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?

  • $\begingroup$ I can't answer these questions but I will give examples of fields without coproducts in the category of fields. (I haven't seen any such examples in the literature. Wikipedia states that coproducts don't always exist but gives no argument for that assertion) $\endgroup$ May 16, 2012 at 22:54
  • $\begingroup$ @Georges: Thanks! I've added some remarks to the question which might point toward the answer... $\endgroup$
    – Zhen Lin
    May 16, 2012 at 23:16
  • 1
    $\begingroup$ Dear Zhen, What are the morphisms in $\mathbf{Dom}$? You state that it is a full subcategory of $\mathbf{CRing}$, but is this really what you want? E.g. $\mathbb Z \to \mathbb F_p$ is a morphism in the category of commutative rings, whose source and target are domains, but it does not induce a corresponding map of fraction fields. (My point is that the Frac functor is only defined on the category whose objects are integral domains and whose morphisms are injections of integral domains, which is not a full subcategory of $\mathbf{CRings}$.) Regards, $\endgroup$
    – Matt E
    May 17, 2012 at 3:01
  • $\begingroup$ @MattE: Good point. I've corrected my error. $\endgroup$
    – Zhen Lin
    May 17, 2012 at 8:04

2 Answers 2


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.

  • $\begingroup$ Probably the coproduct of $F_1$ and $F_2$ exists iff one of them is algebraic over a subfield of the other ? $\endgroup$
    – user18119
    May 16, 2012 at 23:00
  • 2
    $\begingroup$ Dear @QiL: no, I don't think so. Would you care to look at my Edit? $\endgroup$ May 16, 2012 at 23:47
  • $\begingroup$ Thanks! Looks like only primary fields have coproduct with themselves. $\endgroup$
    – user18119
    May 17, 2012 at 8:18

If $K$ and $L$ are fields, then $K\otimes_{\mathbb Z} L$ is non-zero if and only if $K$ and $L$ are of the same characteristic. Suppose this holds from now on.

Then consider Spec $K\otimes_{\mathbb Z} L$. This is non-empty. Each points $x$ has a residue field $\kappa(x)$, and there is a map $K\otimes_{\mathbb Z} L \to \kappa(x)$. These maps are determined by their kernels (this is the meaning of points in Spec!), and so are incomparable for different $x$. Hence $K$ and $L$ have a coproduct if and only if Spec $K\otimes_{\mathbb Z} L$ has a single point.

Georges's various counterexamples can be regarded as explicitly constructing more than one point of this Spec.

Additional explanation: To give a morphism from $K$ and $L$ to a field $F$ is the same as giving a map $K\otimes_{\mathbb Z} L \to F$. Giving such a map is the same as choosing the kernel, which is a prime ideal $\mathfrak p$, and then giving a map $\kappa(\mathfrak p) \to F$.

If $E$ is the coproduct of $K$ and $L$, then there is some $\mathfrak q$ so that $E = \kappa(\mathfrak q)$. But then for any other $\mathfrak p$, the map $K\otimes L \to \kappa(\mathfrak p)$ would have to factor through the map $K\otimes L \to \kappa(\mathfrak q)$, which is only possible if $\mathfrak p = \mathfrak q$. Thus $K\otimes L $ would have to have a single point in its Spec.

  • 1
    $\begingroup$ Dear Matt, I'm afraid I don't understand why the existence of a coproduct of $K$ and $L$ in the category of fields is equivalent to the fact that Spec $K\otimes_{\mathbb Z} L$ has a single point. Would you be so kind as to enlighten me ? Thanks in advance. $\endgroup$ May 17, 2012 at 12:26
  • $\begingroup$ Dear Georges, I've added some explanation; hopefully I haven't blundered. Regards, $\endgroup$
    – Matt E
    May 17, 2012 at 12:47
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    $\begingroup$ Dear Matt, you write in your "Additional explanation" that there is some $\mathfrak q$ so that $E = \kappa(\mathfrak q)$. How do we know that? Is $\mathfrak q$ the kernel of the canonical map $K\otimes L\to E$ ? Then we would obtain a map $g:\kappa(\mathfrak q) \to E$. In order to conclude that $g$ is surjective , do we then have to consider the morphism $E\to \kappa(\mathfrak q)$ deduced from the morphisms $K\to \kappa(\mathfrak q) $ and $L\to \kappa(\mathfrak q)$ ? $\endgroup$ May 17, 2012 at 13:34
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    $\begingroup$ @Georges: Dear Georges, Yes, to deduce surjectivity we have to apply the coproduct property of $E$ to the maps $K,L \to \kappa(\mathfrak q)$. Regards, $\endgroup$
    – Matt E
    May 17, 2012 at 15:09
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    $\begingroup$ @GeorgesElencwajg: This is almost identical to the proof that the tensor product is generated by the pure tensors - using only the universal property. $\endgroup$ Apr 14, 2015 at 0:11

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