$\phi: \mathbb{R}[X] \to \mathbb{C}$ is a homomorphism such that: $\phi(X) = 1 + i$. What is $\ker \phi$?

Question

$\phi: \mathbb{R}[X] \to \mathbb{C}$ is a homomorphism such that: $\phi(X) = 1 + i$. What is $\ker \phi$?

In my thinking $\ker \phi = \{0\}$ because there is no way to add or multiply $1 + i$ in such way that it would be equal to $0$. Am I right?

Answer 1

Notice that if $p(x)=\sum_{j=0}^n a_jx^j$ then $\phi(p(x))=\sum_{j=0}^n a_j\phi(x)^j=\sum_{j=0}^n a_j(1+i)^j$, so $p(x)\in\ker(\phi)$ iff $1+i$ is a root of $p(x)$. But, since we are working with real polynomials, a complex number is a root of a polynomial iff its conjugate is also a root, so $1+i$ and $1-i$ are roots of $p(x)$. Then $p(x)\in\ker(\phi)$ iff $p(x)=q(x)(x-1-i)(x-1+i)=q(x)(x^2-2x+2)$ for some polynomial $q(x)$ iff $p(x)$ is in the ideal generated by $x^2-2x+2$. This proves that $\ker(\phi)=\langle x^2-2x+2\rangle$.

This is the most elementary solution one can get, but as suggested in the comments this is connected with the minimal polynomial. In general, if $\phi(x)=z\in\mathbb{C}$ the kernel will be the ideal generated by the minimal polynomial of $z$.