$ \newcommand{\cn}{\colon} \newcommand{\<}{\leqslant} $ Let $k$ be a field.
Def. Algebra $A$ is associative ring with unity and a ring homomorphism $f\cn k\to A$ such that $f(1_k)=1_A$ and $f(a)x=xf(a)$ for all $x\in A$.
Def. $A$-module $V$ is a vector space over k with a bilinear map $(A,V)\to V,~(a,v)\mapsto av$, such that $1v=v$ and $a(bv)=(ab)v$ for all $a,b\in A$ and $v\in V$.
Let $V_1$ be $A_1$-module and $V_2$ be $A_2$-module. I want to prove or disprove that any $(A_1\times A_2)$-module $V_1\times V_2$ is such that $(a_1,a_2)(v_1,v_2)=(a_1v_1,a_2v_2)$ (exercise 3 after chapter 1 from Kassel's Quantum Groups). Structures on products are defined elementwise here.
Using bilinearity I obtain decomposition $(a_1,a_2)(v_1,v_2)=(a_1,0)(v_1,0)+(a_1,0)(0,v_2)+(0,a_2)(v_1,0)+(0,a_2)(0,v_2)$. So I've come to four simplier cases. Another thing I have is that $V=V_1\times V_2=U_1\oplus U_2$, where $U_1=\left\{v\in V\cn (1,0)v=v\right\},~U_2=\left\{v\in V\cn (0,1)v=v\right\}$ are subspaces of $V$. Also I see that $(a_1,0)v\in U_1$ and $(a_2,0)v\in U_2$ for all $v\in V,~a_1\in A_1,~a_2\in A_2$. Any hints would be appreciated.
Edit. The exact statement is
Let $A_1,A_2$ be some algebras, $V_1$ an $A_1$-module, and $V_2$ an $A_2$-module. Show that the equality $(a_1,a_2)(v_1,v_2)=(a_1v_1,a_2v_2)$ defines the $A_1\times A_2$-module structure at $V_1\times V_2$. Prove that any $A_1\times A_2$-module has this form.