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 $R\longrightarrow F$.)

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    $\begingroup$ That's a good question. It holds for $F=\mathbb Q$ as well $\endgroup$ May 15, 2020 at 7:00
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    $\begingroup$ @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. $\endgroup$ May 15, 2020 at 7:27
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    $\begingroup$ @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 $F/Cring$ $\endgroup$ May 15, 2020 at 8:00
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    $\begingroup$ @Georges: I meant it the way Maxime says it. $\endgroup$ May 15, 2020 at 8:10

2 Answers 2


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.

  • $\begingroup$ This is so clever! How did you come up with this approach? $\endgroup$ May 16, 2020 at 2:26
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    $\begingroup$ 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. $\endgroup$
    – 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...


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