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 $R$ 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.
 A: 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."
A: 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. 
