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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.

Vikas Sharma
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$A=\mathbb{C\times C}$ 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 ?

Maxime Ramzi
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  • If i take p=q=1 and p'=q'=2 then is it so that Apq has elements of form (z,z) while Ap'q' has elements of form (z,z¯) and then if id is the identity homomorphism then id((1+2i,1+2i)+(2+3i,2+3i))=id((3+5i,3+5i))=(3+5i,3+5i) but (3+5i,3+5i) is not in Ap'q' so that id is not the homomorphism? Moreover what are those C-module structures and how to find them? Thanks for the help. – Vikas Sharma Jul 06 '19 at 16:01
  • No, the elements of $A_{pq}$ and $A_{p'q'}$ are the same (they are the same as those of $A$) and in fact the identity is a group morphism between them, so you have to use the $\mathbb C$-module structure to show that it's not a morphism. The first paragraph of my answer describes these structures ! – Maxime Ramzi Jul 06 '19 at 16:13
  • Perhaps i came to know the correct thing,as if f(2i•(1+i,2-i))=f((2i,2i)(1+i,2-i))=f((2i-2,-4))=(2i-2,-4). But 2i•f((1+i,2-i))=2i•(1+i,2-i)=(-2i,-2i)(1+i,2-i)=(-2i+2,4) which shows that f is not a C-module homomorphism. We can have many as such examples,if it is correct. Am i done correctly? – Vikas Sharma Jul 06 '19 at 17:29
  • Yes, you are correct. In principle you would have to do the same for other cases for $pq$ and $p'q'$ (you only did $p=q=1,p'=q'=2$) but you should by now convinced that the result holds and that you could find an example if you wanted to – Maxime Ramzi Jul 06 '19 at 17:35
  • Ok! Yes, thanks a lot for the help. – Vikas Sharma Jul 06 '19 at 17:37