Let $K$ be a field having two field extensions $L\supseteq K$ and $M\supseteq K$. Does there exist a field $N$ along with embeddings $L\to N$ and $M\to N$, such that the diagram
$$
 \require{AMScd}
\begin{CD}
 K @>>> L\\
 @V V V @VV V\\
 M @>>> N
 \end{CD}
$$
is commutative? To put it less formally, *do $L$ and $M$ have a common field extension $N$* (with $K$ lying in the intersection)?

**If yes, consider this side question (but do leave an answer even if you can only answer the main question!):** Does the above property of fields generalise to the following, stronger property? Let $p$ be either $0$ or a prime number. Does there exist a sequence of fields
$$ \mathbb F\_p = L\_0\subseteq L\_1\subseteq L\_2\subseteq L\_3\subseteq L\_4\subseteq\cdots$$
(where I use the convention $\mathbb F\_0 = \mathbb Q$) such that any field $K$ of characteristic $p$ has an extension field among the $L\_\alpha$? This sequence should be understood as enumerated by ordinal numbers. In other words, what I need is a function that assigns to each ordinal number $\alpha$ a field $L\_\alpha$ containing all $L\_\beta$ with $\beta < \alpha$. (Intuitively, I suspect that this might depend on the Axiom of Choice.)

If the second property holds, it shows that you can essentially only extend a field in one "direction." It also shows that the *class* (it is obviously not a *set*) $\mathbb M\_p = \bigcup\_\alpha L\_\alpha$ is a field (class). We can define all the usual field operations (addition, multiplication, division) here since all pairs of elements lie in $L\_\alpha$ for a sufficiently large $\alpha$. This "monster field" of characteristic $p$ then contains all other *set* fields of that characteristic.