Is the Kernel of Trace Equal to the $F$-Subspace of $V_\sigma$?

[mathmari]

Hey!

Let $K/F$ be a finite Galois extension and let $G= \operatorname{Gal}(K/F)$.

1. For each $\sigma\in G$ we define $V_{\sigma}=\{\sigma (b)-b:b\in K\}$. Show that $V_{\sigma}$ is $F$-subspace of $\ker \operatorname{Tr}_{K/F}$.
2. Show that $K/\mathcal{F} (\langle \sigma \rangle) \cong V_{\sigma}$ (isomorphic as $F$-linear spaces).
3. Show that $\ker \operatorname{Tr}_{K/F}=V_{\sigma} \ \iff \ G=\langle \sigma \rangle$.

I have done the following:

1. • $V_{\sigma}$ must contain $0$ :
     
     The Galois group acts transitively on the roots of an irreducible polynomial. We have that $\sigma^i(b)=b$, so $\sigma^i(b)-b=b-b=0$, so $0\in V_{\sigma}$.
     
     $$$$
     
   • $V_{\sigma}$ must be closed under addition:
     
     $\left (\sigma(a)-a\right )+\left (\sigma(b)-b\right )=[\sigma(a)+\sigma(b)]-[a+b]=\sigma(a+b)-(a+b)\in V_{\sigma}$
     
     $$$$
     
   • $V_{\sigma}$ must be closed under scalar multiplication:
     
     Let $\lambda\in \ker \operatorname{Tr}_{K/F}$ and $\sigma(b)-b \in V_{\sigma}$, we have that $\lambda \cdot \left (\sigma(b)-b\right )=\lambda \sigma(b)-\lambda b=\sigma(\lambda b)-(\lambda b)\in V_{\sigma}$.
   
   Do we conclude in that way that $V_{\sigma}$ is $F$-subspace of $\ker \operatorname{Tr}_{K/F}$ ? (Wondering)
   
   $$$$
   
2. I have shown that $K/\mathcal{F} (\langle \sigma \rangle)$ is isomorphic to $V_{\sigma}$. $\checkmark$
   
   $$$$
   
3. Do we have to show at the one direction that $\ker Tr_{K/F}$ and $V_{\sigma} $ have the same dimension? (Wondering)

 

[I like Serena]

$\DeclareMathOperator{Tr}{Tr}$

Let $K/F$ be a finite Galois extension and let $G= \operatorname{Gal}(K/F)$.

1. For each $\sigma\in G$ we define $V_{\sigma}=\{\sigma (b)-b:b\in K\}$. Show that $V_{\sigma}$ is $F$-subspace of $\ker \operatorname{Tr}_{K/F}$.

I have done the following:

• $V_{\sigma}$ must contain $0$ :
  
  The Galois group acts transitively on the roots of an irreducible polynomial. We have that $\sigma^i(b)=b$, so $\sigma^i(b)-b=b-b=0$, so $0\in V_{\sigma}$.

Hey mathmari! (Wave)

What do you mean that 'the Galois group acts transitively on the roots of an irreducible polynomial'?
What is $\sigma^i$? Is it a power? It is not given that $G$ is cyclic is it?

We do have that $0\in K$, so that $\sigma(0)-0=0\in V_\sigma$ for any $\sigma\in G$, don't we? (Thinking)

• $V_{\sigma}$ must be closed under addition:
  
  $\left (\sigma(a)-a\right )+\left (\sigma(b)-b\right )=[\sigma(a)+\sigma(b)]-[a+b]=\sigma(a+b)-(a+b)\in V_{\sigma}$
• $V_{\sigma}$ must be closed under scalar multiplication:
  
  Let $\lambda\in \ker \operatorname{Tr}_{K/F}$ and $\sigma(b)-b \in V_{\sigma}$, we have that $\lambda \cdot \left (\sigma(b)-b\right )=\lambda \sigma(b)-\lambda b=\sigma(\lambda b)-(\lambda b)\in V_{\sigma}$.

Shouldn't we have $\lambda \in F$ if we want $V_\sigma$ to be an $F$-subspace? (Wondering)

Do we conclude in that way that $V_{\sigma}$ is $F$-subspace of $\ker \operatorname{Tr}_{K/F}$ ?

A subspace must also be a subset of the space mustn't it?

Is $V_\sigma \subseteq \ker\Tr_{K/F}$? (Wondering)

 

[mathmari]

$\DeclareMathOperator{Tr}{Tr}$

What do you mean that 'the Galois group acts transitively on the roots of an irreducible polynomial'?

I mean that an element of the Galois group maps a root of an irreducible polynomial to an other root. Is this not correct? (Wondering)

What is $\sigma^i$? Is it a power? It is not given that $G$ is cyclic is it?

Yes, I mean a power. So, the way I am thinking is not correct here, is it? (Wondering)

We do have that $0\in K$, so that $\sigma(0)-0=0\in V_\sigma$ for any $\sigma\in G$, don't we? (Thinking)

Ahh yes (Malthe)

Shouldn't we have $\lambda \in F$ if we want $V_\sigma$ to be an $F$-subspace? (Wondering)

Ahh ok. If we take $\lambda \in F$ is the remaining part correct as I did it? (Wondering)

A subspace must also be a subset of the space mustn't it?

Is $V_\sigma \subseteq \ker\Tr_{K/F}$? (Wondering)

So we have that $\dim V_\sigma \leq \dim \ker\Tr_{K/F}$. To show that $V_\sigma = \ker\Tr_{K/F} \iff G=\langle \sigma\rangle$ we have to show that $\dim V_\sigma = \dim \ker\Tr_{K/F} \iff G=\langle \sigma\rangle$, or not?

From the second question we have that $V_{\sigma}\cong K/\mathcal{F} (\langle \sigma \rangle) $. From that we get \begin{equation*}\dim V_{\sigma}=\dim \left (K/\mathcal{F} (\langle \sigma \rangle)\right ) \Rightarrow \dim V_{\sigma}=\dim K -\dim\mathcal{F} (\langle \sigma \rangle) \end{equation*}
So we want to show that \begin{equation*}\dim \ker\Tr_{K/F}=\dim K -\dim\mathcal{F} (\langle \sigma \rangle)\Leftrightarrow G=\langle \sigma\rangle\end{equation*}

Let $[K:F]=n$. Does this mean that $\dim K=n$ ? Which is the dimension of the kernel? (Wondering)

 

[I like Serena]

I mean that an element of the Galois group maps a root of an irreducible polynomial to an other root. Is this not correct?

Yes, I mean a power. So, the way I am thinking is not correct here, is it?

$\DeclareMathOperator{Tr}{Tr}\DeclareMathOperator{id}{id}$Yes, a root is mapped to a root.
$\sigma^i$ does not necessarily iterate through all the roots though.
For instance $\sigma=\id$ doesn't. $\id$ doesn't even map to 'another' root.

To verify that $0\in V_\sigma$ we want to find an element $b$ such that $\sigma(b)-b=0$.

As yet I don't see how $\sigma^i$ can help us.
$\sigma^i(b)-b$ is not necessarily an element of $V_\sigma$ is it?
After all, it it not of the form $\sigma(b)-b$. (Nerd)

Ahh ok. If we take $\lambda \in F$ is the remaining part correct as I did it?

Yep. (Nod)

So we have that $\dim V_\sigma \leq \dim \ker\Tr_{K/F}$. To show that $V_\sigma = \ker\Tr_{K/F} \iff G=\langle \sigma\rangle$ we have to show that $\dim V_\sigma = \dim \ker\Tr_{K/F} \iff G=\langle \sigma\rangle$, or not?

From the second question we have that $V_{\sigma}\cong K/\mathcal{F} (\langle \sigma \rangle) $. From that we get \begin{equation*}\dim V_{\sigma}=\dim \left (K/\mathcal{F} (\langle \sigma \rangle)\right ) \Rightarrow \dim V_{\sigma}=\dim K -\dim\mathcal{F} (\langle \sigma \rangle) \end{equation*}
So we want to show that \begin{equation*}\dim \ker\Tr_{K/F}=\dim K -\dim\mathcal{F} (\langle \sigma \rangle)\Leftrightarrow G=\langle \sigma\rangle\end{equation*}

Let $[K:F]=n$. Does this mean that $\dim K=n$ ? Which is the dimension of the kernel?

Not so fast. I think we first need to complete the first question. (Worried)
You have proven that $V_\sigma$ is a subspace of $K/F$, but we don't have any relation to $\ker\Tr_{K/F}$ yet, do we?
The dimensions only become relevant if $V_\sigma \subseteq \ker\Tr_{K/F}$.
That is the last part we need to prove that $V_\sigma$ is an $F$-subspace of $\ker\Tr_{K/F}$.

What does it mean that an element is in $\ker\Tr_{K/F}$? (Wondering)

 

[I like Serena]

$\DeclareMathOperator{Tr}{Tr}$Don't we have that:
$$\Tr_{K/F}(\alpha)=\sum_{g\in G}g(\alpha)$$
(Thinking)If $x\in V_\sigma$, then there must be a $b\in K$ such that $x=\sigma(b)-b$ and:
$$\Tr_{K/F}(x)=\sum_{g\in G}g(\sigma(b)-b) = \sum_{g\in G}g(\sigma(b)) - \sum_{g\in G}\sigma(b) = \Tr(\sigma(b))-\Tr(b)$$
Since the trace is not dependent on the choice of the basis, this is always $0$ isn't it? (Thinking)