# What is the theory of local rings and local ring homomorphisms?

It is well-known that the category of local rings and ring homomorphisms admits an axiomatisation in coherent logic. Explicitly, it is the coherent theory over the signature $0, 1, -, +, \times$ with the usual axioms for rings, plus the axioms $$0 = 1 \vdash \bot$$ $$\top \vdash (\exists b . \; a \times b = 1) \lor (\exists b . \; (1 - a) \times b = 1)$$ See, for example, [Sheaves in Geometry and Logic, Ch. VIII, §6]. Unfortunately, because homomorphisms are only required to commute with the various things in the signature, the homomorphisms here are just ring homomorphisms and need not be local. It appears to me that the neatest way to fix this is to introduce a unary relation symbol $(\quad) \in \mathfrak{m}$, with the intention that $\mathfrak{m}$ is interpreted as the unique maximal ideal of the local ring. Then, by the usual rules for homomorphisms of models, a homomorphism $R \to R'$ must map elements of $\mathfrak{m}$ to elements of $\mathfrak{m}'$. But is there a way to axiomatise the theory so that

1. we get a coherent, or at least geometric theory, and

2. the category of models in $\textbf{Set}$ is indeed the category of local rings and local ring homomorphisms, and

3. the structure sheaf homomorphism $f^\ast \mathscr{O}_{Y} \to \mathscr{O}_{X}$ of morphism of locally ringed spaces $X \to Y$ is a homomorphism in the category of models for this theory?

Ideally, we'd like to define $\mathfrak{m}$ to be the subsheaf of nowhere invertible sections defined by $$\{ s \in \mathscr{O} : \nexists t . \; s \times t = 1 \}$$ but unfortunately $\nexists t . \; s \times t = 1$ is not a geometric formula. (The formula $\forall t . \; s \times t \ne 1$ is equivalent to the previous one but has the same defect.) We can salvage one half of the biimplication as the geometric sequent $$(a \in \mathfrak{m}) \land (\exists b . \; a \times b = 1) \vdash \bot$$ which merely expresses the requirement that "$a$ is not in $\mathfrak{m}$ if $a$ is invertible", but we also need to express the requirement that "$a$ is in $\mathfrak{m}$ if $a$ is not invertible". One possibility is the following: $$\top \vdash (\exists b . \; a \times b = 1) \lor (a \in \mathfrak{m})$$ These two axioms appear to give the correct characterisation of $\mathfrak{m}$ in intuitionistic first order logic: it is easy to derive from these axioms that $$a \in \mathfrak{m} \dashv \vdash \nexists b . \; a \times b = 1$$ so the interpretation of $\mathfrak{m}$ is completely determined by the axioms, at least in a topos.

But does every local ring object (in the sense of the first paragraph) admit an $\mathfrak{m}$ satisfying these axioms? The answer appears to be no, for the reason that these axioms assert that a every section of a sheaf of a local ring admits an open cover of the space by open sets on which the restriction is either invertible or nowhere invertible – and this is certainly not true in the contexts of interest. Can this idea be rescued with a more clever approach?

• Isn't this question equivalent to asking whether the category of local rings and local ring homomorphisms is coherent, as in ncatlab.org/nlab/show/coherent+category? – Andrej Bauer Feb 28 '12 at 21:57
• Is it? I can't say I know enough about categorical algebra to see why this is plausible/implausible. – Zhen Lin Feb 29 '12 at 0:17
• @Andrej: No, that doesn't seem to be the case. The theory of abelian groups is algebraic, hence coherent, but the category of abelian groups is not coherent (since coherent categories have a strict initial object, while $\textbf{Ab}$ has a zero object). – Zhen Lin Mar 2 '12 at 7:46
• You may want to read about the concept of a "geometry" introduced in Lurie's DAG V. It was invented precisely to get around this problem. – David Carchedi May 18 '13 at 1:16
• David is right, see here ncatlab.org/nlab/show/… (and notice that the condition imposed there is reaLLy simple and has as such nothing much to do with the oo-category theory in which it is formulated). – Urs Schreiber Jun 11 '13 at 22:59

If you know the objects of a geometric theory then you also know its morphisms because $\mathbb{T}(\mathbf 2,\mathcal E)\simeq [\mathbf 2,\mathbb{T}(\mathcal E)]$. This is Lemma 4.2.3 in Chapter B of Sketches of an Elephant. Hence, it is impossible for the two theories to have the same objects, but different morphisms as you request.

• More explicitly, the local ring corresponding to $\operatorname{Spec} A$ for a discrete valuation ring $A$ corresponds to the ring homomorphism $A \to \operatorname{Frac} A$, which is not a local ring homomorphism. Hence not every local ring in $[\mathbb{2}, \mathbf{Set}]$ comes from a local ring homomorphism in $\mathbf{Set}$. – Zhen Lin Jun 11 '13 at 17:19
• This answer is very misleading. It relies on $\mathbb{T}$ being fixed, and the whole point is that we are looking for a distinct $\mathbb{T}$. OP asked for a topos whose $\mathbf{Set}$-models are local rings and local ring homs. Your answer doesn't exclude this. The theory of decidable objects has a distinct classifying topos from the theory of objects; the models of both of these theories in $\mathbf{Set}$ are precisely sets, but the maps of the former are monomorphisms, whereas the maps of the latter are arbitrary functions. – Morgan Rogers Feb 19 at 13:10

The following should give some insight to this question, which was more or less the inspiring example of Diers' work on locally multipresentable categories. This involves the "multi" versions of usual universal constructions involved in Gabriel-Ulmer duality.

Locally (finitely) multipresentable categories are (finitely) accessible categories with multi-colimits - or equivalently, (finitely) accessible categories with connected limits commuting with (finitely) filtered colimits.

It is known that any locally finitely multipresentable category can be axiomatized by a small finite-limit/coproduct sketch, equivalently by a disjunctive geometric theory - which often happens to be finitely disjunctive, as in the example below.

In particular, Diers proposed in his thesis a process to construct locally finitely multipresentable categories as follows. If you start with a locally (finitely) multipresentable category $$\mathcal{A}$$ - hence in particular with a locally (finitely) presentable category - and a choice of a small family $$\Gamma$$ of cones of finitely presented maps, then you can take the category $$\mathcal{A}_\Gamma$$ of objects that are injective relatively to cones in $$\Gamma$$, and morphisms that are right orthogonal to each arrows involved in the cones of $$\Gamma$$ (in general arrows in the cones of $$\Gamma$$ are chosen in a left class in an orthogonal factorization system, so the morphisms in $$\mathcal{A}_\Gamma$$ are in particular in the corresponding right class).

Then $$\mathcal{A}_\Gamma$$ can be shown to be locally multipresentable itself, and we have a (finitely) accessible functor $$\mathcal{A}_\Gamma \hookrightarrow \mathcal{A}$$ which is a right multi-adjoint and only has to be relatively full and faithful, but not necessarily full. Multireflexiveness involves a kind of small object argument returning a factorization of morphisms with local codomain.

In the example of local rings, take as $$\Gamma$$ as consisting of the single cone made of the following two finitely presented localizations of the free ring on one generator as in Zariski topology

$$\mathbb{Z}[X] \twoheadrightarrow \mathbb{Z}[X,Y]/(XY-1)$$ and $$\mathbb{Z}[X] \twoheadrightarrow \mathbb{Z}[X,Y]/((X-1)Y-1)$$

Then as expected the objects of $$CRing_\Gamma$$ are the local rings, but moreover conservativity of a ring homomorphisms just has to be tested relatively to either one of those localizations. Hence $$CRing_\Gamma$$ is the category with the good choice of morphisms. In this case relative fullness comes from the fact that conservative morphisms are a right class in the (localization, conservative) factorization system in $$CRing$$ and hence have right cancellability. And the result above says that $$CRing_\Gamma = LocRing^{Cons}$$ is a locally finitely multipresentable category. Hence there must be a small finite-limit/coproduct sketch axiomatizing it, and hence a disjunctive theory in some convenient signature (though I must admit I don't know which one).

However I don't think that the finite limit part of this sketch is the same as the finite limit sketch of commutative rings, because there is a result by Adamek-Rosicky suggesting that the category of models of a finite limit-coproduct sketch should be a full multireflective subcategory of the category of models of the underlying finite limit sketch.

There is also an interesting paper from Johnstone about disjunctive theories and their link with Diers multipresentability that may be of interest relatively to this question.

• Benvenue su MO, Axel. Il était temps. – Ivan Di Liberti Jan 21 at 17:59

You can define $\mathfrak{m}_R := \{x ~|~ \forall y : 1 - xy \in R^*\}$. Then a homomorphism $R \to S$ of local rings is a map which is compatible with the ring structure and maps $\mathfrak{m}_R$ maps into $\mathfrak{m}_S$. However, this is not equivalent to the usual condition that images of non-units are non-units: In general it is not true that $R = R^* \cup \mathfrak{m}_R$. This is proven by Thierry Coquand in a remark about the theory of local rings. The counterexample is as follows: Consider the Zariski topos $C$ over $\mathbb{Z}$ and the structure sheaf $\mathcal{O}$ of $\mathrm{Spec}(\mathbb{Z})$. Then $\mathcal{O}$ is a local ring in $C$, one verifies that $\mathfrak{m}=\{0\}$ on global sections, so that in particular $2 \in \mathcal{O}^* \vee 2 \in \mathfrak{m}$ is not satisfied.

• Martin, this is a useful comment, but not an answer to his question: is it possible to axiomatize the theory of local rings as a coherent theory or at least a geometric theory in the technical sense of the word. – Zoran Skoda Feb 28 '12 at 10:23
• You are right. I just wanted to mention that with this axiomatization, 1+2 are fulfilled, but 3 is not. Zhen already asked specifically about the equation $R = R^* \cup \mathfrak{m}_R$ (in logical language), therefore I've added this not just as a comment. I hope it's ok ... – Martin Brandenburg Feb 28 '12 at 10:36
• I'm not sure your definition of $\mathfrak{m}$ is geometric/coherent. Certainly, $\exists z . \; z - x y z = 1$ is a coherent formula, but the problem is that what we want is $(\forall y. \; \exists z . \; z - x y z = 1) \vdash x \in \mathfrak{m}$, and this is not a coherent sequent. – Zhen Lin Feb 28 '12 at 23:25