Galois Theory HT25, Quartic equations

Created: April 15, 2025 | Updated: June 07, 2025 | About these notes | View in context | Study these flashcards

An (almost) complete characterisation of the Galois group in terms of the roots

This is not covered in too much detail in the course notes, but there is almost a complete characterisation of the Galois group of any quartic just in terms of expressions in the roots. This comes from [[Algebra, Artin]]N.

Suppose $f(x)$ has roots $\alpha _ 1, \alpha _ 2, \alpha _ 3, \alpha _ 4$ and let $\Delta$ be the discriminant of $f(x)$.

Let

  • $\beta _ 1 = \alpha _ 1 \alpha _ 2 + \alpha _ 3 \alpha _ 4$
  • $\beta _ 2 = \alpha _ 1 \alpha _ 3 + \alpha _ 2 \alpha _ 4$
  • $\beta _ 3 = \alpha _ 1 \alpha _ 4 + \alpha _ 2 \alpha _ 3$

and define the resolvent cubic as $g(x) = (x - \beta _ 1)(x - \beta _ 2)(x - \beta _ 3)$. Then we have the characterisation:

  $\Delta$ square $\Delta$ not a square
$g$ reducible $G = D _ 4$ $G = D _ 4$ or $G = C _ 4$
$g$ irreducible $G = A _ 4$ $G = S _ 4$

Flashcards

Suppose:

  • $f$ is a separable, irreducible degree 4 polynomial
  • $f$ has roots $\lbrace \alpha _ 1, \ldots, \alpha _ n \rbrace$ in a splitting field $K$
  • $G = \text{Gal}(K/F)$ is its Galois group
  • The following all lie in $F$:
    • $x = \alpha _ 1 \alpha _ 2 + \alpha _ 3 \alpha _ 4$
    • $y = \alpha _ 1 \alpha _ 3 + \alpha _ 2 \alpha _ 4$
    • $z = \alpha _ 1 \alpha _ 4 + \alpha _ 2 \alpha _ 3$

@Prove that then $G$ is isomorphic to $V _ 4$.

Let $g(t) = (t - x)(t - y)(t - z)$. Then we have

  • $x - y = (\alpha _ 1 - \alpha _ 4)(\alpha _ 2 - \alpha _ 3)$
  • $y - z = (\alpha _ 1 - \alpha _ 2)(\alpha _ 3 - \alpha _ 4)$
  • $z - x = (\alpha _ 1 - \alpha _ 3)(\alpha _ 4 - \alpha _ 2)$

Thus $\Delta _ g = \Delta _ f$.

Since $x, y, z \in F$, it follows $\Delta _ f$ is a square in $F$, therefore $G \le A _ 4$. But as $f$ is irreducible, $4 \mid \vert G \vert $, and so $G = A _ 4$ or $G = V _ 4$.

Suppose for a contradiction that $G = A _ 4$. Then $G$ contains the three cycle $(123)$.

But $(123)\cdot x = \alpha _ 2 \alpha _ 3 + \alpha _ 1 \alpha _ 4 = z$. Since $x \in F$, it follows $x = z$ as $F = K^G$.

Thus $z - x = (\alpha _ 1 - \alpha _ 3)(\alpha _ 2 - \alpha _ 4) = 0$, so either $\alpha _ 1 = \alpha _ 3$ or $\alpha _ 2 = \alpha _ 4$, contradicting the separability of $f$.

@exam~



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