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I’ve read the following proof-less claim: there is no operad such that the algebras over it are fields. We can make that precise by asking whether there’s an operad $\mathcal{P}$ in abelian groups such that the category $\mathcal{P}Alg$ is equivalent to the category of fields.

Intuitively this seems true: every axiom but the existence of inverses would be fine (and indeed there’s an operad for commutative rings). But this isn’t a proof, as there could be a presentation of field axioms that would translate to an operad.

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I recognize that you’ve already answered this question, but I think there’s still more to say here.

Intuitively this seems true: every axiom but the existence of inverses would be fine (and indeed there’s an operad for commutative rings).

Okay, but what would you do about inverses? All an operad can describe is operations and equations between them where no variable is repeated. The inverse is not defined on every element of a field so it doesn’t obviously give an operation. You could say, okay, let’s consider the operation which either takes the inverse or returns $0$ (or whatever), and try to axiomatize that (you get what I think is called a “meadow”). But you can’t state the key property of this operation operadically, namely that $a^{-1} a = 1$ when $a \neq 0$, for *two* reasons:

- It only holds on the nonzero elements, and operads can’t describe equations that only hold some of the time.
- $a$ appears twice, and operads can’t describe equations that involve two appearances of the same variable. (This is the same reason operads can’t describe groups.)

The argument you’ve given is fine but it fails to apply once you restrict to, say, characteristic $0$ fields, which you might hope to be the category of algebras of an operad in $\mathbb{Q}$-vector spaces. So here is an argument which also applies in this case. In a nice category like vector spaces, the forgetful functor from algebras over an operad down to the underlying category admits a left adjoint, so there are free algebras. If $P(n)$ are the objects in the operad, the free functor takes the form

$$X \mapsto \bigsqcup_{n \ge 0} P(n) \otimes_{S_n} X^{\otimes n}.$$

Claim:The forgetful functor from fields over $\mathbb{Q}$ to $\mathbb{Q}$-vector spaces does not have a left adjoint.

*Proof.* We will show that the free field over $\mathbb{Q}$ on one element (equivalently, on $\mathbb{Q}$) does not exist. This would be an object representing the forgetful functor, and so would be a field $K$ over $\mathbb{Q}$ such that morphisms $K \to L$, $L$ another field, can be identified with elements of $L$ by evaluation on a “universal element” $k \in K$.

But $k$ is necessarily nonzero, since it needs to be able to map to nonzero elements of $L$, and hence it has an inverse and cannot map to $0 \in L$. $\Box$

The point of writing out this argument is to show exactly what the thing that goes wrong has to do with inverses. There is no *equational* obstruction preventing the universal element $k \in K$ from mapping to $0$, but there is an obstruction which is not equational, namely that $k$ being nonzero implies that it is invertible.

From the above argument it follows that fields over $\mathbb{Q}$ not only fails to be the category of algebras for an operad in $\mathbb{Q}$-vector spaces, it even fails to be the category of algebras for a monad. This is a response to Baby Dragon’s comment: while operads in sets can’t describe groups, monads in sets *can*. So the problem here is not that there are inverses at all but that the behavior of the inverse in fields is fundamentally tied to the non-equational condition “$a \neq 0$.”

The proof is actually not that hard, and I can’t believe I didn’t think about it.

There is no initial object in the category of fields: no field can embed at the same time in $\mathbb{Q}$ and $\mathbb{F}_2$. However, if $P$ is an operad in abelian groups, the free operad $P(0)$ on the trivial abelian group is initial in the category of $P$-algebras, because a morphism of $P$-algebras $P(0) \to A$ is equivalent to a morphism of abelian groups $0 \to A$, which is unique.

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