Lecture 24
Galois Groups of Quartic Polynomials
 


Proposition 24.5  

The resolvent cubic of a separable irreducible quartic has a root in F if and only if $ G_f\subseteq D_4$.


Proof.
Let $ t_1\in F$. Then $ G(E/F(t_1))=G_f=G_f\cap H_1\Rightarrow G_f\subseteq H_1.$ Conversely if $ G_f\subset H_i$ for some $ i$ say $ i=1,$ then each $ \sigma\in G_f$ fixes $ t_1$ and hence $ t_1\in E^{G_f} = F$.


Theorem 24.6  

Let $ f(x)$ be an irreducible separable quartic over a field F of char $ F \neq 2$ and $ E = F(r_1,r_2,r_3,r_4)$ be a splitting field where $ r_1, \ldots,r_4$ are the roots of $ f(x).$ Let $ r(x)$ denote resolvent cubic of $ f(x).$
  1. If $ r(x)$ is irreducible in $ F[x]$ and $ \:$disc$ \;(r(x)) \notin F^2$ then $ G_f\simeq S_4.$
  2. If $ r(x)$ is irreducible in $ F[x]$ and $ \:$disc$ \;(r(x)) \in F^2$ then $ G_f\simeq A_4.$
  3. If $ r(x)$ splits completely in $ F[x]$ then $ G_f\simeq V.$
  4. Let $ r(x)$ have one root in $ F.$ Then
    (a) If $ f(x)$ is irreducible over $ F(\underline{t})$ then $ G_f\simeq D_4.$
    (b) If $ f(x)$ is reducible over $ F(\underline{t})$ then $ G_f\simeq C_4.$