Example 6.18. from the book of Algebra by Aluffi
The question is from (Chapter VII) Aluffi's Algebra:
Example 6.18. : Let $k$ be any field of characteristic zero. The splitting field of $x^n−1$ over $k$ is the composite $k(ζ)$ of $k$ and $mathbb{Q}(ζ)$. By Proposition 6.17 the extension $k ⊆ k(ζ)$ is Galois, and $Aut_k(k(ζ))$ is isomorphic to a subgroup of the group of units of $mathbb{Z}/nmathbb{Z}$.
Proposition 6.17. Suppose $k ⊆ F$ is a Galois extension, and $k ⊆ K$ is any finite extension. Then $K ⊆ KF$ is a Galois extension, and $Aut_K(KF) cong Aut_{F∩K}(F)$.
I am confused how Proposition 6.17. is used in Example 6.18. :
i. First of all, about using different notations: $k$ as a field of characteristic zero in Example 6.18 is $K$ is Proposition 6.17.; and probably (?) $mathbb{Q}$ in Example 6.18 is $k$ is Proposition 6.17.; what about the rest?
ii. If I am right with notations, $KF$ corresponds to $k(ζ)$, but how $kmathbb{Q}(ζ)=k(ζ)$? What does "the composite $k(ζ)$ of $k$ and $mathbb{Q}(ζ)$" mean?
iii. Letting the considered field of characteristic zero be $mathbb{Z}$. Could you re-explain Example 6.18. for this specific case $k=mathbb{Z}$? (esp how $mathbb{Q}$ can be extended to $mathbb{Z}$?(!)
abstract-algebra group-theory galois-theory proof-explanation
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The question is from (Chapter VII) Aluffi's Algebra:
Example 6.18. : Let $k$ be any field of characteristic zero. The splitting field of $x^n−1$ over $k$ is the composite $k(ζ)$ of $k$ and $mathbb{Q}(ζ)$. By Proposition 6.17 the extension $k ⊆ k(ζ)$ is Galois, and $Aut_k(k(ζ))$ is isomorphic to a subgroup of the group of units of $mathbb{Z}/nmathbb{Z}$.
Proposition 6.17. Suppose $k ⊆ F$ is a Galois extension, and $k ⊆ K$ is any finite extension. Then $K ⊆ KF$ is a Galois extension, and $Aut_K(KF) cong Aut_{F∩K}(F)$.
I am confused how Proposition 6.17. is used in Example 6.18. :
i. First of all, about using different notations: $k$ as a field of characteristic zero in Example 6.18 is $K$ is Proposition 6.17.; and probably (?) $mathbb{Q}$ in Example 6.18 is $k$ is Proposition 6.17.; what about the rest?
ii. If I am right with notations, $KF$ corresponds to $k(ζ)$, but how $kmathbb{Q}(ζ)=k(ζ)$? What does "the composite $k(ζ)$ of $k$ and $mathbb{Q}(ζ)$" mean?
iii. Letting the considered field of characteristic zero be $mathbb{Z}$. Could you re-explain Example 6.18. for this specific case $k=mathbb{Z}$? (esp how $mathbb{Q}$ can be extended to $mathbb{Z}$?(!)
abstract-algebra group-theory galois-theory proof-explanation
asked 12 hours ago
72D
572116
add a comment |
The question is from (Chapter VII) Aluffi's Algebra:
Example 6.18. : Let $k$ be any field of characteristic zero. The splitting field of $x^n−1$ over $k$ is the composite $k(ζ)$ of $k$ and $mathbb{Q}(ζ)$. By Proposition 6.17 the extension $k ⊆ k(ζ)$ is Galois, and $Aut_k(k(ζ))$ is isomorphic to a subgroup of the group of units of $mathbb{Z}/nmathbb{Z}$.
Proposition 6.17. Suppose $k ⊆ F$ is a Galois extension, and $k ⊆ K$ is any finite extension. Then $K ⊆ KF$ is a Galois extension, and $Aut_K(KF) cong Aut_{F∩K}(F)$.
I am confused how Proposition 6.17. is used in Example 6.18. :
i. First of all, about using different notations: $k$ as a field of characteristic zero in Example 6.18 is $K$ is Proposition 6.17.; and probably (?) $mathbb{Q}$ in Example 6.18 is $k$ is Proposition 6.17.; what about the rest?
ii. If I am right with notations, $KF$ corresponds to $k(ζ)$, but how $kmathbb{Q}(ζ)=k(ζ)$? What does "the composite $k(ζ)$ of $k$ and $mathbb{Q}(ζ)$" mean?
iii. Letting the considered field of characteristic zero be $mathbb{Z}$. Could you re-explain Example 6.18. for this specific case $k=mathbb{Z}$? (esp how $mathbb{Q}$ can be extended to $mathbb{Z}$?(!)
abstract-algebra group-theory galois-theory proof-explanation
asked 12 hours ago
72D
572116
The question is from (Chapter VII) Aluffi's Algebra:
Example 6.18. : Let $k$ be any field of characteristic zero. The splitting field of $x^n−1$ over $k$ is the composite $k(ζ)$ of $k$ and $mathbb{Q}(ζ)$. By Proposition 6.17 the extension $k ⊆ k(ζ)$ is Galois, and $Aut_k(k(ζ))$ is isomorphic to a subgroup of the group of units of $mathbb{Z}/nmathbb{Z}$.
Proposition 6.17. Suppose $k ⊆ F$ is a Galois extension, and $k ⊆ K$ is any finite extension. Then $K ⊆ KF$ is a Galois extension, and $Aut_K(KF) cong Aut_{F∩K}(F)$.
I am confused how Proposition 6.17. is used in Example 6.18. :
i. First of all, about using different notations: $k$ as a field of characteristic zero in Example 6.18 is $K$ is Proposition 6.17.; and probably (?) $mathbb{Q}$ in Example 6.18 is $k$ is Proposition 6.17.; what about the rest?
ii. If I am right with notations, $KF$ corresponds to $k(ζ)$, but how $kmathbb{Q}(ζ)=k(ζ)$? What does "the composite $k(ζ)$ of $k$ and $mathbb{Q}(ζ)$" mean?
iii. Letting the considered field of characteristic zero be $mathbb{Z}$. Could you re-explain Example 6.18. for this specific case $k=mathbb{Z}$? (esp how $mathbb{Q}$ can be extended to $mathbb{Z}$?(!)
abstract-algebra group-theory galois-theory proof-explanation
abstract-algebra group-theory galois-theory proof-explanation
asked 12 hours ago
72D
572116
asked 12 hours ago
72D
572116
edited 11 hours ago
asked 12 hours ago
72D
572116
asked 12 hours ago
72D
572116
asked 12 hours ago
72D
572116
572116
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