Let A be the direct product ring C×C. Let τ1 denote the identity map on C and let τ2 denote complex conjugation. For any pair p, q ∈ {1, 2} define fp,q : C → C × C by fp,q(z) = (τp(z), τq(z)). Prove that if fp,q≠fp',q' then the identity map on A is not a C-module homomorphism, from A (considered as a C-module via fp,q), to A (considered as a C-module via fp',q'). Also prove that A has four distinct C-module structures.
I have been stucked here as i am not knowing how to define the identity map using fp,q and fp',q' and proving that it is not a homomorphism. Also I don't know what's the use of mentioning that A is a C module via fp,q or fp',q' , may be it is useful in proving the result.
$A=\mathbb{C\times C}$ is a ring, and so is $\mathbb C$. Now when you are given a ring $A$ and a ring morphism $f:\mathbb C\to A$, you get a $\mathbb C$-module structure on $A$ via $z\cdot a := f(z)a$ where I omit the symbol for multiplication in $A$ and $\cdot$ denotes the action.
Here you are given two distinct morphisms $\mathbb C\to A$ : $f_{pq}$ and $f_{p'q'}$, so you get two a priori distinct $\mathbb C$-module structures on $A$: let's denote them $A_{pq}$ and $A_{p'q'}$. As abstract abelian groups, they are the same, but they have a different $\mathbb C$-module structure.
Then, on an "unrelated" note, you have a map $id_A : A\to A$. This map is also a map $A_{pq}\to A_{p'q'}$ (as sets, even as groups, $A_{pq}=A, A_{p'q'}=A$ so that makes sense).
The exercise asks you to show that this map is not a $\mathbb C$-module morphism. Is that clearer now ?
Can you have a try with these explanations ?