Let f be an irreducible polynomial of degree $5$ in $\mathbb{Q}[x]$. Suppose that in $\mathbb{C}$, $f$ has exactly two nonreal roots. Then the Galois group of the splitting field of $f$ is isomorphic to $S_5$.

My effort: Let $G$ be the Galois group. The complex conjugation map $\mathbb{\sigma} (x+iy)=x-iy$ is non-trivial element of $G$ of order $2$. Let $\alpha_1,\alpha_2, \alpha_3$ be the real roots of $f(x)$ and $\beta_1$ and $\beta_2$ be the nonreal roots. Then, $\sigma$ sends $\beta_1$ to $\beta_2$ while keeping other roots fixed. Now if we can construct an element of order $5$ in $G$, then we can proceed in the direction of showing that $G \cong S_5$.

I know that any such element $\tau$ of $G$ (if exists) will permute all the roots of $f$. But I am not sure which permutation of roots of $f$ will give me the correct candidate for the element $\tau$?


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    $\begingroup$ What's the question? $\endgroup$ – Angina Seng Jul 14 '18 at 16:41
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    $\begingroup$ Demanding others solve a problem rather than answer an actual question makes it look like this is a homework problem. In fact it is a standard type of problem for students learning Galois theory. $\endgroup$ – KCd Jul 14 '18 at 16:48
  • $\begingroup$ @KCd please have a look. I have put my efforts along with question. $\endgroup$ – Shubham Namdeo Jul 14 '18 at 16:57
  • $\begingroup$ The Galois group of an irreducible polynomial over a field of characteristic zero will act transitively on its zeros. $\endgroup$ – Angina Seng Jul 14 '18 at 17:01
  • $\begingroup$ @LordSharktheUnknown after that how to proceed? $\endgroup$ – Shubham Namdeo Jul 14 '18 at 17:04

Proof 1(Direct approach)

Theorem Let $p$ be prime. If a transitive subgroup $H$ of $S_p$ contains a transposition $(1 \ 2)$, then $H=S_p$.

Define a relation $\sim$ on $\{1,2,\ldots, p\}$ by $$ a \sim b \ \Longleftrightarrow (a \ b)\in H. $$

This relation is an equivalence relation (reflexive, symmetric, transitive).

Let $[a]_{\sim}$ be the equivalence class containing $a$. Since $(1 \ 2)\in H$, we have $1 \sim 2$.

Since $H$ is transitive, we can prove that there is a bijection between any two equivalence classes.

Then the cardinality of $[1]_{\sim}$ must divide $p$. Since $\{1,2\}\subseteq [1]_{\sim}$, we must have $|[1]_{\sim}|=p$. This tells us that all transpositions containing $1$ are in $H$. Hence, $H=S_p$.

Proof 2(Galois Theory)

If $G$ is Galois group of an irreducible polynomial $f(x)\in\mathbb{Q}[x]$ of degree $p$, then $p$ divides the order of $G$. Also, $G$ is transitive and it contains a transposition. Since $G$ must have an element of order $p$, it contains a $p$-cycle. Hence $G\simeq S_p$.

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