Question about a homomorphism of a cyclic extension

Question

I came across the following problem, and I'm unsure about how to prove a specific part regarding a particular homomorphism of a cyclic extension of a field. Here's the problem:

Let $K_{/F}$ be a cyclic extension of fields of degree $n$ with $\text{Gal}(K_{/F}) = \langle \sigma \rangle$. Prove the following:

a) If $f(X) =a_m X^m + a_{m-1}X^{m-1} + \dots + a_0 \in F[X] $, define the map $f(\sigma): K \longrightarrow K$ by $b \mapsto a_m \sigma^m(b) + a_{m-1} \sigma^{m-1}(b) + \dots + a_0$ (where $\sigma^i$ denotes $\sigma$ composed with itself $i$ times). Prove $f(\sigma)$ is an $F$ module homomorphism.

b) Prove if $f(\sigma) \equiv 0$, and $\deg(f) < n$, then $f \equiv 0$.

c) Prove there exists $b \in K$ such that $\{b,\sigma(b),\dots,\sigma^{n-1}(b)\}$ is basis of $K$ over $F$.

I have two questions about this problem.

1. I don't get how $f(\sigma)$ is an $F$ module homomorphism if $a_0 \neq 0$. We require $f(\sigma)(0) = 0$, but $f(\sigma)(0) = a_m \sigma^{m}(0) + \dots + a_0 = a_0 = 0$ since $\sigma(0)=0$.

2. I don't get how to prove part b without using part c. Part b is immediate if part c is true, and so I tried to figure out a way to prove c without using b, but it's clear to me from the problem that part b is there to establish some sort of linear independence argument for part c to hold.

Here's my work for part b (not using part c): Suppose for contradiction that $f \not\equiv 0$ (so there exists $\alpha \in K$ with $f(\alpha) \neq 0$). Then not all of the coefficients on $f$ are $0$. I know that for the $\alpha$ such that $f(\alpha) \neq 0$, we also get that $f(\sigma)(\alpha) = 0$, and if $Y = \{\beta_1,\beta_2,\dots\beta_n\}$ is a basis of $K$ over $F$, expanding out $f(\sigma)(\alpha)$ gives you something like: $$f(\sigma)(\alpha) = \sum_{k=0}^m \sum_{i=1}^n a_k t_{i,k} \beta_i = 0$$ where $\sigma^j(\alpha) = \sum_{i=1}^n t_{i,j}\beta_i$ for each $1 \leq j \leq m$ ($t_{i,j} \in F$). Since $Y$ is a basis you get $a_{k}t_{i,k} = 0$ for all $i$. This looks promising since not all the coefficients on $f$ are $0$, so it looks like it might be a non-trivial LC of the elements of $Y$ equating to $0$, which is a contradiction. However, this somehow depends on $\deg(f) =m < n$, and the problem i see here is that it may be possible that $a_kt_{i,k} = 0$ for all $i$, but this doesn't imply that $t_{i,k}=0$ for all $i$, since it may be that $a_k=0$ whenever $t_{i,k} \neq 0$ and vice versa.

What do you do from here?

Answer 1

Parts of my answer are hidden to encourage you to try it yourself first. You can click on those hidden blocks to reveal them.

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Part (a)

In your statement, the map $f(\sigma)(b)$ is defined as $$ f(\sigma)(b)=a_m\sigma^m(b)+\cdots+a_1\sigma(0)+a_0, $$ but actually it should be $$ f(\sigma)(b)=a_m\sigma^m(b)+\cdots+a_1\sigma(b)+a_0\color{red}{\sigma^{0}(b)}, $$ where $\sigma^0$ is the identity map, so $\sigma^0(x)=x$ for any $x\in K$, and in particular $\sigma^0(0)=0$. Then we have $f(\sigma)(0)=0$, as expected.

An actual proof should not be difficult now:

Since $F$ is a field, an $F$-module homomorphism is simply an $F$-linear map. For all $c\in F$ and all $v,w\in K$, $$\begin{align*}f(\sigma)(cv+w) &= \sum_{i=0}^m a_i\sigma^i(cv+w) \\&= \sum_{i=0}^m a_i (c\cdot \sigma^i(v) + \sigma^i(w)) \tag{$\sigma\in\text{Gal}(K/F)$} \\&= \sum_{i=0}^m ca_i\sigma^i(v) + \sum_{i=0}^m a_i\sigma^i(w) \\&= c\cdot f(\sigma)(v) + f(\sigma)(w).\end{align*}$$ This shows that $f(\sigma)$ is an $F$-linear map.

Part (b)

It suffices to show that $\{\text{id},\sigma,\dots,\sigma^{n-1}\}$ is $F$-linearly independent. We can proceed by induction on $n$.

Hint:

Use the fact that $\sigma\neq \sigma^{n-1}$ to find an element $x_0\in K$ such that $\sigma(x_0)\neq \sigma^{n-1}(x_0)$. Plug in $x_0x$ into the equation and try to cancel out $\sigma^{n-1}(x)$ term.

Solution:

This is trivial for $n=1$. For $n>1$, assume that $\{\text{id},\sigma,\dots,\sigma^{n-2}\}$ is linearly independent. Suppose $$\sum_{i=0}^{n-1} a_i \sigma^i(x)=0$$ for all $x\in K$. Now, there exists $x_0\in K$ such that $\sigma(x_0)\neq \sigma^{n-1}(x_0)$. We have two equations, for all $x\in K$: $$\begin{align*}\sum_{i=0}^{n-1} a_i \sigma^i(x_0)\sigma^i(x) &=0 \tag{Plug in $x_0x$}\\\sum_{i=0}^{n-1} a_i \sigma^{n-1}(x_0)\sigma^i(x) &= 0 \tag{Multiply by $\sigma^{n-1}(x_0)$}\end{align*}$$ By a substraction, we have $$\begin{align*}\sum_{i=0}^{n-2} a_i (\sigma^{n-1}(x_0)-\sigma^i(x_0))\sigma^i(x) &=0\end{align*}$$ for all $x\in K$. By the linear independence of $\{\text{id},\dots,\sigma^{n-2}\}$ and the fact that $\sigma^{n-1}(x_0)-\sigma^i(x_0)\neq 0$, we have $a_i=0$ for all $1\le i\le n-2$, which implies $a_{n-1}=0$ as well.

Part (c)

We have already shown that $\{\text{id},\dots,\sigma^{n-1}\}$ is $F$-linearly independent. It remains to show that the set spans $K$. This is equivalent to proving that $K$, when viewed as an $F[X]$-module via $\sigma$, is a cyclic module.

Hint:

Use (or prove) that if the minimal polynomial of $\sigma$ equals the characteristic polynomial of $\sigma$, then the list of invariant factors has length $1$.

Solution:

Note that $\sigma^n-\text{id}=0$, so $X^n-1$ is divisible by the minimal polynomial of $\sigma$, but by part (b), no polynomial with degree less than $n$ can annihilate all of $K$, so the minimal polynomial is $X^n-1$. Now, the characteristic polynomial is divisible by $X^n-1$ and has degree $[K:F]=n$, so the characteristic polynomial equals the minimal polynomial. Since the characteristic polynomial is the product of all invariant factors, and the minimal polynomial is the largest invariant factor, the only possibility is that there is only one invariant factor, which shows that $K$ is a cyclic module.