I am working on the following problem :
Let $K$ denote the algebraic closure of the 5 element field $F_5$. Let $f: F_5[x] \longrightarrow K$ be the ring homomorphism determined by $f(x) = \omega$, where $\omega$ is a primitive 6th root of 1 in $K$.
a) What is $f(3x^{50} - 2x^{45} +2)$ ?
b) Find the dimension, over $F_5$, of the image of $f$.
c) Find the kernel of $f$.
I think I've got a good grasp on part a). We use basic properties that any ring homomorphism satisfies:
$f(3x^{50} - 2x^{45} + 2) = f(3x^{50}) - f(2x^{45}) + f(2)$
= $3[f(x)]^{50} - 2[f(x)]^{45} + f(2)$
= $3(\omega^{50}) - 2(\omega^{45}) + f(2)$
= $3(\omega^2) - 2(\omega^3) + f(2)$ (using that $\omega^6 = 1$)
= $3 \cdot \omega^2 - 2 \cdot \omega^3$
My only question here is, am I justified in saying $f(2) = 0$, since the homomorphism is determined completely by $f(x) = \omega$ ?
b) I'm not sure about this part. I thought of using the rank-nullity theorem, but since $F_5[x]$ is not finite-dimensional, I don't know of a clever way to use this to determine the dimension of the image. Is there some other way here to determine the dimension of the image?
c) I know of the identity $1 + \omega + \omega^2 + ... + \omega^{n-1} = 0$, where $\omega$ is an nth root of unity. But, how can I take $f$ of an element in $F_5[x]$ and obtain $1 + \omega + ... + \omega^5$, if I don't know what $f(1)$ is ? I can only deduce what $f(0)$ is, since a ring homomorphism sends identity elements to identity elements. Also, after, if possible, I can find what element in $F_5[x]$ gets sent to $1 + \omega + ... + \omega^5 = 0$, is there any other element of $F_5[x]$ that can get send to $0$ by $f$, besides $0$ ? Or is the kernel only consisting of $0$ and the element mapping to $1 + \omega + .. + \omega^5$?
Thanks!
You seem to be a little confused when it comes to what $f$ does. $f$ does not send any element of $\mathbb{F}_5$ to $0$. What $f$ does, is take a polynomial in $\mathbb{F}_5[X]$ and evaluate that polynomial in $\omega$.
The first thing to do when you solve this kind of exercise, is to find the minimal polynomial of $\omega$ over $\mathbb{F}_5$. Since $\omega$ is a primitive $6$'th root, this must be the sixth Cyclotomic polynomial, $x^2-x+1$.
From the fact that the degree of this minimal polynomial is $2$, we may conclude that the powers of $\omega$ can be written as linear combinations of $1$ and $\omega$. Since $\omega\not\in\mathbb{F}_5$, these two are clearly linearly independent and the dimension of $\operatorname{im}f=\mathbb{F}_5(\omega)$ as a vectorspace over $\mathbb{F}_5$ is $2$. (In general, for algebraic field extensions $K\subset K(\alpha)$, the dimension of $K(\alpha)$ as a vectorspace over $K$ is the degree of the minimal polynomial of $\alpha$ over $K$)
The kernel of $f$ are all polynomials which are zero when evaluated in $f$, so all a polynomials with $\omega$ as a root. These are precisely the polynomials divisible by the minimal polynomial of $\omega$, so $$\ker f = (x^2-x+1),$$ the principal ideal generated by $x^2-x+1$.