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I'm reading Manin's article on formal groups and I have a problem with Lemma 1.1. Consider $k$ a prefect ring of characteristic $p$ and $(A,m,k)$ a noetherian complete local ring of the same characteristic such that $X=\operatorname{spf} A$ is a formal group. Manin wants to prove $X$ is a direct limit of finite group schemes over $k$.

Define $m^{(p^n)}$ as the ideal generated by $$\{x^{p^n}|x\in m\}.$$ (Manin uses the notation $m^{p^n}$ but I don't understand why $m^{p^n}$ is generated by these elements.) He proves that $\frac{A}{m^{(p^n)}}$ are Hopf algebras and because $A$ is complete we have $A=\varprojlim \frac{A}{m^{(p^n)}}$ as rings.

I don't understand if $m^{(p^n)}$ is really equal to $m^{p^n}$ and, if not, why $A$ and $\varprojlim \frac{A}{m^{(p^n)}}$ can be identified as topological rings?

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    $\begingroup$ But surely the ideal generated by the $x^{p^n}$ contains an ideal of form $m^N$ and is contained in $m^{p^n}$. Isn’t that enough? $\endgroup$
    – Lubin
    Commented Oct 15, 2019 at 0:52
  • $\begingroup$ @Lubin I don't understand why it most contains such an ideal?I know it's true for pid but for general complte ring and ideals it is false $\endgroup$
    – ali
    Commented Oct 15, 2019 at 5:57
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    $\begingroup$ As you observe, $m^{(p^n)} \neq m^{p^n}$ if $m$ is not principal. However, since $m$ is finitely generated, the two sequences of ideals are "cofinal": $$ m^{p^N} \subseteq m^{(p^n)} \subseteq m^{p^n} $$ for some $N$ depending on $n$. Indeed, if $m = (x_1, \ldots, x_r)$, then in any monomial of degree $\geq N=r(p^n-1) +1$ in the $x_i$, one factor will appear with exponent $\geq p^n$. So the inverse limits of quotient rings are the same. $\endgroup$ Commented Oct 15, 2019 at 9:37
  • $\begingroup$ @piotr achinger you are right thanks $\endgroup$
    – ali
    Commented Oct 15, 2019 at 14:10
  • $\begingroup$ Ah, @PiotrAchinger, thanks. I searched my leaky brain for the word “cofinal” and it wasn’t there. $\endgroup$
    – Lubin
    Commented Oct 15, 2019 at 14:30

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