I mean $(a,b)^{z} := (a^z, b^z)$ or $(a^z, b^w) =: (a,b)^{(z,w)}$ the latter is a $\Bbb{Z}^2 = \Bbb{Z}\times \Bbb{Z}$ group action on an abelian group, so you have a an abelian group or also a $\Bbb{Z}$-module same thing
Because for $K$ it's $\Bbb{Z}$ but for $\Bbb{R}^2$ it's usuall $\Bbb{R}$. So they may be isomorphic $\Bbb{Z}$-modules but not real-vector spaces
"over the reals"
What I mean is the ring of scalars is different, and $\Bbb{Z}$ is not a field, but $\Bbb{R}^2$ is definitely a $\Bbb{Z}$-module by restricting scalars from $\Bbb{R}$ to $\Bbb{Z}$.
because if you take the infinite series in $\Bbb R^2_*$ you can actually extend it to the complex plane
so you can extend an infinite series in $\Bbb R^2$ and obtain some function $f(s)$ defined for complex numbers, you can extend an infinite product in $\Bbb R^2_*$ and obtain $f(s)$ as well
but....if you take an infinite series in $\Bbb R^2_*$ you can get a completely new function $g(s)$ on the complex plane
