Exercise 7.3.2 (Automorphism over prime subfield)

Question

Using Lemma 7.3.1, show that every automorphism of a field is an automorphism over its prime subfield. In other words,
$Aut(M) = Gal(M:K)$ whenever $M$ is a field with prime subfield $K$.

Hassaan Naeem · Accepted Answer

extbf{Solution:}
By 	extbf{Lemma 7.3.1} we have that $\forall S \subseteq Aut(M), \ Fix(S) \subseteq M$. Since $K$ is the prime subfield of $M$ we have that $K \subseteq Fix(S) \subseteq M$, and we also have that $K \subseteq Fix(S) \subseteq M:K$. Hence we have that:
\begin{equation*}
    \begin{aligned}
        Aut(M) &= \{ S: Fix(S) \subseteq M\}\
        &= \{S: K \subseteq Fix(S) \subseteq M \}\
        &= \{S: K \subseteq Fix(S) \subseteq M:K\}\
        &= \{S: Fix(S) \subseteq M:K\}\
        &= Gal (M:K) \quad \square
    \end{aligned}
\end{equation*}

Richard Ganaye · Answer

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\begin{proof} Let $\sigma \in \mathrm{Aut}(M)$ be an automorphism of $M$. We must prove that $\sigma$ fixes every element of the prime subfield $K$.

If $S = \{\sigma\}$, we know that  $\mathrm{Fix}(S)$ is a subfield of $M$ by Lemma 7.3.1. By definition, the prime subfield $K$ is included in every subfield of $M$, so
$$K \subset \mathrm{Fix}(S).$$
Since $\mathrm{Fix}(S) = \{\alpha \in M \mid \sigma(\alpha) = \alpha\}$, this inclusion means that $\sigma(\alpha) = \alpha$ for every $\alpha \in K$.

This proves that every automorphism $\sigma \in \mathrm{Aut}(M)$ is in $\Gal(M:K)$. Moreover, by definition, $\Gal(M:K) \subset \mathrm{Aut}(M)$, thus
$$\mathrm{Aut}(M) = \Gal(M:K).$$
\end{proof}

Exercise 7.3.2 (Automorphism over prime subfield)

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Exercise 7.3.2 (Automorphism over prime subfield)

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