Recall that a commutative ring is coherent if every finitely generated ideal is finitely presented, or equivalently if every submodule of every finitely generated module is finitely presented.

Let $A \colon= K[[X_1,X_2,\ldots]]$ be an infinitely-many-variable formal power series ring over a field $K$. The following isomorphism holds$\colon$

$$A \colon= \underset{n \geq 1}{\varprojlim} K[[X_1,\ldots,X_n]].$$

We remark that $\Sigma_{n = 1}^{n = \infty} X_n = X_1 + \ldots + X_n + \ldots \in A$. $A$ is a huge non-noetherian complete local ring and is already proved to be a UFD by Hajime Nishimura.

$A$, however, is not yet known to be coherent, so we are interested in whether the whole solutions of

$$(\sharp) \quad a_1Y_1 + a_2Y_2 + a_3Y_3 = 0$$

will be a finitely-generated $A$-module or not. If Claim in the below should hold, I can prove the finite generation of the whole $A$-solutions of $(\sharp)$ or more generally the coherence of $A$.

Let $M \subset A^{\oplus 3}$ be an $A$-module such that $M$ satisfies the following 2 conditions$\colon$

Condition 1. $M$ is an infinitely-generated $A$-module.

Condition 2. $M$ turns into a finitely-generated $B$-module after the variables change

$$X_1 \to X_1$$ $$X_2 \to X_1S_2$$ $$X_3 \to X_1S_3$$ $$\ldots$$ $$X_i \to X_1S_i$$ $$\ldots$$

with $B \colon= K[[X_1,S_2,\ldots]]$, where we naturally view $A$-module $M$ as $B$-module by way of the inclusion $A \subset B$.

${\bf Claim.}$ The module $M$ cannot be the set of whole solutions of $(\sharp)$ for any triple $a_1,a_2,a_3$.

My conviction why ${\bf Claim.}$ holds is that when $M$ simply is an ideal of $A$, the only possible case for $A$-module $M$ satisfying Condition 1 and Condition 2 seems that $M = (X_1,X_2,\ldots)$, in which case $M = (X_1)$ as $B$-module, hence finitely-generated. That is, both Conditions impose a very heavy constraint on the structure of $M$ as an $A$-module, which should ban $M$ from being the set of whole solutions of $(\sharp)$ for any $a_1, a_2, a_3 \in A$.

  • $\begingroup$ What do you mean by the sentence "we naturally view the $A$-module $M$ as a $B$-module via the inclusion $A\subset B$" ? What is the "natural" $B$-module structure? $\endgroup$ – Mark Spivakovsky Jan 20 '18 at 17:36

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