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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
  • $\begingroup$ (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

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