# Galois group's element and roots of polynomial

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.

• 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.
– user403337
Jan 23, 2020 at 7:30
• As far as I know, elements of the galois group fix the base field. So we can't have $\alpha_1\to\alpha_2$.
– user403337
Jan 23, 2020 at 7:39
• 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? ($1 \leq i \leq 2$) Jan 23, 2020 at 9:17

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 $$\sigma(\alpha_i)=\alpha_j$$.
• 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. Jan 23, 2020 at 9:13