Exercise 9.1.3 ((Radical) subfield)

Question

Check that the intersection of any family of subfields
of $\mathbb{C}$ satisfying (9.2) is again a subfield of $\mathbb{C}$ satisfying (9.2). (That any
intersection of subfields is a subfield is a fact we met back on p. 29;
the new aspect is (9.2).)

Hassaan Naeem · Accepted Answer

extbf{Solution:}
Firstly, we know that the prime subfield of $\mathbb{C}$ is $\mathbb{Q}$, and as such every subfield of $\mathbb{C}$ cointains $\mathbb{Q}$.
Hence if we have $\mathbb{Q}^{rad}_a$ and $\mathbb{Q}^{rad}_b$ fullfilling (9.2) we can always find a $\mathbb{Q}^{rad}_c = \mathbb{Q}^{rad}_a \cap \mathbb{Q}^{rad}_b$
satisfying (9.2) since we can always have that $\mathbb{Q}^{rad}_c = \mathbb{Q}$, which satisfies (9.2).

Richard Ganaye · Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

\begin{proof} 
Let $(K_i)_{i\in I}$ be a family of subfields of $\C$ satisfying (9.2).

For all $i \in I$, and for all $\alpha \in \C$, 
\begin{align}
 (\exists n \geq 1,\ \alpha^n \in K_i) \Rightarrow \alpha \in K_i.
\end{align}
Write $K = \bigcap\limits_{i\in I} K_i$ the intersection of all $K_i$. Let $\alpha$ be a complex number, and assume that there exists some $n\geq 1$ such that $\alpha^n \in K$.

Then for all $i \in I$, $\alpha^n \in K_i$. Therefore, by (1), $\alpha \in K_i$ for all $i \in I$, so $\alpha \in K$. This shows that for all $\alpha \in \C$,
$$(\exists n \geq 1,\ \alpha^n \in K) \Rightarrow \alpha \in K.$$
To conclude, the intersection of any family of subfields
of $\mathbb{C}$ satisfying (9.2) is again a subfield of $\mathbb{C}$ satisfying (9.2)
\end{proof}

Exercise 9.1.3 ((Radical) subfield)

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Exercise 9.1.3 ((Radical) subfield)

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