There are convincing arguments that support the claim that universal algebra is essentially the theory of $\lambda$-accessible monads $T$ over Set.

Now, given two equivalent categories of algebras for two different monads $$ \text{Alg}(T) \cong \text{Alg}(S)$$

it is not possible to say that $T$ and $S$ are essentially the same funtor, in fact this is false. But, is there a relation between the two?

Of course, using the Gabriel-Ulmer duality one can build an equivalence of categories between:

$$\text{Pres}_{\lambda}(\text{Alg}(T)) \cong \text{Pres}_{\lambda}(\text{Alg}(S)).$$

Which is already telling us that the limits theories that describe them are equivalent.

Q: Given a category of algebras for a finitary monad $\text{Alg}(T)$ is it possible to give a description of the class of accessible monads describing the same category of models?

In a way I am asking for a sort of Nullstellensatz theorem where on one side the category of algebras for a monad is the zero locus of the limit theory individuated by the monad and on the other the ideals are the monads. In this context the answer to my question is all the ideals corresponding to the same variety have the same radical.

Q: Is it even possible to make this analogy more precise, locating the Nullstellensatz in a Morita-theoretic framework?!

Maybe the only possible analogy is that the radical extraction $\sqrt{\_} $ and the cauchy completion are both closure operators induced by a Galois-like correspondence.

I have the feeling that this question is just a reference request and I do believe that it was already answered. I just to not understand how to google the question.

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    $\begingroup$ Once you fix the category of algebra $C$, a monad $M$ representing it is entirely specified by by the data of the algebra $M(1) \in C$. Indeed, Specifying $M(1)$ give you the forgetful functor $C \rightarrow Set$ hence its adjoint, hence the monad. Conversely, any object $X \in C$ define a monad on $Set$ (its endomorphism monad), which will be $\lambda$-accessible if and only if $X$ is $\lambda$-presentable. So you just have to determine condition on $X \in C$ so that the functor $Hom(X, \_ )$ is monadic. Beck monadicity theorem give you such condition (not very nice though) $\endgroup$ – Simon Henry May 30 '18 at 16:31
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    $\begingroup$ (this is analogous to the fact that rings $R'$ morita equivalent to a given ring $R$ are in correspondance with full finite projective $R$-module, by $M \mapsto R' = End(M) $) $\endgroup$ – Simon Henry May 30 '18 at 16:33

This is an elaboration on Simon Henry's comments. The following precise characterization of categories monadic over $\text{Set}$ is known, closely paralleling Gabriel's theorem characterizing categories of modules over rings:

Theorem: A category $C$ is equivalent to the category of algebras over a monad $M$ on $\text{Set}$ iff it is Barr-exact and there exists an object $P$ which is projective, such that all coproducts $\bigsqcup_I P$ exist, and which is a generator in the sense that every object $X \in C$ admits an epimorphism $\bigsqcup_I P \to X$. In this case the monadic functor $C \to \text{Set}$ is given by $\text{Hom}(P, -)$, which has a left adjoint, and the monad $M$ is the monad associated to this adjunction.

I first saw this theorem in Borceux, and don't know who it's due to. For an easier-to-find reference see On the characterization of monadic categories over $\text{Set}$ by Vitale.

So, if you already know that a category $C$ is the category of algebras over some monad, it's already Barr-exact and the problem reduces to characterizing the ($\lambda$-compact, I guess) projective generators in it.

This problem can be further reduced as follows. Fix such a generator $P$. Any other projective object $Q$ admits some epimorphism $i : \bigsqcup_I P \to Q$, and by projectivity the identity $Q \to Q$ lifts along this epimorphism, from which it follows that $Q$ is a retract of $\bigsqcup_I P$ (a direct generalization of "every projective module is a retract of a free module"). So it suffices to determine which such retracts are generators (which is equivalent to checking whether the functor $\text{Hom}(Q, -)$ is faithful).

All of this produces the usual abelian results when applied to categories of modules over rings, so here is a "nonabelian" example. Take $C = \text{Grp}$. Every projective object is a retract of a free group, and since subgroups of free groups are free, the projective objects are precisely the free groups, all of which are generators. The monad associated to the free group on $I$ generators is the monad on $\text{Set}$ which sends a set $X$ to the coproduct (in $\text{Grp}$) of $I$ copies of the free group on $I \times X$ (analogous to an "$I \times I$ matrix algebra"). I'm not quite sure how to think about this as an algebraic theory.

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  • $\begingroup$ I should say, all the free groups on at least one generator are generators. $\endgroup$ – Qiaochu Yuan May 14 '19 at 8:20

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