$\det(\varphi + \psi)$ of 2 A-endomormphisms $\varphi,\psi: M \to M$ with A: commutative ring with unit, M: free A-module or rank 2

Question

As is said in the title, we consider A to be a commutative ring with unit, M a free A-module of a finite rank r = 2, and $\varphi$,$\psi: M \to M$ A-endomorphisms. With this data I would like to show that

$$det(φ+ψ) + det(φ−ψ) = 2(det(φ) + det(ψ))$$

However the only thing related to this that came to my mind was the fact that $det(λφ) =λ^2det(φ)$ for $\lambda \in A$, but I'am not sure it could be usefull here.

Also, after some research I think that this problem can be shown using external product, but I have not seen it. Is there another way to show this? I just want to find where to starts, and know wether I have seen enough in class to solve this problem or not.

Thanks for your help!

Answer 1

Let $u_1,u_2$ be the free generators of $M$, then every element of $M$ can be uniquely expressed as $\alpha_1u_1+\alpha_2u_2$ with $\alpha_1,\alpha_2\in A$.

Also, any homomorphism from $M$ is determined by the images of $u_1$ and $u_2$ which can be arbitrary.

So, an endomorphism $\varphi:M\to M$ can be encoded in a $2\times 2$ matrix with entries in $A$, namely let $$[\varphi]=\pmatrix{\alpha&\beta\\ \gamma&\delta}$$ mean that $\varphi(u_1)=\alpha u_1+\gamma u_2$ and $\varphi(u_2)=\beta u_1+\delta u_2$.

Then $\det\varphi=\alpha\delta-\beta\gamma$, and addition of endomorphisms simply correspond to addition of matrices.

We simply open up the brackets here: $$\det\pmatrix{\alpha+\alpha'&\beta+\beta'\\ \gamma+\gamma'&\delta+\delta'}\ +\ \det\pmatrix{\alpha-\alpha'&\beta-\beta'\\ \gamma-\gamma'&\delta-\delta'}\ = $$ $$=\ (\alpha+\alpha')(\delta+\delta')-(\beta+\beta')(\gamma+\gamma')\ +\ (\alpha-\alpha')(\delta-\delta')-(\beta-\beta')(\gamma-\gamma')\ =\ \dots $$