Why does $\rho_{\mathbf{Z}^t\otimes P}=\rho_R$ imply the isomorphsim of $\mathbf{Z}^t\otimes P\cong R$?

Question

so I have happened upon a thesis regarding the calculations of various $\mathbf{Z}D_6$ modules and their isomorphisms and came across a technique which is bothering me. Let me give an example. Let $A=\mathbf{Z}D_6$ where $D_6=<1,x,y\:|\:x^3=y^2=1, yx=x^2y>$ and let $\mathbf{Z}^t$ be the module where $x$ acts trivially and $y$ acts by -1. Further, let $R=\mathbf{Z}[\omega]=\mathbf{Z}[x]/(x^2+x+1)$, i.e. the Eisenstein integers, and let $P=(1-\omega)R$. I want to show $\mathbf{Z}^t\otimes P\cong R$. The argument is as follows: Since $\rho_{\mathbf{Z}^t\otimes P}(g)=\rho_R(g) \forall g\in D_6$, it follows that $\mathbf{Z}^t\otimes P\cong R$. I want to know why the equality of representations implies an isomorphism since we are working over the integers and not an algebraically closed field.