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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!

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    Some hints that might help describe the kernel better: $R[x]$ is a PID, so any ideal is generated by a single element. Thus, if you can find an element in the kernel of smallest possible degree, you have your generator.2012-09-03
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    @ Tobias: I forgot to include that part: I came up with the conclusion that the kernel of $f$ should be generated by $(x^3-x^2+x-1)$.2012-09-03

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