I need clarification about a claim with reference to the dimension of a free module that is summand of a free module.

Question

I am trying to understand the claim that dim$_k(P)= 4|I|$ in the following proof.

Answer 1

It boils down to the following statement:

If $R$ is a $k$-vector space with a compatible multiplication that makes it a ring (altogether, a $k$-algebra), and if $M$ is a free $R$-module of rank $r$, then $M$ is a $k$-vector space, and its dimension as such is $$\dim_k(M)=r\dim_k(R)$$

which is more or less obvious from the definitions - you can set up a $k$-basis for $M$ into a rectangular fashion.

---

Edit: whenever an $R$-module $M$ is free, it means that it has a basis - that is, a family of elements of the module, say $ \{e_i\}_{i\in I}$, such that any $x\in M$ is an $R$-linear combination of the $e_i$'s in a unique way.

If $\{f_j\}_{j=1}^d$ is a $k$ basis of $R$ (as a $k$-vector space), you can then easily prove that $\{f_je_i\}$ is a $k$-basis of $M$, with $d|I|$ elements.