Let R and C be the field of real numbers and of complex numbers, respectively, and $\displaystyle \theta$:R[X]->C be a ring homomorphism defined by $\displaystyle \theta$(f)=f(i). I want to show that the kernel is the ideal generated by $\displaystyle x^2 + 1$. I can prove that there is no linear polynomial in the kernel and that $\displaystyle x^2 + 1$ is the only quadratic in the kernel (up to associates of course), but I don't know how to show the kernel is actually $\displaystyle <x^2 + 1>R[X]$.

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 $\displaystyle f(x) = (x^2 + 1)g(x) + r(x)$. This gives that r(i)=0 and so r is in the kernel. But $\displaystyle deg(r) < deg(x^2 + 1)$ 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)

Any help is appreciated

EDIT: apparently in my book it says the division algorithm can be used in Euclidean Domains (with Euclidean function deg(f)), but wikipedia says it can be done in PIDs already. Can someone clarify? I know that for K a field K[X] is Euclidean, so this already solves my problem.