Ytukyg

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 $mathbb{Z}$?(!)

i. To be clear, given the conflicts in notation, let me rewrite Proposition 6.17 as follows:

Proposition 6.17. Suppose $L ⊆ F$ is a Galois extension, and $L ⊆ K$ is any finite extension. Then $K ⊆ KF$ is a Galois extension, and $Aut_K(KF) cong Aut_{F∩K}(F)$.

In that notation, from context I would guess $K=k$, $F=mathbb Q(zeta)$, $L=mathbb Q$. This is doesn't match Proposition 6.17 as quoted, since $k/mathbb Q$ need not be a finite extension (e.g. if $k=mathbb R$), but the proposition still holds for any extension I believe; see the comment between Examples 5.6 and 5.7 in these notes by Keith Confrad.

ii. The composite (or compositum, or tensor product) $KF$ of two field extensions $K$ and $F$ of a is roughly the smallest field containing both $K$ and $F$. In this case, $kmathbb Q(zeta)$ is the smallest field containing $k$, $mathbb Q$, and $zeta$. Since $mathbb Q$ is contained in $k$, this simplifies to the smallest field containing $k$ and $zeta$, which is of course the field extension $k(zeta)$.

iii. As mentioned in a comment, it doesn't make sense to set $k=mathbb Z$, as the integers don't form a field. If your question is to how we can guarantee any characteristic zero field is an extension of $mathbb Q$, this is shown as follows.

Let $k$ be a field with characteristic $0$. Let $1_k$ be the identity of $k$, $0_k$ be the zero element of $k$, and for $nin mathbb Z$ let $n_k=ncdot 1_k$.
Note that by definition, for $a,bin mathbb Z$, $a_k+b_k=(a+b)_k$ and $a_kb_k=(ab)_k$. In particular, note that $f:mathbb Zrightarrow k$ given by $f(a)=a_k$ is a ring homomorphism. Moreover, since the characteristic of $k$ is $0$, $n_kneq 0_k$ if $nneq 0$. Then the map $f$ is an injection, so $f(mathbb Z)$ is isomorphic to $mathbb Z$ as a ring, hence we may as well identify the two. Since $mathbb Q$ can be defined as the smallest field containing $mathbb Z$ and $k$ is a field containing $mathbb Z$, it must be that $mathbb Qsubseteq k$.