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It is not right exact. Assume that $k$ is algebraically closed. If the map $Res_{X/k}B\to Res_{X/k}C$ was surjective as a map of sheaves for the fppf topology, then in particular, the map on sections …

$\newcommand{\Spec}{\mathrm{Spec}}\newcommand{\oL}{\overline{L}}\newcommand{\bG}{\mathbb{G}}\newcommand{\bZ}{\mathbb{Z}}\newcommand{\cL}{\mathcal{L}}$Not in general. The sequence of $p$-divisible grou …

$\newcommand{\bP}{\mathbb{P}}\newcommand{\bQ}{\mathbb{Q}}$Definition. A Belyi function is a non-constant rational function $f:\bP_{\bQ}^1\to \bP^1_{\bQ}$ such that the image of any of its critical poi …

1)Pick a sequence of elements $p^{1/p^n}\in \mathcal{O}_{\overline{K}}$ such that $(p^{1/p^{n+1}})^p=p^{1/p^n}$. The ideal $\ker\phi$ is in fact principal and is generated by the element $p^{\flat}:=( …

$\newcommand{\bQ}{\mathbb{Q}}\newcommand{\cO}{\mathcal{O}}\newcommand{\bC}{\mathbb{C}}\newcommand{\bZ}{\mathbb{Z}}\newcommand{\bF}{\mathbb{F}}$The open subgroup $Gal(\overline{\bQ}_p/\bQ_p(\mu_p))$ ac …

Here is a way to argue without showing directly that the polynomial must have degree $2$. It was explained to me by Borys Kadets (all further mistakes are, of course, my contribution). Lemma. If a se …

It is false that $\mathfrak{S}^{un}\cap pA_{cris}=p\mathfrak{S}^{un}$. For example, $E(u)\in\mathfrak{S}\subset\mathfrak{S}^{un}$ is not divisible by $p$ in $\mathfrak{S}^{un}$ but gets mapped to $\va …

$\def\bQ{\mathbb{Q}}\def\bF{\mathbb{F}}\def\bZ{\mathbb{Z}}$This is false as stated because of the following important difference between $K$ and $\mathbb{C}$: the former is not algebraically closed. F …

If a crystalline representation has finite image when restricted to inertia, then this restriction has to be trivial. Indeed, suppose that $K$ is a discretely valued extension of $\mathbb{Q}_p$ and …

$\newcommand{\Gal}{\mathrm{Gal}}\newcommand{\Z}{\mathbb{Z}}$Fix a compatible system $(t_n)$ of roots of $t$. It provides us with a section of $\rho$ thus giving an isomorphism between $\Gal(\overline{ …

$\newcommand{\eps}{\varepsilon}$It seems that for any $y$ the number of such $x$ is infinite. First of all, let's fix a prime $p$ and compute $v_p(\frac{(x^2)!}{(x!)^x})$ -- the exponent of $p$ in th …

Let $K$ be a finite extension of $\mathbb{Q}_p$ with the ring of integers $\mathcal{O}_K$ and the residue field $k$. By a theorem of Berthelot and Ogus(https://link.springer.com/article/10.1007%2FBF0 …

(I assume that $K$ is a field) Decompose each $f_m$ into a product of irreducible series $f_m=g^1_m\dots g^{i_m}_m$. For each $m$ we get $f_m=\overline{g^{1}_{m+1}}\dots \overline{g^{i_{m+1}}_{m+1}}$ …

No. If $E_p$ is a supersingular elliptic curve and $p>3$ then trace of Frobenius on $E_p$ is zero, so $\theta_E(p)=\pi/2$. By a result of Elkies any elliptic curve over $\mathbb{Q}$ has supersingular …

If $f$ is not required to be a morphism of $A$-algebras, a stupid counteraxmple exists. For instance, $A=k[x_1,x_2,\dots]$(infinitely many variables), $d=0$ and $$f:A\to A/(x_1,x_2,\dots)\cong k\subse …