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I know the 1965 version of Lang's book is not the latest but it is the one available at my library so I'm using it for now. In the section on rings, page 73 exercise 2 he asks me to prove "If $A$ is a local ring and $S$ is a multiplicative subset, prove that $S^{-1}A$ is a local ring". I tried for quite a while and couldn't manage to prove it. I searched here on Math Stack Exchange and found this but it talks about localizations at prime ideals and doesn't assume that $A$ is local so it is not quite the same. I looked in the newest version of Lang's book and the exercise isn't there anymore.

Is the statement true? If it isn't can you show me a counter-example? Thanks in advance

Bowei Tang
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Arthur Farias Zaneti
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  • My intuition would be that this is not true. If you can construct a local ring $A$ with two incomparable prime ideals, i.e. $I_1, I_2\subset A$ prime neither contained in the other, then we can pick $S=A\setminus (I_1\cup I_2)$. In this case, the ideals $S^{-1} I_1, S^{-1} I_2$ would be maximal ideals of $S^{-1} A$ and hence it would not be local. – Severin Schraven Jul 11 '25 at 00:19
  • Incomparable is not quite the right condition. We would need something slightly stronger, namely that $I_1\cup I_2$ is not prime. – Severin Schraven Jul 11 '25 at 00:29
  • The maximal ideal of such a local ring cannot be principal (https://math.stackexchange.com/questions/704613/local-ring-with-principal-maximal-ideal). – Severin Schraven Jul 11 '25 at 00:38
  • The ring constructed here https://mathoverflow.net/questions/426917/an-example-of-a-local-integral-domain-with-special-spectrum might work. – Severin Schraven Jul 11 '25 at 02:03

1 Answers1

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In this post https://mathoverflow.net/questions/426917/an-example-of-a-local-integral-domain-with-special-spectrum a local integral domain $(D, m)$ is constructed such that the spectrum looks like $$ \mathrm{Spec}(D)=\{(0), m\}\cup \{ P_i\}_i$$ with $$ \bigcup_i P_i \neq m$$ and for $i\neq j$ we have $$P_i\not\subset P_j, P_j\not\subset P_i.$$ Furthermore, $\{ P_i\}_i$ contains infinitely many primes (two would be enough for us).

Now pick an element $x\in m\setminus \bigcup_i P_i$ and define $S=\{ 1, x, x^2, \dots\}$ to be the multiplicative set generated by $x$. Then $S^{-1} P_i$ are all maximal in $A=S^{-1} D$ (they are prime as they don't intersect $S$ by construction and they are maximal as the map on the spectrum is inclusion-preserving and $S^{-1} m=A$ is not prime). In particular, $A=S^{-1} D$ is no longer local.

Easier version: Pick $R=\mathbb{C}[x,y]_{(x,y)}$ (i.e. $\mathbb{C}[x,y]$ localized at the multiplicative set $\mathbb{C}[x,y]\setminus (x,y)$). Then $R$ is a local ring and $$P_1:=(x)_{(x,y)}, P_2:=(y)_{(x,y)}$$ are two distinct prime ideals in $R$. Pick $$S=R\setminus (P_1\cup P_2)$$ as our multiplicative set. The idea is that we want to force $S^{-1}P_1, S^{-1}P_2$ to be maximal ideals in the localization. One way of doing so is to make everything outside $P_1, P_2$ invertible (and so kill those larger prime ideals).

Furthermore, we have using the strategy in Localizing a localization at a prime ideal $$ S^{-1} R \cong T^{-1} \mathbb{C}[x,y] $$ with $T=\mathbb{C}[x,y]\setminus ((x)\cup (y))$. Now, the prime ideals in $T^{-1} \mathbb{C}[x,y] $ are in correspondence with the prime ideals in $\mathbb{C}[x,y]$ not intersecting $T$ (see How to show that there is a bijective correspondence between two sets of prime ideals), i.e. which are contained in $(x)\cup (y)$. Hence, the maximal ideals in $T^{-1} \mathbb{C}[x,y] $ are $T^{-1} (x), T^{-1}(y)$ and therefore it is not a local ring.

Severin Schraven
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  • Nice answer, but do you need such a complicated ring? Would your construction with $R = k[X, Y]{\langle X, Y \rangle}, I{1} = \langle X \rangle, I_{2} = \langle Y \rangle$ not work? – Alex Wertheim Jul 11 '25 at 02:23
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    @AlexWertheim Sorry, I misunderstood your comment. You mean the construction I indicated in the comments, right? Indeed, I think that would work too. Tomorrow I'll rewrite my answer. – Severin Schraven Jul 11 '25 at 02:37
  • Yes, exactly - no rush, and nice update! – Alex Wertheim Jul 11 '25 at 03:20
  • Thank you for your answer (I mainly looked at the "easier version", just skimmed the first counter-example). Some things you used I think are quite standard but I've never seen them, for instance the bijection between prime ideals that don't intersect a multiplicative set and the prime ideals in the localization by such set. – Arthur Farias Zaneti Jul 11 '25 at 13:56
  • @ArthurFariasZaneti Sorry, I assumed that you are familiar with this. I can try and see whether I find the corresponding statement on this site. – Severin Schraven Jul 11 '25 at 16:53
  • @SeverinSchraven No problem, it took some effort but I managed to prove it. If the result is fundamental enough for you to assume that I'm familiar with it already I guess it is good I learned it sooner rather than latter – Arthur Farias Zaneti Jul 12 '25 at 00:25
  • @ArthurFariasZaneti Indeed, I feel knowing how the spectrum behaves under localization and under taking a quotient are useful things to know. – Severin Schraven Jul 12 '25 at 11:21