Let $\mathcal{V}$ be a monoidal closed (complete, cocomplete, reasonable...) category.


Let $\mathsf{T}$ be an enriched monad over $\mathcal{V}$. The forgetful functor $\mathsf{U}: \mathsf{Alg}(\mathsf{T}) \to \mathcal{V}$ is tautologically monadic in $\mathcal{V}$-Cat. If we pass to the [underlying categories](https://ncatlab.org/nlab/show/underlying+ordinary+category) $\mathsf{U}\_0: \mathsf{Alg}(\mathsf{T})\_0 \to \mathcal{V}\_0$, do we still get a monadic functor?


$\mathsf{U}\_0$ is still a right adjoint because $(-)\_{0}: \mathcal{V}\text{-Cat} \to \text{Cat}$ is a $2$-functor, but
I expect a negative answer to my question. Yet, I can't find a counterexample.



Is monadicity preserved by the underlying functor?
Following up on my comment, I think the direct proof is easiest - it’s just an exercise in translating classical definitions into the enriched setting. Let’s call the underlying monad $T\_0$.


For objects, observe that an algebra of $T$ is given by $a: I \to V(TA, A)$ (which is a map $TA \to A$ in the underlying category) so that $\rho^{-1};(\eta\_A \otimes a);m = j\_A$ (which is exactly $\eta\_A;a = id $ in the underlying category) and $\rho^{-1};((a;T) \otimes a);m= \rho^{-1};(\mu\_A \otimes a);m$ (which is precisely $T\_0(a);a = \mu\_A;a$ in the underlying category).


Let $(A,a), (B,b)$ be algebras if T. A map $f:a \to b$ is a point $f:I \to V(A,B)$ so that $T\_0(f)b = af$, that is $f;\rho^{-1};(T \otimes b)m = f;\lambda^{-1};(a \otimes id )m$. This is equivalent to saying $f$ is a point of $eq(\rho^{-1};(T \otimes b)m, \lambda^{-1};(a \otimes id )m) = eq(T;V(TA,b), V(a,B))$ which is the hom-object between the algebras.


