0
$\begingroup$

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.

$\endgroup$
2
  • 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

3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.