This is a slightly more advanced version of another question here.
Let $\textbf{CRing}$ be the category of commutative rings with unit. Let $\textbf{Dom}$ be the category of integral domains – by which I mean a non-trivial commutative ring with unit such that the zero ideal is prime. Let $\textbf{Fld}$ be the category of fields – by which I mean an integral domain such that every non-zero element is invertible. (Homomorphisms preserve the unit, etc.; in the case of $\textbf{Dom}$ we only allow injective homomorphisms.)
There is an evident inclusion of categories $\textbf{Fld} \hookrightarrow \textbf{Dom}$, and it has a left adjoint $\operatorname{Frac} : \textbf{Dom} \to \textbf{Fld}$. Since $\operatorname{Frac}$ is a left adjoint, it preserves coproducts, and it is clear that $\operatorname{Frac} K \cong K$ if $K$ is a field. Thus, coproducts in $\textbf{Fld}$, if they exist, must be the same as coproducts in $\textbf{Dom}$, if they exist.
Question 1. What are necessary and sufficient conditions for the coproduct of two integral domains / fields to exist?
Now, since $\textbf{Fld}$ is a full subcategory of $\textbf{Dom}$ and a full subcategory of $\textbf{CRing}$, by more general nonsense, any $\textbf{CRing}$-coproduct (or $\textbf{Dom}$-coproduct) of fields that happens to be a field is also the coproduct in $\textbf{Fld}$
Question 2. Is it possible for a coproduct to exist in $\textbf{Dom}$ without being the coproduct in $\textbf{CRing}$?
(As a concrete example of why this matters, observe that $\textbf{CRing}^\textrm{op}$ is a full subcategory of $\textbf{Sch}$, but limits computed in $\textbf{CRing}$ in general differ from colimits computed in $\textbf{Sch}$.)
Some remarks. For every pair of fields $K$ and $L$, we can take their $\textbf{CRing}$-coproduct $K \otimes_{\mathbb{Z}} L$. For each prime $\mathfrak{p}$ in $\operatorname{Spec} K \otimes_{\mathbb{Z}} L$, there is an integral domain $(K \otimes_\mathbb{Z} L) \mathbin{/} \mathfrak{p}$, and we can take the fraction field $\operatorname{Frac} ((K \otimes_\mathbb{Z} L) \mathbin{/} \mathfrak{p})$ to get a quasi-coproduct. There is only a set of these quasi-coproducts, and if $F$ is a field, every pair of maps $(K \to F, L \to F)$ must factor through at least one (or exactly one…?) of these quasi-coproducts. So the category of fields equipped with a homomorphism from $K$ and a homomorphism from $L$ has a "weakly initial set", and a genuine coproduct exists if and only if this category has an initial object.
Thus, the failure of $K \sqcup L$ to exist can be quantified in terms of the structure of the subcategory of quasi-coproducts. What are the possibilities?
Best Answer
Here are examples in all characteristics of fields with the same prime field but without a coproduct.
Let $k=\mathbb F_p$ or $\mathbb Q$ and let $ k(x)$ and $ k(y)$ be two copies of the field of rational functions over $k$.
I claim they have no coproduct in Fld.
Indeed, let $C$ be a hypothetical coproduct $C=k(x) \sqcup k(y)$.
Consider the identity morphisms $i_1:k(x)\to k(t) $ and $i_2:k(x)\to k(t)$ (where $k(t)$ is yet another copy of the rational function field over $k$)
It gives rise to a morphism $i_1\sqcup i_2:C\to k(t)$.
Since this morphism is injective, we necessarily have $trdeg_k C=1$.
However by considering the obvious morphisms $j_1:k(x)\to k(x,y) $ and $j_2:k(y)\to k(x,y) $ we obtain a field morphism $j_1\sqcup j_2:C\to k(x,y)$ whose image must contain $x$ and $y$.
In other words the (automatically injective) morphism $j_1\sqcup j_2:C\to k(x,y)$ is surjective and thus is an isomorphism, which contradicts the preceding assertion that $trdeg_k C=1$.
This implies that $k(x)\otimes_k k(y)$is not a field, which of course can also proved directly. (The slickest way being to use a theorem of Grothendieck implying that $k(x)\otimes_k k(y)$ has Krull dimension one: see here)
Edit Let me show, as an answer to a comment below, that $\mathbb Q(\sqrt 2)$ and $\mathbb Q(\sqrt 2)$ have no coproduct.
Let $C=\mathbb Q(\sqrt 2)\sqcup \mathbb Q(\sqrt 2)$ be their hypothetical coproduct in Fld.
The identities $i_1,i_2:\mathbb Q(\sqrt 2) \to \mathbb Q(\sqrt 2)$ lead to a field morphism $i_1 \sqcup i_2:C\to \mathbb Q(\sqrt 2)$.
This proves that $C=\mathbb Q(\sqrt 2)$.
Let now $j_1,j_2:\mathbb Q(\sqrt 2)\to Q(\sqrt 2)\sqcup \mathbb Q(\sqrt 2)=C=\mathbb Q(\sqrt 2)$ be the structural morphisms for the coproduct ($j_1,j_2\in Gal(\mathbb Q(\sqrt 2)/\mathbb Q))$.
Consider then the morphisms $u_1=j_1:\mathbb Q(\sqrt 2)\to \mathbb Q(\sqrt 2)$ and $u_2:\mathbb Q(\sqrt 2)\to \mathbb Q(\sqrt 2)$ where $u_2$ is fiendishly chosen as the automorphism of $Gal(\mathbb Q(\sqrt 2)/\mathbb Q))$ different from $j_2$.
It is now impossible to choose $u:C=\mathbb Q(\sqrt 2)\to \mathbb Q(\sqrt 2)$ with $u\circ j_1=u_1$ and $u\circ j_2=u_2$.
Hence the coproduct $C$ actually cannot exist.