Let $k$ be a field and $V$ a $k$-vector space of numerable infinite dimension.
Let $E=End_k(V)$. For each $n\geq 1$, $E\simeq E^n$
Let $E=End_k(V)$. For each $n\geq 1$, as $E$-module, $E$ has a basis of cardinality $n$.
It's clear for me that $E$ is a $k$-vector space, but I can't see why the basis of $E$ will be infinite numerable. What I want to use is the fact that $V\simeq V^n$ when $\dim(V)$ is infinite numerable.
I know this, but I can't finish:
$$End_k(V)=Hom(V,V)\cong Hom(V,V^2)\cong Hom(V,V^n)$$
For the other case the problem says that If is an $E$-module, it is a free module of rank $n$, but that seems clear to me by definition of $E$-module. What I have to prove there so?
Could anyone give me a hint or explain this to me?
The idea is that the abelian group isomorphism $\hom(V_k,V_k\oplus V_k)\cong \hom(V_k, V_k)\oplus \hom(V_k,V_k)$ induced by an isomorphism $V_k\to V_k\oplus V_k$ turns out to be $E$ linear. The following proof appears in Lam's Lectures on Modules and Rings page 4 (I cannot really improve upon it):
If you still feel uncertain about it, you can try to compute the basis of $R$ with two elements using the isomorphism of $V\to V^2$ that identifies a copy of $V$ as generated by the even basis elements of $V$ and another copy generated by the odd basis elements.
Obviously by induction, once you have $E_E\cong E^2_E$, then you get $E_E\cong E^n$ for any positive integer $n$.