Kernel of trace is equal to V

In summary, the conversation discusses the properties of $V_\sigma$, which is defined as the set of elements in $K$ that are fixed by $\sigma\in G$. It is shown that $V_\sigma$ is an $F$-subspace of $\ker\Tr_{K/F}$ by proving that it contains $0$, is closed under addition and scalar multiplication. It is also shown that $V_\sigma$ is isomorphic to $K/\mathcal{F}(\langle\sigma\rangle)$. The conversation then continues to discuss the relationship between the dimensions of $V_\sigma$ and $\ker\Tr_{K/F}$, and whether $G=\langle\sigma\rangle$ is necessary for
  • #1
mathmari
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Hey! :eek:

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:

    • $V_{\sigma}$ 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)

    $$$$
  1. I have shown that $K/\mathcal{F} (\langle \sigma \rangle)$ is isomorphic to $V_{\sigma}$. $\checkmark$

    $$$$
  2. Do we have to show at the one direction that $\ker Tr_{K/F}$ and $V_{\sigma} $ have the same dimension? (Wondering)
 
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  • #2
$\DeclareMathOperator{Tr}{Tr}$
mathmari said:
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? :confused:

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)
mathmari said:
  • $V_{\sigma}$ 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)

mathmari said:
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)
 
  • #3
$\DeclareMathOperator{Tr}{Tr}$
Klaas van Aarsen said:
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)
Klaas van Aarsen said:
What is $\sigma^i$? Is it a power? It is not given that $G$ is cyclic is it? :confused:

Yes, I mean a power. So, the way I am thinking is not correct here, is it? (Wondering)
Klaas van Aarsen said:
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)
Klaas van Aarsen said:
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)
Klaas van Aarsen said:
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)
 
  • #4
mathmari said:
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)

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

Yep. (Nod)

mathmari said:
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)
 
  • #5
$\DeclareMathOperator{Tr}{Tr}$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)
 

1. What is the "Kernel of trace is equal to V" theorem?

The "Kernel of trace is equal to V" theorem, also known as the trace-nullity theorem, states that for a linear transformation T on a finite-dimensional vector space V, the dimension of the kernel of T (nullity) plus the dimension of the range of T (rank) is equal to the dimension of V.

2. What does the theorem tell us about the relationship between the kernel and the range of a linear transformation?

The theorem tells us that the dimensions of the kernel and the range are complementary, meaning that if one increases, the other decreases and vice versa.

3. How is the "Kernel of trace is equal to V" theorem useful in linear algebra?

The theorem is useful in linear algebra because it provides a way to determine the dimension of the kernel and the range of a linear transformation without having to explicitly find a basis for each. This can save time and effort in solving problems involving linear transformations.

4. Can the "Kernel of trace is equal to V" theorem be extended to infinite-dimensional vector spaces?

No, the theorem only holds for finite-dimensional vector spaces. In infinite-dimensional spaces, the dimensions of the kernel and the range may not be finite, making the theorem invalid.

5. Are there any real-world applications of the "Kernel of trace is equal to V" theorem?

Yes, the theorem has applications in fields such as physics, engineering, and computer science. It is used in solving problems involving linear systems and transformations, such as in circuit analysis, signal processing, and image compression.

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