I have originally posted this on math.SE and been suggested to post this here. I'm merely an undergraduate student and it is the first time for me to ask questions here. I'm sincerely sorry if these questions are too trivial or silly for this site.

I've been working on understanding the proof of Fermat's last theorem and now focusing on the patching technique for modularity lifting. I found that the patching technique described in the paper Modularity lifting beyond the Taylor–Wiles method, i.e. Proposition 2.3 is helpful. However, I have come up with some difficulties on understanding certain parts of the proof.

My confusions mainly come from the following paragraph (At the bottom of Page 13 of the linked paper above): enter image description here

Question 1: Why the image of $S_{\infty}$ in $\mathrm{End}_{\mathcal{O}}(X_{\infty})$ is contained in the image of $R_{\infty}$? Though the authors explained in the parenthese, I still found that hard to understand, and hoping for more details.

Question 2: Why does the inclusion of image (as in the Question 1) implies that $X_{\infty}$ is a finite $R_{\infty}$-module?

Question 3: Why the fact that "$S_{\infty}$ is formally smooth over $\mathcal{O}$" implies that we can choose a homomorphism $\iota: S_{\infty} \rightarrow R_{\infty}$ in $\mathsf{C}_{\mathcal{O}}$? Where does such a homomorphism come from. I have gone through Chapter 11 (Formal Smoothness) of Matsumura's book Commutative Algebra, yet haven't found similar arguments.

Sorry for such a post with so detailed questions and the lack of explanations in the post (although the contexts are clear in the linked paper). Thank you all for answering and commenting!

Any more detailed references are welcome as well!

  • 2
    $\begingroup$ The finiteness statement presumably follows from $X_{\infty}$ being a finite $S_{\infty}$ module. For the last question, formal smothness implies that $S_{\infty}$ is a formal power series ring over $\mathcal{O}$ so you can define a map by defining it on generators (and using the inclusion). $\endgroup$
    – naf
    Apr 12, 2021 at 4:22
  • 1
    $\begingroup$ For question 1, $X_\infty$ is an inverse limit of some collection $X_n$ of modules which all have this "S action factors through R action" property, and the bracketed sentence is just saying that nothing funny happens as you pass to the limit over $n$. $\endgroup$ Apr 12, 2021 at 13:12
  • $\begingroup$ @naf Thank you for your comment! For the last question, I'm wondering why the group algebra $S_{\infty}:= \mathcal{O}[(\mathbb{Z}_p)^q]$ is formally smooth over $\mathcal{O}$? I guess that there might be an isomorphism $\mathcal{O}[(\mathbb{Z}_p)^q] = \mathcal{O}[[x_1, \ldots , x_q]]$ directly but unfortunately failed to "prove" it. Or could you provide some reference on why formal smoothness implies $S_{\infty}$ is a power series ring over $\mathcal{O}$. (Since may be this is the key for the isomorphism?) $\endgroup$
    – Hetong Xu
    Apr 13, 2021 at 4:26
  • $\begingroup$ @naf And for the finiteness statement, $X_{\infty}$ is a finite $S_{\infty}$-module (it seems that the finiteness is preserved when taking limit?), and do you mean that by this finiteness and the inclusion of image (image of $S_{\infty}$ in $\mathrm{End}_{\mathcal{O}}(X_{\infty})$ is contained in the image of $R_{\infty}$), we can deduce that $X_{\infty}$ is a finite $R_{\infty}$-module? (So sorry for these trivial questions!) $\endgroup$
    – Hetong Xu
    Apr 13, 2021 at 4:30
  • $\begingroup$ @DavidLoeffler Thank you! It's great to interpret that inclusion in this way and it's much more clearer to me now! It seems that the arguments in the bracket are using the fact that "hom-functor preserves limits" step by step passing to the limit. Thank you so much! $\endgroup$
    – Hetong Xu
    Apr 13, 2021 at 4:33


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