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This question says the following: Let $R$ and $S$ be local rings with the maximal ideals $M$ and $N$, respectively. Assume that $R\subset S$ and that $S$ is a finitely generated $R$-module. If there exists a proper ideal $I$ of $R$ such that $I=IS \cap R$ and the canonical image of $R/I$ in $S/IS$ equals $S/IS$, then prove that $R=S$.

Now, I am confused about the above result, since it seems that the following is a counterexample: $R=\mathbb{C}[x(x-1)]_{\langle x(x-1) \rangle}$, $S=\mathbb{C}[x]_{\langle x \rangle}$, $I=M=Rx(x-1)$. Notice that $IS=N=Sx$ and then $I=IS \cap R$. If I am not wrong, since $R \subseteq S$ is algebraic (actually, $S$ is finitely generared as an $R$-module by $1,x$), we obtain that $R/I \subseteq S/IS$ is algebraic. Moreover, this extension is actually $R/M \subseteq S/N$ and it is a field extension of degree two (it is enough for us that it is algebraic field extension). Also, $R/M = \mathbb{C}$ and $S/N = \mathbb{C}$. Since there are no proper algebraic extensions of $\mathbb{C}$, we get that $R/M = S/N$. Therefore, all the conditions of the theorem are satisfied, hence the conclusion $R=S$, but here clearly $R \neq S$.

Question: Where is my error?

Any hints and comments are welcome! Thank you.

user237522
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  • Just one thing I noticed: I don't see how $S$ is generated by $1,x$ as an $R$-module. In fact, it seems like $S$ isn't even finitely generated as an $R$-algebra. In $S$ we need infinitely many denominators like $x-a$ with $a\neq 0,1$, but how do we get these from an $R$-action? – Dave Apr 22 '21 at 00:31
  • @Dave, thank you very much for your comment! I had in mind the non-local case $\mathbb{C}[x(x-1)] \subseteq \mathbb{C}[x]$, which is free with basis ${1,x}$ and I thought that a set of generators of the non-local case is also a set of generators for the local case. Apparantly, I was wrong... since I did not take into account the denominators $x-a$, $a \neq 0$ (elements outside the maximal ideal). – user237522 Apr 22 '21 at 00:39
  • I'm not sure what the $R$-algebra structure would be. It can't be the usual one inherited from $\mathbb C[x(x-1)]\to\mathbb C[x]_{\langle x\rangle}$, because when we localize the domain at $x(x-1)$ we make this element invertible, but this is not invertible in $S$. – Dave Apr 22 '21 at 01:16
  • $R$ is not a local ring, this is not a counterexample. – Hank Scorpio Apr 22 '21 at 04:56
  • @HankScorpio, thank you. I meant in the comments that $R$ is still as I have taken in the question, the localization of $\mathbb{C}[x(x-1)]$ at the maximal ideal $\langle x(x-1) \rangle$. It is just that I missed the point the from moving from 'usual' polynomial rings to their localizations, we should not forget the denominators. – user237522 Apr 22 '21 at 05:41
  • @Dave, thank you again. I am curious what can be said about the structure of the $R$-algebra $S$. Anyway, I am happy that at least you corrected my error. (Isn't $x(x-1) \in M$ so it is not invertible in $R$, and in $S$ it is an associate of $x$, since $x-1$ is invertible in $S$). – user237522 Apr 22 '21 at 05:49
  • That's not what you've written: you wrote $\Bbb C[x(x-1)]{x(x-1)}$, not $\Bbb C[x(x-1)]{(x(x-1))}$. You should fix that! – Hank Scorpio Apr 22 '21 at 06:10
  • @HankScorpio, oh, sorry, now I see. It is really a misprint. I will fix it now. Thank you very much! – user237522 Apr 22 '21 at 06:19
  • And now I see what I did not understand in the discussion with Dave: He thought like you thought (of course, it was my mistake forgetting the brackets) that the multiplicatively closed set in $R$ is ${1,x(x-1),x^2(x-1)^2,\ldots}$, while I meant that it would be $R-\langle x(x-1) \rangle$. (In my second comment above it should be $a \in \mathbb{C}-{0}$, as in my first comment). – user237522 Apr 22 '21 at 06:32
  • If I am not wrong, actually, the generators of $S$ over $R$ are: ${ \frac{1}{x-a} }_{a \in \mathbb{C}-{0}}$ and $x$. – user237522 Apr 22 '21 at 06:55
  • In second thought, I think that $S$ is finitely generated as an $R$-algebra, by the following two generators: $\frac{1}{x-1}$ and $x$. (But it is not finitely generated as an $R$-module, as was requiered in the theorem, so the theorem is not applicable, and there is no contradiction). – user237522 Apr 22 '21 at 08:02
  • @Dave, please is $\mathbb{C}[x(x-1)]{\langle x(x-1) \rangle} \subset \mathbb{C}[x]{\langle x \rangle}$ unramified? https://stacks.math.columbia.edu/tag/024L The first condition is satisfied, but I am not sure about the other two. Thank you. – user237522 Apr 24 '21 at 23:11
  • Concerning condition 3: $\mathbb{C}[x]$ is finitely generated $\mathbb{C}[x(x-1)]$-algebra, with one generator $x$. But $\mathbb{C}[x]$ is not finitely generated $\mathbb{C}[x(x-1)]_{\langle x(x-1) \rangle}$-algebra (again because of the denominators). – user237522 Apr 24 '21 at 23:27

1 Answers1

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This is based on @user237522's comment.

The extension $R \subset S$ is not finite. The extension $A:=C[x^2-x] \subset B:=C[x]$ is finite (of degree two). In $B$, $(x)$ and $(x-1)$ are the maximal ideals of $B$ containing $x^2-x$. With $W = A- (x^2-x)A$, we have the induced inclusion of locaizations $W^{-1}A \subset W^{-1}B$. By definition $W^{-1}A = R$. However, since $W^{-1}B$ has two maximal ideals (extensions of $(x)$ and $(x-1)$), it is not equal to $S$. Indeed, $S$ is the localization of $W^{-1}B$ at the extension of $(x)$. Note that $W^{-1}B \to S$ is not finite (this can be seen as there is no prime ideal in $S$ lying over the maximal ideal $(x-1)W^{-1}B$). Thus, $R \subset S$ is not finite.

Youngsu
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  • Please, is my extension unramified? https://stacks.math.columbia.edu/tag/024L There are three conditions that should be satisfied. The first one is satisfied $MS=N$. However, I am not sure about the other two conditions. Sould I ask this as a separate question? Thank you. – user237522 Apr 24 '21 at 23:09
  • I think that condition 2 is not satisfied. – user237522 Apr 25 '21 at 00:43
  • Otherwise, if the residue field extension, denote it by $k_1 \subseteq k_2$, was finite dimensional, then since each $k_i$ is isomorphic to $\mathbb{C}$ and $\mathbb{C}$ does not have finite field extensions, we would get $k_1=k_2$. But then, according to the answer to the following question, we would obtain that $R=S$. But clearly, those two local rings are different ($x$ does not belong to $R$). https://math.stackexchange.com/questions/4092716/definition-of-homomorphism-of-local-rings/4093922#4093922 – user237522 Apr 25 '21 at 00:50
  • What happens if we replace the base field $\mathbb{C}$ by $\mathbb{R}$ or $\mathbb{Q}$? – user237522 Apr 25 '21 at 00:58
  • @user237522: I believe once you have that $B \cong A[Y]/(Y^2-Y - (x^2-x)$, you use the module of differentials of $B$ over $A$ to answer your questions. Let me know if you need more details. – Youngsu Apr 25 '21 at 05:00
  • Thank you for the hints. It would be nice to have more details. Sould I post it as a new question that you can answer? – user237522 Apr 25 '21 at 08:28
  • @user237522 Let me try. From the presentation you can compute the relative module of differentials $\Omega_{B/A}$. As this module localizes, we have $\Omega_{W^{-1}B/R} = 0$ (it boils down to checking that $2Y-1$ is a unit). I believe that Stacks Project has a characterization of an unramified ring map in terms of modules of differentials. You can also check unramifiedness directly. You already showed condition 2 in your post, and Condition 3 follows from the first part of this comment as $S$ is a localization of $W^{-1}B$. – Youngsu Apr 25 '21 at 23:49
  • Thank you for your explanation. I will think about it and if something will still not be clear, I will post it as a separate question (and let you know here). Thank you for your help! – user237522 Apr 25 '21 at 23:56