= Solution
On the <monoidal category> of <modules> over the <commutative ring> $k$, consider the <monad> $T=H\otimes_k-$ coming from the unit and multiplication of the <bialgebra> $H$. Its <opmonoidal functor> structure has comparison maps
$$
T_2\bigl(h\otimes(x\otimes y)\bigr)=\sum(h_{(1)}\otimes x)\otimes(h_{(2)}\otimes y),\qquad T_0=\varepsilon:H\otimes_k k\cong H\to k.
$$
Here and below <Sweedler notation> abbreviates the <comultiplication> $\Delta h=\sum h_{(1)}\otimes h_{(2)}$. The opmonoidal associativity and unit axioms are the coassociativity and counit laws of the <coalgebra>. The <unit and multiplication of a monad> are <opmonoidal natural transformations> because the <bialgebra> axioms say
$$
\Delta(hg)=\sum h_{(1)}g_{(1)}\otimes h_{(2)}g_{(2)},\qquad\Delta(1)=1\otimes1,\qquad\varepsilon(hg)=\varepsilon(h)\varepsilon(g),\quad\varepsilon(1)=1.
$$
The <Eilenberg-Moore category> of this <opmonoidal monad> is the category of left $H$-<modules>: a monad-algebra map $H\otimes_kM\to M$ is exactly a unital associative action.
Applying the preceding construction gives the <diagonal bialgebra action> and the unit action
$$
\boxed{h\cdot(u\otimes v)=\sum(h_{(1)}\cdot u)\otimes(h_{(2)}\cdot v),\qquad h\cdot a=\varepsilon(h)a\quad(a\in k).}
$$
The usual <associators> and <unitors> for the <tensor product of modules> are $H$-linear, and the underlying tensor product is exactly $\otimes_k$. \b[The forgetful functor into $k$-modules is strict monoidal.] The base need not be a <field>; the <modules> need be neither <flat modules> nor <finitely generated modules>.
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