Let $j$ be the modular invariant and let $\tau$ be a point in the upper half-plane. Let $\mathfrak o_\tau$ consist of all $f\in \mathbf Q(j)$ which are defined at $\tau$. Let $\mathfrak O_\tau$ consist of all $f\in F_N$ defined at $\tau$. Here $F_N$ is a the field of modlar functions of level $N$ rational over $\mathbf Q(\zeta_N)$, see this question for a description. Is it true that $\mathfrak O_\tau$ is the integral closure of $\mathfrak o_\tau$ in $F_N$? Alternatively, we may ask the question for the extension $\mathbf C(j)\subset \mathbf C \cdot F_N$.
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1$\begingroup$ I think maybe you wanted to ask a slightly different question, because as it is right now, the answer is "no" for trivial reasions: $\mathfrak{o}_\tau$ is uncountable, because it contains $\mathbb{C}$ and $\mathfrak{O}_\tau$ is countable, because it is contained in $F_N$ (which is countable per the description in the linked question). Thus $\mathfrak{O}_\tau$ is not even an extension of $\mathfrak{o}_\tau$, let alone the integral closure. $\endgroup$– Johannes HahnCommented Oct 17, 2019 at 12:27
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$\begingroup$ @JohannesHahn, thanks, I corrected the question. $\endgroup$– ShimrodCommented Oct 17, 2019 at 13:51
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2$\begingroup$ I think the integral closure will be the subring of modular functions of level N which are regular at every point of the preimage of \tau (so the whole orbit of \tau under the SL_2(Z/NZ) action). It is a semi-local ring. $\endgroup$– François BrunaultCommented Oct 17, 2019 at 16:52
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$\begingroup$ Why the downvote? I think this is a reasonable Research Question. $\endgroup$– ShimrodCommented Oct 18, 2019 at 5:20
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