# If $f(x)$ irreducible degree $p$, over $\mathbb{Q}$ with exactly two nonreal roots in $\mathbb{C}$ Then the Galois group of $f(x)$ is $S_p.$

Let $$f(x)$$ be an irreducible polynomial over $$\mathbb Q$$ of degree $$p$$, $$p$$ prime. Suppose that $$f(x)$$ has exactly two nonreal roots. Then the Galois group of $$f(x)$$ over $$\mathbb{Q}$$ is the symmetric group $$S_p.$$

I have the following ideas (written in a non-rigorous way):

Let $$K$$ the splitting field of $$f(x)$$ over $$\mathbb{Q}$$. By hypothesis, exists $$\alpha\in\mathbb{C}\setminus\mathbb{R}$$ root of $$f(x)$$. Further, $$\beta=\overline{\alpha}$$ is the other root of $$f(x$$) in $$\mathbb{C}\setminus\mathbb{R}$$.

Let $$\sigma\in Gal(K/\mathbb{Q})$$ given by $$\sigma(\alpha)=\beta$$, $$\sigma(\beta)=\alpha$$ and $$\sigma(k)=k$$ for any $$k\in K$$, $$k\neq \alpha,\beta.$$

Then $$\sigma$$ "behaves like a permutation" in $$Gal(K/\mathbb{Q})$$.

On the other hand, $$p\mid |Gal(K/\mathbb{Q})|$$. By Cauchy's theorem, there exists an element of order $$p$$, i.e., a permutation of order $$p$$.

Therefore, $$Gal(K/\mathbb{Q})$$ contains a permutation of order 2 and a permutation of order $$p$$.

A permutation of order $$2$$ and a permutation of order $$p$$ generates $$S_p$$. Therefore, $$Gal(K/\mathbb{Q})=S_p$$.

But, there is something that is wrong or I do not understand. By hypothesis, $$p=[K:\mathbb{Q}]=|Gal(K/\mathbb{Q})|$$ but if $$Gal(K/\mathbb{Q})=S_p$$ then $$|Gal(K/\mathbb{Q})|=p!\neq p$$...

Actualization 1. Now I have better understood the situation. I have the general idea but the proof is not written rigorously.

Let $$K$$ the splitting field of $$f(x)$$ over $$\mathbb{Q}$$ i.e. $$K=\mathbb{Q}(r_1,\ldots, r_p)$$ with $$r_i$$ the roots of $$f(x)$$.

Let $$Gal(K/\mathbb{Q})\times \left\{r_1,\ldots, r_p\right\}\to \left\{r_1,\ldots, r_p\right\}$$ an action. Then exists $$\psi:Gal(K/\mathbb{Q})\to Aut(\left\{r_1,\ldots, r_p\right\})\simeq S_p$$ homomorphism.

Therefore $$\psi(Gal(K/\mathbb{Q}))$$ subgroup of $$S_p$$.

$$f(x)$$ is the minimal polynomial of $$r_i$$ over $$\mathbb{Q}$$ for any $$i=1,\ldots, p$$ then $$[\mathbb{Q}(r_i):\mathbb{Q}]=p$$. Because $$[K:\mathbb{Q}]=[K:\mathbb{Q}(r_i)][\mathbb{Q}(r_i):\mathbb{Q}]$$ then $$p\mid [K:\mathbb{Q}]=|Gal(K/\mathbb{Q})|$$.

Therefore exists $$g\in Gal(K/\mathbb{Q})$$ of order $$p$$. Then $$\psi(g)$$ has order $$p$$ in $$S_p?$$

Therefore $$\psi(g)$$ is a $$p$$-cycle in $$S_p$$.

By the other hand, Let $$\sigma\in Gal(K/\mathbb{Q})$$ given by $$\sigma(\alpha)=\beta$$, $$\sigma(\beta)=\alpha$$

Then $$\sigma$$ "behaves like a permutation" in $$Gal(K/\mathbb{Q})$$. Therefore $$\psi(\sigma)$$ is a permutation in $$S_p$$

We also know that $$S_p$$ is generated by a permutation and $$p$$-cycle.

Therefore $$Gal(K/\mathbb{Q})\simeq S_p$$.

• Why do you think $[ K : \mathbf{Q} ] = p$? The proof is mostly correct, but your $\sigma$ isn't a homomorphism as you've defined it.
– bzc
Commented Mar 9, 2021 at 6:48
• To add to that, the degree of an extension is different from the degree of the corresponding splitting field. Galois theory has a lot of concepts that link the same idea for polynomials, fields and Galois groups (e.g. solvability), but the degree is not one of them. This confused me as well for some time. Commented Mar 9, 2021 at 7:43
• Thanks. I have already fixed that confusion I had. I have updated the theme. Commented Mar 10, 2021 at 5:27
• "$S_p$ is generated by a transposition and $p$-cycle" is math.stackexchange.com/questions/2792165 and math.stackexchange.com/questions/64848 Commented Nov 14, 2023 at 20:19

Now you know that the Galois group has elements of two special orders. Both of them must have a very specific cycle structure, when you consider them as elements of the symmetric group acting on the $$p$$ roots. What can you say about a subgroup of $$S_p$$ containing two such elements?