Modular functions and integral closure of a valuation ring

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 $$\mathbf C(j)\subset \mathbf C \cdot F_N$$.

• 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. – Johannes Hahn Oct 17 '19 at 12:27
• @JohannesHahn, thanks, I corrected the question. – Shimrod Oct 17 '19 at 13:51
• 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. – François Brunault Oct 17 '19 at 16:52
• Why the downvote? I think this is a reasonable Research Question. – Shimrod Oct 18 '19 at 5:20