To what extent is $\textbf{Fld}$ not a filtered category?

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EDIT: Basically, I want to see if one can construct examples of fields $F,K$ such that they don't have a common extension field.

Apologies in advance if this is obvious, but for example, is there a field containing isomorphic copies of both $\mathbb{C}$ and $\mathbb{Q}_p$ for any prime $p$? If not, what's a good field invariant that could tackle a lot of these kinds of questions? I was thinking of using Galois groups somehow, but not sure where to begin.

This comes from the observation that the underlying set of the geometric realisation (as a locally ringed space) $|F|$ of a functor $F: \textbf{CRing} \to \textbf{Set}$ can be given by $\text{colim}_\textbf{Fld} F(k)$ (limit taken over the functor sending a field $k$ to $F(k)$ and sending $f: k \to K$ to $F(f): F(k) \to F(K)$), where $\textbf{Fld}$ is the full subcategory of $\textbf{CRing}$ where the objects are fields.

If you want to see what this actually looks like, you can consider $\bigsqcup_{\textbf{Fld}} F(k)/\sim$, where $\sim$ is the equivalence relation generated by the relation $x \sim y$ for $x \in F(k)$ and $y \in F(K)$ if there exists $L$ and $f: k \to L$ and $g: K \to L$ such that $F(f)(x) = F(g)(y)$. Now, if such an $L$ existed always, then this would be an equivalence relation, but if not, then you have to basically force transitivity. So I'm wondering to what extent you really are doing this.