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Pointed out on famous disbelief, I know now that there is an embedding

$\iota_n \colon {\Bbb F}_p[[T_1,...,T_n]] \hookrightarrow {\Bbb F}_p[[X,Y]]$

for any $n \leq \infty$. Then I would like to ask

Q. Is any complete local sub-algebra $\iota_A \colon A \hookrightarrow {\Bbb F}_p[[X]]$ noetherian of Krull dimension $1$?

That is, I guess that complete local ring $A$ having an embedding $\iota_A \colon A \hookrightarrow {\Bbb F}_p[[X]]$ must be a noetherian local ring of Krull-dimension $1$.

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  • $\begingroup$ Laurent Moret-Bailey teaches us the following. That is, choose the unique maximal ideal $J$ of $A$. Then $A$ is $J$-adically complete and separated. Let $M$ be an $A$-module. If $M$ is $J$-adically complete and separated and $M/JM$ is finitely generated, then $M$ is finitely generated. We apply this to $M={\Bbb F}_p[[X]]$, and we have always $M/JM$ is a finite-dimensional ${\Bbb F}_p$-vector space. This means that $\iota_A$ is ''finite'', which (perhaps) means that $A$ has the Krull-dimension one without noetherian assumption of $A$. Is it obvious that $A$ is noetherian? $\endgroup$ Jun 17, 2016 at 13:02

1 Answer 1

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No : just take $A = \mathbb{F}_p$.

This is the only counterexample : any complete local sub-algebra $\iota_A \colon A \hookrightarrow {\Bbb F}_p[[X]]$, with $A \neq \mathbb{F}_p$, is noetherian of Krull dimension 1.

Indeed, if $A \neq \mathbb{F}_p$, then $\mathfrak{m}_A \neq 0$ (if not, then $A$ would be a field, and the existence of the morphism $A \rightarrow {\Bbb F}_p[[X]] \rightarrow \mathbb{F}_p$ would imply $A = \mathbb{F}_p$). Then :

  • $\mathfrak{m}_A {\Bbb F}_p[[X]]$ is a nonzero ideal of ${\Bbb F}_p[[X]]$, so that ${\Bbb F}_p[[X]] / \mathfrak{m}_A {\Bbb F}_p[[X]]$ is a finite dimensional $\mathbb{F}_p$-vector space.

  • $\mathfrak{m}_A {\Bbb F}_p[[X]]$ is contained in $(X)$ (in particular, the $A$-module ${\Bbb F}_p[[X]]$ is $\mathfrak{m}_A$-adically complete, since it is $X$-adically complete). If it wasn't, then we would have some equation $$ 1 = \sum_i a_i f_i, $$ with $a_i \in \mathfrak{m}_A$ and $f_i \in {\Bbb F}_p[[X]]$. The element $1 - \sum_i a_i f_i(0)$ of $A$ would then be an invertible element of $A$, and thus of ${\Bbb F}_p[[X]]$, which belongs to $(X)$ ; a contradiction.

As noticed by the OP in his comment, these two facts imply that $\iota_A$ is finite. Now, ${\Bbb F}_p[[X]]$ has Krull dimension $1$, and $\iota_A$ is integral and injective, so that $A$ has Krull dimension $1$ by Cohen-Seidenberg theorem. Moreover, $A$ is noetherian by Eakin's theorem (Eakin, P.M. "The Converse to a Well Known Theorem on Noetherian Rings." Mathematische Annalen 177 (1968): 278-282).

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  • $\begingroup$ In the above answer it is written, ''these two facts imply that $\iota_A$ is finite''. Why? I, however, heartily thank js21 for the kind comment. Anyhow the only counter-example to my question is simply ${\Bbb F}_p$. Finally what is 'OP' ? $\endgroup$ Jun 17, 2016 at 23:44

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