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You suspect correctly.Could you please specify if you mean:
$\text{End}_K(K^2)$
$\text{End}_{\Bbb Z}(K^2)$ or:
$\text{End}_A(K^2)$
as all of these are different structures....
I suspect you mean the latter, that is you want $A$-linear maps.
Hi Deveno, I have been following out a thought of my own (for once) and tried to show that the map $f:K->End_{A}(M)$ given by $f(t)=w_{t}$ where $w_{t}(x,y)=t(x,y)$ is an isomorphismHere is my idea (I haven't worked through all the details):
1) Show that any $A$-linear map is a $K$-linear map by using the fact that if:
$(x,y) \in K^2, r\in K, L \in \text{End}_A(K^2)$
$L(r(x,y)) = L(\alpha.(x,y)) = \alpha.L(x,y) = r(L(x,y))$
where $\alpha = \begin{bmatrix}r&0\\0&r \end{bmatrix} \in A$.
This means we can represent an $A$-linear map as multiplication by a 2x2 matrix over $K$.
2) Show such a matrix must lie in $A$.
(hint: it suffices to show the map $(x,y) \mapsto (0,kx)$ (for $k \in K \setminus \{0\}$) is not $A$-linear).
3) Use what you are given in the hint concerning $Z(A)$:
If $L(x,y) = \beta.(x,y)$ for some $\beta \in A$ then $A$-linearity means that:
$\alpha.L(x,y) = L(\alpha.(x,y))$ for ANY $\alpha \in A$
$\alpha.(\beta.(x,y)) = \beta.(\alpha.(x,y))$
$(\alpha\beta).(x,y) = (\beta\alpha).(x,y)$
so that $\alpha\beta = \beta\alpha$, that is: $\beta \in Z(A)$.
Note that you wind up with something *isomorphic* to $K$, not equal to it.
what has the centraliser got to do with surjectivity?Establishing surjectivity is essentially the "whole problem", because: we don't know much specifically about $\text{End}_A(K^2)$.
That is, it is hard to say what a "typical" $A$-linear endomorphism of $K^2$ looks like.
The fact that an isomorphic copy of $K$ lies within $Z(A)$ is trivial, because $A$ is a $K$-algebra.