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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 $f.$


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:


$\sum_{k=0}^{\lfloor{n}\rfloor}(-1)^ka_{2k}=0$ $\hspace{1cm}$ and


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. – Tobias Kildetoft Sep 3 '12 at 7:05
@ Tobias: I forgot to include that part: I came up with the conclusion that the kernel of $f$ should be generated by $(x^3-x^2+x-1)$. – Alexander Sep 3 '12 at 7:32

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.

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@ Brian M. Scott: for the kernel I found essentially what you found after a little more work that is $\ker f=(x^3-x^2+x-1)$. For the ran of f i have already shown the equality too. – Alexander Sep 3 '12 at 7:36
@Alexander: I wasn’t sure, since you didn’t mention either in the question. In that case you have a complete solution to the problem. – Brian M. Scott Sep 3 '12 at 7:40
Thank you for your suggestions...they definitely helped me write a neater solution! – Alexander Sep 3 '12 at 7:46

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 $I = \mathbb{R}\times\mathbb{C}$.

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