Let $R = \mathbb{Z}[X,X^{-1}]$, and let $\mathbb{G}_m = \textrm{Spec } R$. If $A$ is a ring, the set of $A$-rational points $$\textrm{G}_m(A) = \textrm{Hom}_{\textrm{Sch}}(\textrm{Spec } A, \mathbb{G}_m) = \textrm{Hom}_{\textrm{Ring}}(R,A)$$
can be identified with $A^{\ast}$. Fix an integer $N \geq 1$, and consider the morphism of schemes $f: \mathbb{G}_m \rightarrow \mathbb{G}_m$ corresponding to the ring homomorphism $R \rightarrow R, X \mapsto X^N$. Then for any ring $A$, $f(A): A^{\ast} \rightarrow A^{\ast}$ is the map $x \mapsto x^N$, so $f$ is a homomorphism of group schemes.
If $e: \textrm{Spec } \mathbb{Z} \rightarrow \mathbb{G}_m$ is the identity section (the morphism of schemes corresponding to homomorphism $R \rightarrow \mathbb{Z}, X \mapsto 1 $), then the scheme-theoretic kernel $K$ of $f$ is the fiber product
$$\mathbb{G}_m \times_{\mathbb{G}_m} \textrm{Spec } \mathbb{Z}$$ via the maps $f$ and $e$. If $\pi: K \rightarrow \mathbb{G}_m$ is the projection morphism, then for each ring $A$, the image of $\pi(A): K(A) \rightarrow A^{\ast}$ is the kernel of the homomorphism $f(A)$. This follows by universal property of the fiber product.
I have two questions about this:
1 . Is there is a nicer description of the ring $\mathcal O_K(K)$ whose spectrum is $K$? What we have is that $\mathcal O_K(K)$ is the ring $\mathbb{Z}[X,X^{-1}] \otimes_{\mathbb{Z}[X,X^{-1}]} \mathbb{Z}$, where $\mathbb{Z}[X,X^{-1}]$ is a module over itself via $f(X) \cdot g(X) = f(X^N)g(X)$, and $\mathbb{Z}$ is a module over $\mathbb{Z}[X,X^{-1}]$ via $f(X) \cdot n = f(1)n$.
2 . Is $\pi(A): K(A) \rightarrow A^{\ast}$ an injection for each ring $A$? Moreover, we usually think of a kernel as being a closed subgroup. Is the projection map $K \rightarrow \mathbb{G}_m$ a closed immersion?