homomorphisms and product rings The problem is this: Let $f:\mathbb{R}[x]\rightarrow \mathbb{C}\times \mathbb{C}$ be the homomorphism defined by $f(x)=(1,i)$ and $f(r)=(r,r)$, for $r\in \mathbb{R}.$ Determine the kernel and the image of $f.$
Solution:
Let $K=\ker f=\{p(x)=\sum_{k=0}^na_kx^k\in R[x]: f(p(x))=(0,0)\}$
After a few steps I got:  $f(p(x))=\left(\sum_{k=0}^na_k,\sum_{k=0}^n(i)^ka_k\right)=(0,0)$
So then $p(x)$ is in the kernel of $f$ if its coefficients satisfy the following conditions:
$\sum_{k=0}^na_k=0$
$\sum_{k=0}^{\lfloor{n}\rfloor}(-1)^ka_{2k}=0$    $\hspace{1cm}$ and
$\sum_{k=0}^{\lceil{n}\rceil}(-1)^{k-1}a_{2k-1}=0$.
And the image of $f$ would consist of all the pairs (a,b+ic). That is the real part of the first 'coordinate' would be zero. 
Am I on the right track, or am I missing something?
Any suggestions are appreciated!
 A: Let $\alpha = (1, i)$.
Then $\alpha^2 = (1, -1)$
We first show that $(1, 1), \alpha, \alpha^2$ are linearly independent over $\mathbb{R}$.
Suppose $a(1, 1) + b\alpha + c\alpha^2 = 0$, where $a, b, c \in \mathbb{R}$.
$a + b + c = 0$
$a + bi - c = 0$
Then $b = 0, a = c$.
Hence $a = b = c = 0$ as desired.
Since $\alpha^3 = (1, -i)$, $\alpha^3 = (1, 1) - \alpha + \alpha^2$
Hence $x^3 - x^2 + x - 1$ is the minimal polynomial of $\alpha$ over $\mathbb{R}$.
So the kernel is $(x^3 - x^2 + x - 1)$.
The image of the honomorphism is $I = \{a(1, 1) + b\alpha + c\alpha^2; a, b, c \in \mathbb{R}\}$.
We show that $I = \mathbb{R}\times\mathbb{C}$.
Clearly $I \subset \mathbb{R}\times\mathbb{C}$.
So it suffices to prove the other inclusion.
Let $(x, z) \in \mathbb{R}\times\mathbb{C}$.
Suppose $(x, z) = a(1, 1) + b\alpha + c\alpha^2 = (a + b + c, a + bi -c)$.
Then $x = a + b + c, z = a + bi -c$
Suppose $z = p + qi$, where $p, q \in \mathbb{R}$.
Then $p = a - c, q = b$.
Hence $x = a + c + q$
The following equations determine $a$ and $c$ by $x$ and $p$ and $q$.
$p = a - c$
$x - q = a + c$
Hence we are done.
Remark
Since dim$_\mathbb{R}(\mathbb{R}\times\mathbb{C}) = 3$, it is also clear that $I = \mathbb{R}\times\mathbb{C}$.
A: Note that $f(p)=\langle p(1),p(i)\rangle$, so $p\in\ker f$ iff $p(1)=p(i)=0$, where I’m treating $f$ as a polynomial over $\Bbb C$ with real coefficients. This is essentially what you found, expressed a little more compactly, but it lets you go a little further. Suppose that $p\in\ker f$. Since $p(1)=0$, $x-1$ is a factor of $p$ in $\Bbb R[x]$. Moreover, complex roots of real polynomials come in conjugate pairs, so $p(-i)=0$, and it follows that $(x-1)(x-i)(x+i)=(x-1)(x^2+1)$ is a factor of $p$ in $\Bbb C[x]$. But $(x-1)(x^2+1)\in\Bbb R[x]$, so for each $p\in\ker f$ there is a $g\in\Bbb R[x]$ such that $f(x)=(x-1)(x^2+1)g(x)$. Prove the converse of this, and you’ll have a nice characterization of $\ker f$.
Note that while it’s clear that $\operatorname{ran}f\subseteq\Bbb R\times\Bbb C$, it’s not instantly clear that you have equality; you actually need to show that for any $\langle r,s+ti\rangle\in\Bbb R\times\Bbb C$ there is a $p\in\Bbb R[x]$ such that $p(1)=r$ and $p(i)=s+ti$. This isn’t too hard: you can do it with a first-degree polynomial.
