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Let $(R,m_R)$, $(S,m_S)$ be two local rings. By definition, a local homomorphism of local rings is a ring homomorphism $f: R \to S$ such that the ideal generated by $f(m_R)$ in $S$ is contained in $m_S$: $f(m_R) \subseteq m_S$. According to Lemma 10.18.3, the condition $f(m_R) \subseteq m_S$ is equivalent to other conditions.

Question: What if $f(m_R)=m_S$? Does this imply something interesting, some stronger version of Lemma 10.18.3?

This question may be relevant, although I am not assuming thet $S$ is a finitely generated $R$-module.

Thank you very much!

user237522
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    Perhaps you can formulate equivalent conditions to $f(\mathfrak{m}_R) = \mathfrak{m}_S,$ but those conditions probably wouldn't look like the ones in lemma 10.18.3. Additionally, observe that such a condition includes both local ring homomorphisms of the form $(R,\mathfrak{m}_R)\to (R/\mathfrak{m}_R,0)$ as well as field extensions -- these types of morphisms seem rather different, and it's unclear how much could be said so generally about a class of morphisms containing both types. – Stahl Apr 08 '21 at 01:41
  • Thank you for your comment. – user237522 Apr 08 '21 at 01:44
  • An easy observation: Let $f : (R,\mathfrak{m}R)\to (S,\mathfrak{m}_S)$ be a morphism of local rings, and let $\bar{f}$ be the induced map on residue fields. Then $f(\mathfrak{m}_R) = \mathfrak{m}_S$ is equivalent to the following condition: if $s + \mathfrak{m}_S\in\operatorname{Im}(\bar{f}),$ then $s\in\operatorname{Im}(f).$ This is because by the snake lemma, the sequence of $R$-modules $$0\to\operatorname{Coker}\left(\left.f\right|{\mathfrak{m}_R}\right)\to \operatorname{Coker}(f)\to\operatorname{Coker}\left(\bar{f}\right)\to 0$$ is exact. – Stahl Apr 08 '21 at 02:26
  • [continued] Thus, $\operatorname{Coker}\left(\left.f\right|{\mathfrak{m}_R}\right)\to \operatorname{Coker}(f)$ is always injective. Since $\operatorname{Coker}\left(\left.f\right|{\mathfrak{m}R}\right)=0$ if and only if $\left.f\right|{\mathfrak{m}_R}$ is surjective, we have $\operatorname{Coker}(f)\to\operatorname{Coker}\left(\bar{f}\right)$ is an isomorphism if and only if $f(\mathfrak{m}_R) = \mathfrak{m}_S.$ – Stahl Apr 08 '21 at 02:28
  • I suppose the condition above is somewhat of a complement to condition (4) in lemma 10.18.3. Condition (4) is about liftability of units from $S^\times$ to $R^\times,$ and this is about liftability of images from $S/\mathfrak{m}_S$ to $S.$ – Stahl Apr 08 '21 at 02:35
  • Very nice, thank you! Could you make those three comments into an answer? It is more or less what I was looking for (indeed, I wondered what condition (4) in Lemma 10.18.3 should be). – user237522 Apr 08 '21 at 06:19
  • (Of course, you can leave those comments as they are, without making them an answer). – user237522 Apr 08 '21 at 06:28
  • I'd be happy to! – Stahl Apr 08 '21 at 07:04

1 Answers1

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Proposition: Let $R$ and $S$ are local rings with maximal ideals $\mathfrak{m}_R$ and $\mathfrak{m}_S$ and residue fields $k_R = R/\mathfrak{m}_R$ and $k_S = S/\mathfrak{m}_S$, respectively, let $\varphi : R\to S$ be a local ring homomorphism, and let $\overline{\varphi} : k_R\to k_S$ be the induced map on residue fields. Then the following are equivalent:

  1. $\varphi$ is a local ring map such that $\varphi(\mathfrak{m}_R) = \mathfrak{m}_S$, and
  2. If $s + \mathfrak{m}_S\in k_S$ is in the image of $\overline{\varphi},$ then $s$ is in the image of $\varphi.$

Proof: For any morphism $\varphi : R\to S$ of local rings, the snake lemma applied to $$ \require{AMScd} \begin{CD} 0 @>>> \mathfrak{m}_R @>>> R @>>> k_R @>>> 0\\ @VVV @V\left.\varphi\right|_{\mathfrak{m}_R}VV @V\varphi VV @V\overline{\varphi}VV @VVV\\ 0 @>>> \mathfrak{m}_S @>>> S @>>> k_S @>>> 0\\ \end{CD} $$ implies that we have a short exact sequence $$ 0\to\operatorname{Coker}\left.\varphi\right|_{\mathfrak{m}_R}\to\operatorname{Coker}\varphi\to\operatorname{Coker}\overline{\varphi}\to 0. $$ The map $\operatorname{Coker}\left.\varphi\right|_{\mathfrak{m}_R}\to\operatorname{Coker}\varphi$ is always injective, so that $\operatorname{Coker}\varphi\to\operatorname{Coker}\overline{\varphi}$ is an isomorphism if and only if $\operatorname{Coker}\left.\varphi\right|_{\mathfrak{m}_R} = 0.$ It follows that $\operatorname{Coker}\varphi\to\operatorname{Coker}\overline{\varphi}$ is an isomorphism if and only if $\left.\varphi\right|_{\mathfrak{m}_R}$ is surjective; i.e., if $\varphi(\mathfrak{m}_R) = \mathfrak{m}_S.$

Now, $\operatorname{Coker}\varphi\to\operatorname{Coker}\overline{\varphi}$ being an isomorphism means that the class of $s + \mathfrak{m}_S$ in $\operatorname{Coker}\overline{\varphi}$ is $0$ if and only if the class of $s$ in $\operatorname{Coker}\varphi$ is. But the class of $s + \mathfrak{m}_S$ being $0$ is equivalent to $s + \mathfrak{m}_S$ being in the image of $\overline{\varphi},$ and similarly, the class of $s$ being $0$ is equivalent to $s$ being in the image of $\varphi.$

Observe that $s\in S$ being in the image of $\varphi$ always implies that $s + \mathfrak{m}_S$ is in the image of $\overline{\varphi}.$ Thus, it follows that $\operatorname{Coker}\varphi\to\operatorname{Coker}\overline{\varphi}$ being an isomorphism is equivalent to the condition that if $s + \mathfrak{m}_S\in\operatorname{Im}(\overline{\varphi}),$ then $s\in\operatorname{Im}(\varphi).$

Putting all of the above together, the proposition follows. $\square$

Loosely, the second condition of the proposition is similar to the fourth condition of lemma 10.18.3. The fourth condition in the lemma implies that units in $S$ lift to units in $R,$ and the second condition of the proposition implies that elements in the image of $\overline{\varphi}$ lift to elements in the image of $\varphi.$ However, observe that the liftability condition in the lemma is "vertical" in the sense of the commutative diagram in the proof of the proposition above, while the liftability condition in the proposition is "horizontal."

Stahl
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  • Thank you very much! – user237522 Apr 08 '21 at 08:14
  • Please, is your above proposition a known result in the literature? – user237522 Apr 27 '21 at 17:00
  • @user237522 I'm not aware of any place where this appears in the literature. To answer your now deleted comment, it seems your application of this result was fine - I would only note that a priori you only know $R\to S$ is a surjection if the extension of residue fields is an isomorphism (although in your case, it would have needed to be an isomorphism since the map of local rings was an injection). – Stahl Apr 27 '21 at 17:28
  • @user237522 In general, if you have a morphism of local rings $f : (R,\mathfrak{m})\to (S,\mathfrak{n})$ such that $f(\mathfrak{m}) = \mathfrak{n}$ and the induced map $R/\mathfrak{m}\to S/\mathfrak{n}$ is an isomorphism, all you can conclude is that $f$ is surjective. You know that every $s + \mathfrak{n}$ is in the image of the induced map, so that every $s$ is in the image of $f,$ but you do not necessarily know that $s$ has a unique preimage. – Stahl Apr 27 '21 at 18:09
  • @user237522 As a counterexample, consider the surjection $(k[x]/(x^2),(x))\to (k,(0))$ for $k$ a field given by $p(x) + (x^2)\mapsto p(0)$. The induced map on residue fields is the identity $k\to k$, but the map on local rings is only a surjection, not an isomorphism. – Stahl Apr 27 '21 at 18:11
  • Thank you very much for the comments! (Yes, I understood the above claim, based on your answer of course). Your last example is very helpful! However, I am confused about something else, which I will try to ask. – user237522 Apr 27 '21 at 18:12
  • Please, to make sure, concerning https://math.stackexchange.com/questions/1400834/finite-ring-extension-of-local-rings?rq=1 If we make two changes in that question: (1) omit the assumption that $S$ is a finitely generated $R$-module (2) assume that $MS=N$ and $R/M \subseteq S/N$ is an isomorphism (therefore, $M$ satisfies the conditions of $I$), then applying your above claim yields $R=S$. This sounds really nice, since we were able to omit the finiteness condition (instead, added a condition on the maximal ideals, $MS=N$, and a condition on the induced residue fields extension, surjectivity). – user237522 Apr 27 '21 at 18:48
  • But truly I am confused about something else, which I will write as a new question and attach a link here. Also, I am not sure, but maybe if I would be able to write a paper applying your proposition (I am working on in recently), would you like to be a co-author or is it enough to quote your proposition from here and thank you in the acknowledgements? Perhaps this depends on what the paper is about, and what is the role of your proposition in the paper? If it is a minor role, then mentioning that it is your result+thanking you is enough, but if it plays a more serious role, then co-authoring? – user237522 Apr 27 '21 at 20:31
  • https://math.stackexchange.com/questions/4119061/properties-of-kxx-1-langle-xx-1-rangle-subseteq-kx-langle-x-ran – user237522 Apr 27 '21 at 23:38
  • I think there was a misunderstanding, and for you $f(m_R)=m_S$ meant simply the image of $m_R$ under $f$, while I meant to consider a much larger set, namely, the ideal in $S$ generated by $f(m_R)$. (If so, then unfortunately I cannot apply your proposition in my paper, since it requires a much stronger condition then I have). – user237522 Apr 28 '21 at 00:26