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Well.. It looks like a obvious for me, But I want to check the below things right or not.

Let $K = SF(f/K)$ for separable polynomial $f(x) \in F[x]$ with its degree, $deg f = n(\geq2)$

(Here the $SF(f/F)$ means splitting field of $f(x)$ over the field $F$)

Say the set $A$ = $\{\alpha_1 \alpha_2,... \alpha_n\}$ is a set of the all roots of the $f(x)$

So my questions begin.

First Question)

For $\forall \sigma \in G(K/F) $, Is the bijective $\sigma : A \to A$ ?

(I.E. Are all the Galois group's elements bijective mapping between all roots of the $f(x)$?)

Second Qeustion)

Let's consider the case the $f(x)$ is not irreducible for a root $\alpha_i \in F$

I would say the $\{\alpha_1, \alpha_2 \}$ $\in F$ and $\{\alpha_3, ..., \alpha_n \} \in K-F$

Then, Is this true?

$\forall \sigma \in G(K/F) $, $\begin{cases} \sigma : \{\alpha_1, \alpha_2 \} \to \{\alpha_1, \alpha_2 \} & \text{} \\ \sigma : \{\alpha_3, ..., \alpha_n \} \to \{\alpha_3, ..., \alpha_n \} & \end{cases}$

(I.E. Should be roots which is elements of the each field "$F$" and "$K\setminus F$" mapped to $F$ and "$K\setminus F$" respectively by the bijective $\sigma : A \to A$?)

p.s.) Related with the second question Does it exist $\sigma(\in G(K/F)) : \alpha_1 \to \alpha_2$?

Thanks.

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  • $\begingroup$ I believe that is correct. The elements of the galois group permute the roots. Permutations are bijections. But this is not my strong suit. Didn't get the second question. $\endgroup$ – Chris Custer Jan 23 at 7:30
  • $\begingroup$ As far as I know, elements of the galois group fix the base field. So we can't have $\alpha_1\to\alpha_2$. $\endgroup$ – Chris Custer Jan 23 at 7:39
  • $\begingroup$ Ah yes. I forgot the Galois group's definition. My second question is nonsense. There should be $\alpha_i \to \alpha_i$(I.E. identity), right? ($1 \leq i \leq 2$) $\endgroup$ – se-hyuck yang Jan 23 at 9:17
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Your answer to the first question is correct. Given a separable polynomial $f\in K[X]$ (not necessarily irreducible), $\operatorname{Gal}(f)$ permutes its roots (bijectively, this is what permuting is).

In case $f$ is reducible, we factor $f$ into irreducible factors over $K[X]$ and then the Galois group permutes the roots of each irreducible factor between themselves. Consider the following simple example: $f=X^4-4\in \mathbf{Q}[X]$. We factor into irreducible factors: $(X^2-2)(X^2+2)$, so $A=\{\sqrt{2},-\sqrt{2},i\sqrt{2},-i\sqrt{2} \}$. Indeed, the Galois group permutes $\sqrt{2},-\sqrt{2}$ between themselves and $i\sqrt{2},-i\sqrt{2}$ between themselves. It is isomorphic in this case to $\mathbf{Z}/2\mathbf{Z}\times \mathbf{Z}/2\mathbf{Z}$.

If $f$ is irreducible over $K[X]$, then the action of $\operatorname{Gal}(f)$ on the roots is transitive. This means exactly that if $A=\{\alpha_1,\ldots,\alpha_n \}$, for all $1\leq i,j\leq n$ there exists $\sigma\in\operatorname{Gal}(f)$ for which $\sigma(\alpha_i)=\alpha_j$.

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  • $\begingroup$ Then, When it comes to my second question if the $f=g \bullet h$ $s.t.$ both of the $g$ and $h$ are irreducible does it true?(Here $g(x) = (x-\alpha_1)(x-\alpha_2)$, $h(x)=\Pi_{i=3}^{n}(x-\alpha_i)$) Because there is a transitive action, $\sigma(\in G(K/Q)) : \alpha_1 \to \alpha_2$ and $\alpha_i \to \alpha_j $ for $3 \leq i,j \leq n$simultaneously I'm so curious my guess is right. $\endgroup$ – se-hyuck yang Jan 23 at 9:13

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