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This is actually true for any ring $S$ of characteristic $p$, and the argument I have in mind requires considering Witt vectors of non-perfect rings(such as $S/t^n$), so let me do it in full generalit …

$\newcommand{\bC}{\mathbb{C}}\newcommand{\bZ}{\mathbb{Z}}$The resolution constructed by Dylan Wilson seems to show that $Ext^1(M,N)$ is non-zero already for $M=\bC[x][\frac{1}{x+n},n\in\bZ]$. The …

In the Russian edition it is $$(\xi_0,\xi_1,\xi_2,\dots)\leftrightarrow f(\xi_0)+pf(\xi_1)+p^2f(\xi_2)+\dots$$ where $f$ is the Teichmuller map.

$\newcommand{\Gal}{\mathrm{Gal}}\newcommand{\Z}{\mathbb{Z}}$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{ …

Let $k$ be a field and take $$R=\{(a_i)\in\prod\limits_{i\in\mathbb{N}}k[t]\mid a_i(0)=a_j(0)\text{ for all }i,j\}$$ An idempotent in this ring has to be sent to $0$ or $1$ under the map $R\xrightarr …

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}:=( …

The connected finite kernel $H$ is not solvable, provided that $n>2$ or $p>2$, see the edit below. $\def\eps{\varepsilon} \def\m{\mathfrak{m}}$Suppose by contradiction that $H$ is solvable and the $m$ …

Suppose that $I$ admits a divided power structure. On the one hand, $\gamma_p(x_1x_2+x_3x_4+x_5x_6)$ has to be equal to zero because the element $x_1x_2+x_3x_4+x_5x_6$ is zero in our ring, but let's e …

In the previous lines it is proven that $U(T)^W$ is integral over $\gamma(Z^{\mathcal G})$. Since $U(T)^W\subset Frac(U(T)^W)=Frac(\gamma(Z^{\mathcal G}))$ the equality follows from $\gamma(Z^{\mathca …

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 …

(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 $f_m=\overline{g^{1}_{m+1}}\dots \overline{g^{i_{m+1}}_{m+1}}$ …

Without any finiteness assumptions on $M$, the converse fails already for $A=k[x, y]$ ($k$ is a field). Take $M=k[x,y^{\pm 1}]\oplus k[x^{\pm 1},y]$ and $a=(x,y)$. The module $M$ is flat over $A$ so …

$\newcommand{\bQ}{\mathbb{Q}}\newcommand{\bZ}{\mathbb{Z}}\DeclareMathOperator{\Tor}{Tor}\newcommand{\Tors}{\mathrm{Tors}}$I tried to write up the computation with some level of details, please let me …