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Let $F$ be a field. Does the category $C_F$ of local rings $R$ equipped with a surjective morphism $R\longrightarrow F$ 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 either $\mathbb{Q}$ or $\mathbb{F}_{p}$ (where $p$ is a prime), but I have no idea how to prove it.

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  • $\begingroup$ I guess that "$F$ of finite type over $\mathbf{Z}$" should be "$F$ is a finitely generated field"? as formulated now it looks like it's a field that is a finitely generated $\mathbf{Z}$-algebra, but this would then be a finite field. Still a first interesting case is that of non-prime finite fields. $\endgroup$
    – YCor
    Commented May 15, 2020 at 12:58
  • $\begingroup$ @YCor thanks, fixed that. $\endgroup$
    – The Thin Whistler
    Commented May 15, 2020 at 13:02
  • $\begingroup$ But I'm unsure what you mean by "finite type". Probably "finitely generated field" is more clear, or "finite field or number field" if you really mean finite type (= finitely generated ring) $\endgroup$
    – YCor
    Commented May 15, 2020 at 13:03

2 Answers 2

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A typical approach is to ask for your universal property among complete DVRs, not among all local rings, because there you have a very nice positive result. Given a perfect field $k$ of characteristic $p$, the Witt ring $W(k)$ is initial among complete DVRs of characteristic $0$ equipped with an isomorphism of residue field with $k$. This is Theorem II.5.4 of Serre's book "Local Fields."

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I will show that this is not the case for $F=\mathbb{F}_9$. The proof generalize to any $\mathbb{F}_{p^k}$, with $k >1$.

I'm starting from the observation that: $\mathbb{F}_9 \simeq \mathbb{Z}[i]/(3) \simeq \mathbb{Z}[\sqrt{2}]/(3)$. where I'm using the isomorphism that identifies $i$ and $\sqrt{2}$ to identify $\mathbb{Z}[i]/(3)$ and $\mathbb{Z}[\sqrt{2}]/(3)$.

I'm now considering $A= \mathbb{Z}[i,\sqrt{2}]$. It has a surjective map $\phi:A \to \mathbb{F}_9$ induced by the two maps $\mathbb{Z}[i] \to \mathbb{F}_9$ and $\mathbb{Z}[\sqrt{2}] \to \mathbb{F}_9$ above.

I can localize $A$ at $\ker \phi$ to make it an object of $C_F$.

So, in the category $C_F$ I have a diagram:

$$ \mathbb{Z}[i]_{(3)} \to A_{\ker \phi} \leftarrow \mathbb{Z}[\sqrt 2]_{(3)} $$

If there was an initial object $B$ in $C_F$, its unique map to $A_{\ker \phi}$ should factors through both $\mathbb{Z}[i]_{(3)} $ and $\mathbb{Z}[\sqrt 2]_{(3)}$ hence, it should factor through their in $A_{\ker \phi}$. But this intersection is reduced to $\mathbb{Z}_{(3)}$, so we should have a map $B \to \mathbb{Z}_{(3)}$ compatible with the map back to $\mathbb{F}_9$, but as the map $\mathbb{Z}_{(3)} \to \mathbb{F}_9$ is not surjective, and the map $B \to \mathbb{F}_9$ needs to be, we have a contradiction.

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  • $\begingroup$ The pullback of $\mathbb{Z}[\sqrt{2}]_{(3)}\longrightarrow\mathbb{F}_{9}$ and $\mathbb{Z}[i]_{(3)}\longrightarrow\mathbb{F}_{9}$ is $\mathbb{Z}_{(3)}$ when viewed as morphisms in the category of rings. As morphisms in the category $C_{F}$, their pullback is $\mathbb{Z}[x]_{(3,x^{2}+1)}$ - and that lies in $C_{F}$. $\endgroup$
    – The Thin Whistler
    Commented May 15, 2020 at 15:04
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    $\begingroup$ The argument is not about pullbacks : it looks at the map $R\to A_{\ker\phi}$ (where $R$ is your initial object). Since this map factors through both of the subrings indicated (by uniqueness), its image (as a map of rings) is contained in their intersection; and that is not possible, as the restriction of the quotient map to said intersection is not surjective. $\endgroup$
    – Maxime Ramzi
    Commented May 15, 2020 at 15:46
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    $\begingroup$ Well, it is about pullback, but pullback over $A_{\ker \phi}$ not pullback over $\mathbb{F}_9$. For a general field $F$ this category has all products (given by fiber product over $F$), but it can be shown by abstract category theory that this category will have an initial object if and only if has all limits, i.e. if and only if it has fiber products. Then I started from an example of a fiber product that do not exists (the fiber product over $A_{\ker \phi}$) and back tracked it to a counter example to the original question. $\endgroup$
    – Simon Henry
    Commented May 15, 2020 at 16:20
  • $\begingroup$ @MyGrandmother'sCobblestone: By the way, your computation of the product in $C_F$ (or equivalently, fiber product over $\mathbb{F}_9$) is not quite right: you also need to quotient out by the kernel of $\mathbb{Z}[X] \to \mathbb{Z}[\sqrt 2] \times \mathbb{Z}[i]$, that is by $(X^2+1)(X^2-2)$. $\endgroup$
    – Simon Henry
    Commented May 18, 2020 at 15:57

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