TITLE: Field extensions with(out) a common extension
QUESTION [5 upvotes]: 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.

REPLY [13 votes]: The $K$-algebra  $A=L\otimes_K M$ has, like all non-zero rings,  a maximal ideal $\mathfrak m\subset  A$ .
The quotient algebra $N=A/\mathfrak m $ is then an extension  field of both $L$ and $M$, via the composed morphisms $L\to A\to N$ and $M\to A\to N$  .
[Recall that any morphism of rings with source a field is injective.]