Lemma Let $R_{1}=\mathbb{R}[x]$ and let $R_{2}=\mathbb{C}$. Define $\phi:R_{1} \to R_{2}$ as follows then $\phi(f(x))=f(i)$(i.e. $\phi$ evaluates $f(x) \in \mathbb{C}[x]$ at $x=i$) then $\phi$ is a [[Homomorphism of Rings]]. **Proof** Clearly $\phi(1)=1$ (i.e you evaluate the polynomial 1 at any complex number you just get the number 1). We also have $\phi(f(x)+g(x))=f(i)+g(i)=\phi(f(x))+\phi(g(x))$ and $\phi(f(x)g(x))=f(i)g(i)=\phi(f(x))\phi(g(x)).$ So $\phi$ is a homorphism