Tutorial 7
Galois Groups of Quartics and Solvability by Radicals
 

  1. Let $ x_1,x_2,x_3$ be indeterminates and let $ s_1, s_2, s_3$ be the elementary symmetric polynomials of $ x_1,x_2,x_3.$ Show that $ E=\mathbb{Q}(x_1,x_2,x_3)$ is not a radical extension of $ F=\mathbb{Q}(s_1,s_2,s_3)$ but $ \mathbb{Q}(\zeta_3)(x_1,x_2,x_3)$ is a radical extension of $ \mathbb{Q}(\zeta_3, s_1,s_2,s_3).$

    Solution: Let $ d$ be the square root of the discriminant of the polynomial $ f(x)=x^3-s_1x^2+s_2x-s_3.$ We know that $ E$ is a splitting field of $ f(x)$ over $ F$ and $ G(E/F)=S_3.$ Moreover $ F(d):F]=2.$ Hence $ F(d)/F$ is a simple radical extension. We show that $ E/F(d)$ is not a simple radical extension. If it were, then $ E$ is a splitting field of an irreducible cubic $ g(x)=x^3-a$ for some $ a \in F(d).$ But $ E/F(d)$ is a Galois extension. Hence all the roots of $ g(x)$ are in $ E.$ If $ r,s$ are any two roots of $ g(x)$ then $ (r/s)^3=1.$ Hence $ r/s$ is a primitive cube root $ w$ over $ \mathbb{Q}.$ This is a contradiction. Similar argument shows that the other cubic extensions of $ F$ in $ E$ are not simple radical extensions. Hence $ E/F$ is not a radical extension.

  2. Let $ G$ be the Galois group of an irreducible quintic over $ \mathbb{Q}.$ Show that $ G=A_5$ or $ S_5$ if $ G$ has an element of order $ 3.$

    Solution: Let $ f$ be an irreducible quintic over $ \mathbb{Q}.$ Then its Galois group $ G$ is a transitive subgroup of $ S_5.$ Hence we see that the order of $ G$ is a multiple of $ 15.$ Thus $ \vert G\vert=15, 30, 60$ or $ 120.$ We must show that it cannot have order $ 15$ or $ 30.$ If the order is $ 15$ then it is cyclic. But there is no permutation of oder $ 15$ in $ S_5.$ Suppose $ \vert G\vert=30.$ Then due to simplicity of $ A_5,$ $ G$ is not a subgroup of $ A_5.$ Hence $ G$ has an odd permutation say $ \sigma.$ Then $ \sigma$ is either a product of two disjoint cycles of order $ 2$ and $ 3$ or a $ 4$-cycle. In the latter case $ 4 \mid \vert G\vert$ which is not possible. In the former case $ \sigma^3$ is a transposition. But such a group is $ S_5.$