# Does the category of local rings with residue field $F$ have an initial object?

Let $$F$$ be a field. Does the category $$C_F$$ of local rings with residue field isomorphic to $$F$$ have an initial object?

This is, for instance, true if $$F=\mathbb{F}_{p}$$ for some prime $$p$$: If $$R$$ is a local ring with residue field $$\mathbb{F}_{p}$$, then any $$x\in\mathbb{Z}\setminus(p)$$ must map to something invertible under the morphism $$\mathbb{Z}\longrightarrow R$$. Hence that morphism factors as $$\mathbb{Z}\longrightarrow\mathbb{Z}_{(p)}\longrightarrow R$$; thus $$\mathbb{Z}_{(p)}$$ is the initial object.

But what happens in the more general case? I guess it should be true at least if $$F$$ is of finite type over $$\mathbb{Z}$$, but I have no idea how to prove it.

(EDIT - To avoid any confusion: I am talking about an initial object in the category of local rings $$R$$ with a fixed surjection $$R\longrightarrow F$$.)

• That's a good question. It holds for $F=\mathbb Q$ as well May 15, 2020 at 7:00
• @GeorgesElencwajg : I guess it depends on how you set up the category $C_F$ exactly : if it's just a full subcategory of rings, then you are right (composing $R\to F\to F$ gives two different morphisms $R\to F$ because $R\to F$ is surjective); but if you set it up as a subcategory of $CRing/F$ instead, it's not that clear. May 15, 2020 at 7:27
• @Georges : I'm not saying $F$-algebras, on the other hand "the category of local fields with residue field $F$" could definitely mean "local rings $R$ with a fixed map $R\to F$ which exhibits $F$ as the residue field". So a subcategory of $CRing/F$, not $F/Cring$ May 15, 2020 at 8:00
• @Georges: I meant it the way Maxime says it. May 15, 2020 at 8:10

Let $$\mathbb{F_4}=\{0,1,w,1+w\}$$ be the field of 4 elements. Suppose $$R$$ is the initial object in the category described in the question for the field $$\mathbb{F_4}$$. Then $$R$$ must contain some element $$x$$ which maps to $$w\in\mathbb{F_4}$$. Thus we have a map $$f\colon S\to R$$, where $$S=\mathbb{Z}[y]_M$$, sending $$y \mapsto x$$. Here $$M$$ is the maximal ideal of $$\mathbb{Z}[y]$$ containing $$2,1+y+y^2$$.

The following composition must be the identity: $$R \to S \stackrel f \to R$$ Thus $$R=S/I$$ for some ideal $$I\subset M$$. Further we know $$I\neq 0$$ as $$S$$ cannot be the initial object: there are multiple distinct maps $$S\to S$$, such as the identity map and the map sending $$y\mapsto y+2$$.

Under the composition $$S \stackrel f \to R\to S$$, we have $$y\mapsto p/q$$, for some $$p,q$$ integer polynomials in $$y$$. We know $$p/q$$ is not a rational number as $$p/q\mapsto w\in\mathbb{F_4}$$. Thus $$p/q$$ is a non-constant rational function in one variable, taking infinitely many values, which cannot all satisfy the same polynomial over the integers.

On the other hand, as $$I\neq 0$$ there must be a polynomial over the integers satisfied by $$p/q$$. This gives us the desired contradiction.

• This is so clever! How did you come up with this approach? May 16, 2020 at 2:26
• Thankyou. It felt like if there was an initial object it ought to be S, but at the same time it couldn't be. Playing those off against each other and using that S is a well behaved, concretely described ring led to the contradiction.
– tkf
May 16, 2020 at 3:55

The category $$C_{F}$$ possesses a weak initial object $$I_{F}$$, i.e. an object that is unique up to not necessarily unique isomorphism.

Let $$F$$ be a field and $$L$$ be its minimal subfield (the smallest subfield contained in $$F$$). Then either $$L=\mathbb{F}_{p}$$ for some prime $$p$$ or $$L=\mathbb{Q}$$.

Assume first that $$F$$ is of finite type over $$L$$. Let $$n\in\mathbb{N}$$ be the smallest natural number so that $$F=L[x_{1},...,x_{n}]/\mathfrak{m}$$ for some maximal ideal $$\mathfrak{m}\subseteq L[x_{1},...,x_{n}]$$. Let $$\overline{x}_{i}$$ be the image of $$x_{i}\in L[x_{1},...,x_{n}]$$ in $$F$$.

Let $$\zeta:R\longrightarrow F$$ be a surjection where $$R$$ is a local ring. Since every $$\overline{x}_{i}$$ has a (not necessarily unique) preimage $$\zeta^{-1}(\overline{x}_{i})\in R$$, there is a (not necessarily unique) morphism $$\kappa:\mathbb{Z}[x_{1},...,x_{n}]\longrightarrow R$$ that fits into a commutative diagram $$\require{AMScd}$$ $$\begin{CD} \mathbb{Z}[x_{1},...,x_{n}]@>{\kappa}>>R\\ @V{\pi}VV @VV{\zeta}V\\ L[x_{1},...,x_{n}] @>>{\chi}>F \end{CD}$$ Let $$\mathfrak{i}:=\chi^{-1}\pi^{-1}(0)=\pi^{-1}(\mathfrak{m})$$. The ideal $$\mathfrak{i}$$ is always prime; it is maximal if and only if $$L=\mathbb{F}_{p}$$ for some prime $$p$$. Since $$R$$ is local, every element of $$\mathbb{Z}[x_{1},...,x_{n}]$$ is mapped by $$\kappa$$ onto something invertible in $$R$$. Hence $$\kappa$$ factors as $$\begin{CD} \mathbb{Z}[x_{1},...,x_{n}] @>>>\mathbb{Z}[x_{1},...,x_{n}]_{(\mathfrak{i})} @>{\lambda}>> R \end{CD}$$ Thus $$I_{F}:=\mathbb{Z}[x_{1},...,x_{n}]_{(\mathfrak{i})}$$ is a weak initial object in the category $$C_{F}$$.

Note that the assignment $$\kappa\longleftrightarrow\lambda$$ is unique in both ways: To each choice of $$\kappa$$ there is a unique $$\lambda$$ and vice-versa.

Assume next that $$F$$ is of infinite type over $$L$$. Then $$F$$ is the direct limit of all morphisms $$F'\longrightarrow F''$$, where $$F',F''$$ are fields of finite type over $$L$$. Since the construction of $$I_{-}$$ is functorial and compatible with direct limits, $$I_{F}$$ can be defined as $$I_{F}:=\lim_{F'\text{ of fin. t.}/L}I_{F'}$$.

The initial object is strong, i.e. unique up to unique isomorphism, if and only if $$F=L$$.

Namely, if $$F=L$$, then $$n=0$$ and the unique morphism $$\kappa:\mathbb{Z}\longrightarrow R$$ induces a unique morphism $$\lambda:\mathbb{Z}_{(\mathfrak{i})}\longrightarrow R$$.

Else, if $$F\neq L$$, then $$n\geq 1$$ and for any $$i\in\{1,...,n\}$$ and any $$s\in\mathfrak{i}\setminus\{0\}$$, the map $$\xi_{i,s}:x_{i}\mapsto x_{i}+s$$ yields a nontrivial automorphism $$I_{F}\longrightarrow I_{F}$$ that commutes with the surjection $$I_{F}\longrightarrow F$$.

My guess is that the $$\xi_{i,s}$$ actually generate the whole group $$\operatorname{Aut}(I_{F})$$, but I have yet to figure out a proof for this...