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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:

Limits

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.

Colimits

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 tr.degree $1$) 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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Martin Brandenburg
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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). –  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 08 '15 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 '15 at 10:03
  • @Arrow: I haven't worked on it. – Martin Brandenburg Nov 03 '15 at 06:58
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    What is the product in the category of smooth proper curves? – Kenta S Aug 13 '21 at 12:46
  • It's worth noting that $\mathbf{Fld}$ is a locally multipresentable category (i.e. is an accessible category with connected limits). – varkor Jul 29 '22 at 12:03

1 Answers1

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Products

We can characterize which products exist. Let $\mathbb{F}_p$ be the prime field of characteristic $p$ (so if $p = 0$ then $\mathbb{F}_0 = \mathbb{Q}$). Then, as a warmup:

Proposition: If $K_i$ are fields of characteristic $p$, the product $\prod K_i$ exists whenever the $K_i$ have the property that they don't all share a common subfield (where by "common subfield" I mean subfields $F_i \subseteq K_i$ which are isomorphic) other than $\mathbb{F}_p$, and it is $\mathbb{F}_p$.

The proof is pretty easy; by hypothesis $\text{Hom}(F, \prod K_i)$ is empty unless $F = \mathbb{F}_p$ in which case it has a single element, and this is the same functor as $\text{Hom}(F, \mathbb{F}_p)$. So in this case the product is acting as a "universal intersection." The projections $\pi_i : \prod K_i \to K_i$ are just the inclusions of $\mathbb{F}_p$ into each $K_i$. An easy class of examples is given by taking the $K_i$ to be finite extensions of $\mathbb{F}_p$ of relatively prime degree.

Can we get products with a non-prime value? If the $|I|$-fold product $K = \prod_{i \in I} K_i$ exists the projections $\pi_i : K \to K_i$ embed the product $K$ as a subfield of each $K_i$, and then the universal property identifies morphisms $L \to K$ with $|I|$-tuples of morphisms $L \to K_i$ which must sit inside each copy of $K$, hence with $|I|$-tuples of morphisms $L \to K$. For $|I| \ge 2$ this is possible iff $\text{Hom}(L, K)$ only ever consists of $0$ or $1$ element, or equivalently, iff $K$ is subterminal.

Non-prime subterminal fields do exist! An example is $\mathbb{Q}(\sqrt[3]{2})$, a non-normal extension of $\mathbb{Q}$ with no automorphisms and no nontrivial subfields. So $\text{Hom}(L, \mathbb{Q}(\sqrt[3]{2}))$ is empty unless $L = \mathbb{Q}$ or $L = \mathbb{Q}(\sqrt[3]{2})$, in which case it consists of one element. More generally "subterminal" is equivalent to the following field-theoretic conditions:

  • No two subfields of $K$ are isomorphic.
  • No subfield of $K$ (including $K$) can have a nontrivial automorphism.

You could say that such fields are "anti-normal." This gives:

Characterization: The product $\prod K_i$ exists and is equal to $K$ iff each $K_i$ contains $K$ as a subfield in such a way that any of their common subfields must lie in $K$, and moreover $K$ is subterminal.

The proof is again pretty easy; we have pretty much just clarified the meaning of the universal property of the product for fields. Intuitively speaking I might describe the situation as, the product exists whenever the category of fields behaves sufficiently as if it were a poset.

Pushouts

Suppose we work in the category of field extensions of $k$ and the coproduct $K = \bigsqcup_{i \in I} K_i$ exists, for $|I| \ge 2$ (this is equivalent to the existence of a certain pushout in fields). This means that morphisms $K \to L$ of fields are identified with $|I|$-tuples of morphisms $K_i \to L$, hence (by the ordinary universal property of the tensor product) with morphisms $\widetilde{K} = \bigotimes_i K_i \to L$ where the tensor product is taken in commutative $k$-algebras. Taking $L = K$ this means that there is a universal map

$$f : \widetilde{K} \cong \bigotimes_i K_i \to K$$

from $\widetilde{K}$ to a field, namely $K$, in the sense that every morphism from $\widetilde{K}$ to a field must factor uniquely through $K$. It follows that $\ker(f)$ must be the unique prime ideal $P$ of $\widetilde{K}$ and $K$ must be the residue field of $P$; in particular, $P$ is also the unique maximal ideal, so $\widetilde{K}$ must be a local ring. Furthermore $P$ must be the nilradical of $\widetilde{K}$, so every element of $\widetilde{K}$ is either invertible or nilpotent. These conditions are necessary and sufficient, so, restating them slightly:

"Characterization": If $k$ is a field and $K_i/k$ are field extensions, then the coproduct $\bigsqcup_i K_i$ exists in the category of field extensions of $k$ iff the tensor product $\bigotimes_i K_i$ has the property that its nilradical is a prime ideal $P$ and the quotient $\left( \bigotimes_i K_i \right)/P$ is a field $K$; in this case the coproduct is $K$.

This happens if the $K_i$ are linearly disjoint over $k$ but it also happens, for example, for the purely inseparable extension $k = \mathbb{F}_p(t), K_1 = K_2 = \mathbb{F}_p(t)[\sqrt[p]{t}]$, where the tensor product is isomorphic to $K_1[\alpha]/\alpha^p$, so has unique prime ideal $(\alpha)$ with quotient $K = K_1$. This extension has the property, dual to the subterminal property above, of being "subinitial" (this is not a very good term but I don't know a better one) in the category of field extensions of $\mathbb{F}_p(t)$, meaning that $\text{Hom}_{\mathbb{F}_p(t)}(K_1, -)$ has either $0$ or $1$ element; this traces back to the fact that the Frobenius map is injective on fields, and implies that the $|I|$-fold coproducts of $K_1$ all exist and are equal to $K_1$ without the need to compute the tensor product.

I don't think this is a good characterization yet because it's not clear to me exactly when this condition actually holds. I guess this condition means "linearly disjoint up to inseparability" or something but it would be nice to say something more precise than this.

That's all I've got so far. I don't see a nice argument about monoidal structures either way but I would be extremely surprised if there was one.

Qiaochu Yuan
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