $\mathbf F_p[x]/(Q)$, so for each $m \geq 1$ the $Q$-adic Hensel's lemma tells us we can lift each of those roots mod $Q$ uniquely to a root …