Let R and C be the field of real numbers and of complex numbers, respectively, and $\theta$:R[X]->C be a ring homomorphism defined by $\theta$(f)=f(i). I want to show that the kernel is the ideal generated by $x^2 + 1$. I can prove that there is no linear polynomial in the kernel and that $x^2 + 1$ is the only quadratic in the kernel (up to associates of course), but I don't know how to show the kernel is actually $R[X]$.
I have to show that for any other polynomial f in the kernel, the quadratic divides f, but how? I know that R[X] is a PID so I'm sure it's of that form. I've tried using division with remainder so that for f in the kernel $f(x) = (x^2 + 1)g(x) + r(x)$. This gives that r(i)=0 and so r is in the kernel. But $deg(r) < deg(x^2 + 1)$ or r=0. Since no linear or constant polynomial is in the kernel, then r=0. But the problem is: am I allowed to use division with remainder? (I'm not sure in what domain I can start using division with remainder)