We shall define the infinitely-many-variable formal power series ring $A = {\Bbb F}_q[[X_1,\ldots,X_{\infty}]]$ over a finite field ${\Bbb F}_q$ as the following$\colon$

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

For example, $\Sigma_{n = 1}^{n = \infty} X_n = X_1 + X_2 + \ldots \in A$. The ring $A$ is a non-noetherian local ring with the unique closed maximal ideal. I would like to pose the following theorem on which please let me know the correctness$\colon$

Theorem. $A$ is coherent.

Proof. Let us fix a positive integer $l \geq 1$ and for an arbitrarily chosen elements $a_1,\ldots,a_l$ consider the linear equation

$(*) \quad a_1Y_1 + \ldots + a_lY_l = 0 \quad (a_1,\ldots,a_l \in A).$

We shall define ${\mathrm M}_{\infty}$ as the set of the whole solutions of $(*)$ in the ring $A$. We shall show the finiteness of the number of generators of ${\mathrm M}_{\infty}$ as an $A$-module.

We define ${\mathrm M}_{n,m}$ as the image of ${\mathrm M}_{\infty}$ in the quotient ring $A_{n,m} \colon= A/((X_1,\ldots,X_n)^m,X_{n+1},X_{n+2},\ldots)$. We have simply $A_{n,m} = {\Bbb F}_q[[X_1,\ldots,X_n]]/(X_1,\ldots,X_n)^m$.

Lemma. For sufficiently large $n, m$, the $A_{n,m}$-module ${\mathrm M}_{n,m}$ has generators whose number is less than or equal to $l$.

Proof. Obviously, ${\mathrm M}_{n,m}$ can be viewed as the subset of the whole solutions of the linear equation $(*)$ in the quotient ring $A_{n,m}$. But the whole solutions of the linear equation $(*)$ in the quotient ring $A_{n,m}$ has its cardinality less than or equal to $|A_{n,m}|^l$. Consequently we have

$|{\mathrm M}_{n,m}| \leq |A_{n,m}|^l. $

When we view ${\mathrm M}_{n,m}$ as a $A_{n,m}$-module and set the number of the generators of ${\mathrm M}_{n,m}$ to be $\delta_{n,m}$, the following inequality holds$\colon$

$|(A_{n,m})^{*}|^{\delta_{n,m}} \leq |{\mathrm M}_{n,m}|. $

In short we have

$|(A_{n,m})^{*}|^{\delta_{n,m}} \leq |A_{n,m}|^l.$

We can see the equality $|(A_{n,m})^{*}| = \frac{(q-1)}{q}|A_{n,m}|$ because $A_{n,m}$ can be divided into $q$ disjoint cosets as $c + {\frak m}$ with $c \in {\Bbb F}_q$ and ${\frak m}$ being the maximal ideal of $A_{n,m}$. Thus we obtain

${\delta_{n,m}} {\mathrm log}_e(\frac{(q-1)}{q}|A_{n,m}|) \leq l\,{\mathrm log}_e(|A_{n,m}|)$.

So, we have

${\delta_{n,m}} \leq l \,{\mathrm log}_e(|A_{n,m}|)/{\mathrm log}_e(\frac{(q-1)}{q}|A_{n,m}|)$.

When $n,m \to \infty$, we have $|A_{n,m}| \to \infty$, which ensures that the positive integer ${\delta_{n,m}}$ must be less than or equal to $l$ for sufficiently large $n, m$. Q.E.D.

Now, by Lemma we can conclude that ${\mathrm M}_{\infty}$ has the set of generators with its cardinality less than or equal to $l$ as $A$-module because we have the equality ${\mathrm M}_{\infty} = \underset{n,m \geq 1}{\varprojlim} {\mathrm M}_{n,m}$, where all natural transition maps ${\mathrm M}_{n',m'} \to {\mathrm M}_{n,m}$ with $n' > n, m' > m$ are surjective. Q.E.D.

  • 3
    $\begingroup$ I do not think that the role of MO is to check the correctness of alleged proofs. $\endgroup$
    – abx
    Jan 6, 2018 at 7:06
  • $\begingroup$ Does it not follow from the answer here mathoverflow.net/questions/117890/… ? $\endgroup$
    – Frank
    Jan 8, 2018 at 13:23

1 Answer 1


I do not see where the inequality $$ |(A_{n,m})^∗|^{δ_{n,m}}≤|M_{n,m}| $$ comes from. Different choices of $δ_{n,m}$-tuples of elements of $(A_{n,m})^∗$ could give rise to the same element of $M_{n,m}$. For this reason I do not understand the proof.


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