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It is said that the category of fields $\mathsf{Fld}$ is ill-behaved, for example it is not an algebraic theory, does not have initial or terminal objects. In particular it is not presentable. On the other hand, it is accessible and quite a few limits and colimits do exist:


Pullbacks and equalizers exist and coincide with the ones for commutative rings. More generally, one can check that the forgetful functor $\mathsf{Fld} \to \mathsf{CRing}$ creates connected limits. Recall that the correct definition of connected includes non-empty, so that we exclude the terminal object here.

So what about products? If $P$ is a prime field, then $P \times P$ exists and is given by $P$. This is because any morphism $F \to P$ is an isomorphism and uniquely determined. Are there any other examples of products which exist?

In general, which diagrams have a limit? If $\{F_i\}_{i \in I}$ is any diagram of fields, let $I_k$ be the connected components of $I$. Then $\lim_{i \in I} F_i$ is the product of the $\lim_{i \in I_k} F_i$, provided that these limits exist. But a priori there is no reason for this.


The forgetful functor $\mathsf{Fld} \to \mathsf{CRing}$ creates directed colimits. In particular, they exist. Colimits only have a chance to exist for fields of the same characteristic. This motivates to pass to the category of field extensions $k \downarrow \mathsf{Fld}$ of a given field $k$. This has an initial object $k$ and probably is not as pathological as $\mathsf{Fld}$.

If $E,F$ are field extensions of $k$, and the coproduct of the underlying $k$-algebras $E \otimes_k F$ is a field, i.e. $E,F / k$ are everywhere linear disjoint, then of course this is also the coproduct in the category of fields. For example, when $F$ is an algebraic extension of $k$, then $k(T) \otimes_k F \cong F(T)$ is a field. But even when the tensor product is not a field, it may happen a priori that the coproduct exists. So which coproducts exist in the category of fields?

If $k$ is a perfect field, then the category of smooth proper curves over $k$ is anti-equivalent to the category of function fields over $k$ (i.e. finitely generated and of $1$) via $X \mapsto k(X)$. Since the geometric category has products, the algebraic category has coproducts. But I think it just equals the tensor product.

Coequalizers only exist in trivial cases, because regular epimorphisms which are monomorphisms are isomorphisms.

Monoidal structure

Is there an interesting monoidal structure on $k \downarrow \mathsf{Fld}$? What about $(E \otimes_{k} F)/\mathfrak{m}_{E,F}$ for some carefully choosen maximal ideal $\mathfrak{m}_{E,F}$? I know that this is quite naive. But a priori I don't see a reason why this category should not be monoidal.

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$F \times F \cong F$ whenever it exists, because the projection has to be surjective (because $F \xrightarrow{\Delta} F \times F \xrightarrow{\pi_1} F$ is surjective), and it is also injective (because all field morphisms are). – Hurkyl Apr 12 '13 at 11:50
You comment that pushouts and filtered colimits exist in $\mathsf{Fld}$. This is equivalent to the existence of cofibered coproducts, or of simply-connected colimits, i.e. colimits of functors whose diagram is a category $\mathcal{D}$ with trivial fundamental groupoid $\Pi_1(\mathcal{D}) = \mathcal{D}[\mathcal{D}^{-1}]$ (See Theorem 1 and Proposition 2 in the linked paper.) – tcamps Jan 8 at 22:16
Dear Martin, have you managed to resolved parts of this question? If so, could you write your insights as an answer? – Arrow Oct 19 at 10:03
@Arrow: I haven't worked on it. – Martin Brandenburg Nov 3 at 6:58

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