$\text{End}_{\mathbb{Z}}\left(\prod_{i\in\mathbb{N}}\mathbb{Z}\right)$ is a free module over itself

Question

Denote the $\mathbb{Z}$-module $\prod_{i\in\mathbb{N}}\mathbb{Z}$ by $M$ and define $R:=\text{End}_{\mathbb{Z}}(M).$ We have two functions $\phi_1,\phi_2\in R$ satisfying $$\phi_1(a_1,a_2,a_3,\dots)=(a_1,a_3,a_5,\dots),\qquad\phi_2(a_1,a_2,a_3,\dots)=(a_2,a_4,a_6,\dots).$$

Claim. $\{\phi_1,\phi_2\}$ is a free basis of the left $R$-module $R$.

Hint. Define maps $$\psi_1(a_1,a_2,a_3,\dots)=(a_1,0,a_2,0,\dots),\qquad\psi_2(a_1,a_2,a_3,\dots)=(0,a_1,0,a_2,\dots)$$ and show $\phi_i\psi_i=1,~\phi_1\psi_2=0=\phi_2\psi_1,~\psi_1\phi_1+\psi_2\phi_2=id$.

Do we mean $\phi_i\psi_i=\phi_i\circ\psi_i$? If yes, then the identities hold! And what is a free basis?

So we need two things to show:

1. For every $f\in R$ we find functions $f_1,f_2\in R$ such that $$f=f_1\phi_1+f_2\phi_2.$$
2. $f_1\phi_1+f_2\phi_2=0\Rightarrow f_1=f_2=0\quad\forall f_1,f_2\in R.$

Then the left $R$-module $R$ has a basis and is free. I guess this is meant by free basis.

Any ideas how I can prove the claim?

Thanks you.

Answer 1

Yes, they mean $f\circ g$ when they write $fg$, and you are correct about what is meant by a "free basis."

For (1): '

Given that $\psi_1\circ\phi_1 + \psi_2\circ\phi_2=\mathrm{id}_M$ then $f=f\circ \mathrm{id}_M = f\circ(\psi_1\circ\phi_1 + \psi_2\circ\phi_2)$.

Proceed from there.

For (2):

$$f_1\circ\phi_1+f_2\circ\phi_2=0$$ implies $$(f_1\circ\phi_1+f_2\circ\phi_2)\circ\psi_i=0$$ Proceed from there.

The point to this exercise is that every ring $R$ has a free basis of dimension $1$ when considered as an $R$-module - namely, $\phi_1=1$.

This exersize shows a difference between vector spaces and modules. Modules are a generalization of vector spaces, and we want to see where modules differ from vector spaces.

We have shown that given an $R$-module, In vector spaces, all bases must have the same cardinality (size,) but in $R$-modules, we see that sometimes we have two bases of different sizes.

(It is also true that sometimes we don't have a basis at all for an $R$-module - that having a basis is equivalent to being a free $R$-module.)