Let $f \in \mathbb{Q}[x]$ be an irreducible polynomial of degree 3. Suppose $f$ has one real root, we want to show that $$\text{Gal}(L/\mathbb{Q}) \cong S_3,$$ where $L$ is the splitting field of $f$.

Since $f(x)$ is irreducible, $f(x)$ has 3 distinct roots $r_1, r_2$ and $r_3$. If we consider the discriminant, we note that $$\Delta = \prod_{i < j} (r_j - r_i)^2.$$ Since the roots $r_2$ and $r_3$ are distinct, we have that $\sqrt{\Delta} \not \in \mathbb{Q}$.The splitting field of $f$ is given by $\mathbb{Q}(r_1, \sqrt{\Delta})$. By the tower law, we have that $$[\mathbb{Q}(r_1, \sqrt{\Delta}) : \mathbb{Q}] = [\mathbb{Q}(r_1, \sqrt{\Delta}) : \mathbb{Q}(\sqrt{\Delta})] \cdot [\mathbb{Q}(\sqrt{\Delta}) : \mathbb{Q}] = 3 \cdot 2 =6.$$ We therefore have that the Galois group of $L/\mathbb{Q}$ is isomorphic to $S_3$.

Is this is a legitimate proof?

Also, how can I generalise this result to prove that the Galois group of an irreducible polynomial of degree $p$ with $p-2$ roots is $S_p$?

  • 1
    $\begingroup$ I think you mean $\sqrt\Delta\notin\mathbb Q?$ $\endgroup$
    – awllower
    Commented Jun 7, 2016 at 2:34
  • $\begingroup$ Why can't it be $\mathbb Z_3 \times \mathbb Z_2$ ? $\endgroup$ Commented Apr 30 at 12:24

1 Answer 1


You might be able to turn it into a proof with a bit more explanation. But, for a more generalizable approach$\ldots$

By irreducibility and the Tower Law, as well as the Galois Correspondence, we know that $p$ divides the order of the Galois group. Using Cauchy's Theorem, what kind of element do we know then exists in the Galois group (think cycles)? We also have exactly $2$ nonreal roots; they must come as a conjugate pair, so we have conjugation in the Galois group. This is a $2$-cycle in $S_p$. Can you show that the first element (which you need to find) and a $2$-cycle generate $S_p$ (I'm assuming $p$ is prime; if it isn't, this isn't true)?


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