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Let $A:= \underset{\lambda \in \Lambda}{\varinjlim} \,A_{\lambda}$ be an inductive limit of geometric regular local ring $(A_{\lambda}, {\frak m}_{\lambda})$, whose transition map $\phi_{\mu\lambda} \colon A_{\lambda} \to A_{\mu}$ for a couple $(\lambda, \mu)$ is not necessarily flat. Geometric means that each $A_{\lambda}$ is a localisation of the smooth algebra of finite type over a field $K$ such that $|K| = \infty$. We denote by ${\frak m}_{A_{\lambda}}$ the unique maximal ideal of the local ring $A_{{\lambda}_0}$.

We consider a polynomial ring $A[X]$ over $A$, and for a fixed index $\lambda_0 \in \Lambda$ also the polynomial ring $A_{\lambda_0}[X]$ over $A_{\lambda_0}$.

Choose finitely many elements $f_1(X),...,f_n(X) \in A_{\lambda_0}[X]$ such that $f_{i}(k) \in {\frak m}_{A_{\lambda_0}}$ for $1 \leq i \leq n$ for an arbitrary element $k \in K$ and consider an arbitrary function $F(X) \in A[X]$ which satisfies the following condition:

$({\sharp})$ For any element $k \in K$, $F(k) \in (f_1(k),\ldots,f_n(k)) \subset {\frak m}_{A_{{\lambda}_0}}$.

Q. For any inductive limit to define $A$, does there always exist a finitely generated ideal ${\cal I}$ of $A[X]$ ($\,{\cal I} \not= A[X]\,$) such that any function $F(X)$ satisfying the condition $({\sharp})$ belongs to ${\cal I}$, i.e., $F(X) \in {\cal I}$?

For a general inductive limit, ${\cal I} =(f_1(X),\ldots,f_n(X))$ might not be sufficient in general, in whichever case I would like to at least bound the size of ${\cal I}$ to be finitely-generated.

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  • $\begingroup$ I probably misunderstood your question. Still, I'll ask: what about $\mathcal{I} \in \{A[X], A[X]F(X)\}$? If $A = A_{\lambda_0} = K$, isn't the condition $(\sharp)$ void? $\endgroup$
    – Luc Guyot
    Sep 27 '17 at 19:57
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In the case $A=K$ condition (#) says that $F$ vanishes at all the common zeroes of $f_1,...,f_n$. This example illustrates the following fact. In general, the answer to the question Q is negative whenever $(f_1(k),...,f_n(k))A$ is the unit ideal of $A$ for all the elements $k\in K$.

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  • $\begingroup$ I must have put the following definition: each $A_{\lambda}$ is a geometric regular local ring with the unique maximal ideal ${\frak m}_{A_{\lambda}}$. Also the condition $(\sharp)$ must be revised as follows: $(f_{1}(k),\ldots,f_n(k)) \subset {\frak m}_{A_{\lambda_0}}$. That is each specialisation $f_i(k)$ with $X = k \in K$ belongs to the maximal ideal ${\frak m}_{A_{\lambda_0}}$. $\endgroup$ Sep 29 '17 at 6:07

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