$(d\mid \cdot)(c\mid \cdot) = (\text{lcm}(d,c) \mid \cdot)$ where $(n\mid x) \in \{0,1\}$ is whether (1) or not (0) $n$ divides $x \in \Bbb{Z}$.
Thus, we can linearly extend all formall $K$-linear sums over a commutative ring $K$, of functions $(n \mid x), n \in \Bbb{N}$.
I know that the monoid ring $K[\Bbb{N}] \approx K[X]$ so what must be $K[(\Bbb{N}, \text{lcm})]$ the monoid ring, be isomorphic to when $\text{lcm}$ is our monoid law in place of usual multiplication.
Attempt. There very map $(n \mid \cdot)$ itself, by definition, induces a ring hom from $\Bbb{K}[(\Bbb{N}, \text{lcm})] \twoheadrightarrow K[\Bbb{N}]$ when we extend it linearly from $\Bbb{N}$. Therefore ?
The ring is isomorphic to $K[X_1, X_2, \ldots]$ mod the ideal $I$ generated by $(X_aX_b - X_{\text{lcm}(a, b)})$. This is generated by symbols of the form $X_{p^n}$ where $p$ is prime, there is no relation between $X_{p^n}$ and $X_{\ell^m}$ for $\ell \neq p$, and we have $X_{p^n}X_{p^m} = X_{p^{\max(m, n)}}$.
I don't know how to simplify the ideal of relations further. The ring is non-noetherian; for instance, for each $p$, the quotient we obtain by setting $X_\ell = 0$ for all $\ell \neq p$ is isomorphic to the monoid-ring coming from the max function.