Notes - Galois Theory HT25, Cubic equations
How can we use Galois theory to derive a formula for the cubic (in terms of radicals)? First we determine the Galois group of the polynomial (i.e. the group of $\mathbb Q$-linear automorphisms of the splitting field). We can show that it is either (isomorphic to) $A _ 3$ or $S _ 3$, depending on the discriminant of the cubic.
To make things concrete, suppose that the splitting field is $K/\mathbb Q$, $\text{Gal}(K / \mathbb Q) \cong S _ 3$ and that the roots are $\alpha _ 1, \alpha _ 2, \alpha _ 3$. Results from [[Notes - Galois Theory HT25, Solvability by radicals]]U tell us that we need to consider the normal series
\[\\{1\\} \trianglelefteq A_3 \trianglelefteq S_3\]By [[Notes - Galois Theory HT25, Main theorems of Galois theory]]U, this corresponds to the set of field extensions
\[K \ge K^{A_3} \ge \mathbb Q\]To get a handle on these field extensions, we adjoin a non-trivial third root of unity $\omega$. This gives us a new sequence of fields
\[K(\omega) \ge K^{A_3}(\omega) \ge \mathbb Q\]This is useful because you can show that $K(\omega)$ is still a Galois extension of $F$, and results from [[Notes - Galois Theory HT25, Kummer extensions]]U say that a field extension that contains a third root of unity has a simple description in terms of an extension of the base field, i.e.
\[K(\omega) = K^{A_3}(\omega)(u)\]for some $u$, and that $u^3 \in K^{A _ 3}(\omega)$. Results from [[Notes - Galois Theory HT25, Kummer extensions]]U say that $u$ can be described by eigenvectors of the cyclic permutation $\sigma$ in $A _ 3$. Therefore the eigenvectors are:
\[u := \frac{\alpha_1 + \omega \alpha_2 + \omega^2 \alpha_3}{3}\]and
\[v := \frac{\alpha_1 + \omega^2 \alpha_2 + \omega a_3}{3}\]So then $u^3, v^3 \in K^{A _ 3}(\omega)$. To calculate these, manipulating symbolic expressions shows that $u^3$ and $v^3$ are the roots of the polynomial
\[z^2 + qz - \frac{p^3}{27} = 0\]And then $\alpha _ 1 = u + v$.
Todo, still don’t really understand this.
Flashcards
Suppose:
\[f(x) := ax^3 + bx^2 + cx + d\]
@State the substitution that can be used to put the cubic in a simpler form.
Let $t = x + \frac{b}{3a}$. Then the equation becomes
\[f(t) = t^3 + pt + q\]for some $p$, $q$ (really this is a specific case of being able to make a substitution that eliminates the next lower-degree term).
Suppose:
- $F$ is a field
- $\text{char}(F) \ne 2, 3$
- $f := t^3 + pt + q \in F[t]$
- $f$ is irreducible
- $K$ is a splitting field for $f$
@State a theorem which completely characterises the Galois group $G = \text{Gal}(K/F)$.
Let $\Delta$ be the discriminant (the square of the product of the differences of roots). Then:
- If $\Delta$ is a square in $F$, then $G = A _ 3$
- If $\Delta$ is not a square in $F$, $G = S _ 3$
@Prove that if:
- $F$ is a field
- $\text{char}(F) \ne 2, 3$
- $f := t^3 + pt + q \in F[t]$
- $f$ is irreducible
- $K$ is a splitting field for $f$
- $\Delta$ is the discriminant of $f$ (the square of the product of the differences of roots)
- $G = \text{Gal}(K/F)$
Then:
- If $\Delta$ is a square in $F$, then $G = A _ 3$
- If $\Delta$ is not a square in $F$, then $G = S _ 3$
$G$ must be either $S _ 3$ or $A _ 3$:
Since $f$ is irreducible over $F$, $m _ {F, \alpha _ 1} = f$ where $\alpha _ 1$ is a root. Hence
\[[F(\alpha_1) : F] = \deg m_{F, \alpha_1} = \deg f = 3\]Hence $[F(\alpha _ 1) : F] \mid [K : F]$ by the tower law and since $ \vert G \vert = [K : F]$, $ \vert G \vert $ is a multiple of three.
The only subgroups of $S _ 3$ with this property are $A _ 3$ and $S _ 3$.
$\Delta$ determines which group:
By the result that says
If:
- $f \in F[t]$ is a polynomial of degree $n$
- $K$ is a splitting field for $f$
- $\{\alpha _ 1, \ldots, \alpha _ n\}$ are the roots of $f$ in $K$
- $G = \text{Gal}(K/F)$, identified with a subgroup of $S _ n$
- $\text{char}(F) \ne 2$
- $f$ has no repeated roots Then: $G \le A _ n$ if and only if $\Delta$ is a square in $F$.
it follows that if $\Delta$ is a square in $F$, $G = A _ 3$, otherwise $G = S _ 3$.
Suppose:
- $\text{char}(F) \ne 3$
- $f = t^3 + pt + q \in F[t]$ is an irreducible cubic with roots $\alpha _ 1, \alpha _ 2, \alpha _ 3$.
- $u := \frac{\alpha _ 1 + \omega \alpha _ 2 + \omega^2 \alpha _ 3}{3}$
- $v := \frac{\alpha _ 1 + \omega^2 \alpha _ 2 + \omega \alpha _ 3}{3}$
@Prove that:
- $uv = -\frac{p}{3}$
- $u^3 v^3 = -\frac{p^3}{27}$
- $u^3 + v^3 = -q$
- $u^3$ and $v^3$ are the roots of a quadratic polynomial with coefficients in $F$
- $\alpha _ 1$ can be written in terms of $u$ and $v$
Equating coefficients in $t^3 + pt + q = (t - \alpha _ 1)(t - \alpha _ 2)(t - \alpha _ 3)$, we have that
- $\alpha _ 1 + \alpha _ 2 + \alpha _ 3 = 0$
- $\alpha _ 1 \alpha _ 2 + \alpha _ 2 \alpha _ 3 + \alpha _ 3 \alpha _ 1 = p$
- $\alpha _ 1 \alpha _ 2 \alpha _ 3 = -q$
Using the fact that $\omega + \omega^2 = -1$, it follows that
\[\begin{aligned} 9uv &= (\alpha_1 + \omega \alpha_2 + \omega^2 \alpha_3)(\alpha_1 + \omega^2 \alpha_2 + \omega \alpha_3) \\\\ &= \alpha_1^2 + \omega \alpha_1 \alpha_2 + \omega^2 \alpha_1 \alpha_3 + \omega^2 \alpha_1 \alpha_2 + \alpha_2^2 + \omega \alpha_2 \alpha_3 + \omega \alpha_2 \alpha_3 + \omega \alpha_1 \alpha_3 + \omega^2 \alpha_2 \alpha_3 + \alpha_3^2 \\\\ &= \alpha_1^2 + \alpha_2^2 + \omega_3^2 - \alpha_1 \alpha_2 - \alpha_2 \alpha_3 - \alpha_3 \alpha_1 \\\\ &= (\alpha_1 + \alpha_2 + \alpha_3)^2 - 3(\alpha_1 \alpha_2 + \alpha_2 \alpha_3 + \alpha_3 \alpha_1) \\\\ &= -3p \end{aligned}\]Since $\text{char}(F) \ne 3$, it follows $uv = - \frac p 3$.
(b) follows from (a).
For (c), note that
\[u^3 + v^3 = (u + v) (\omega u + \omega^2 v) (\omega^2 u + \omega v)\]Then
\[\begin{aligned} 3(u + v) &= 2\alpha_1 + (\omega + \omega^2) \alpha_2 + (\omega^2 + \omega) \alpha_3 = 3\alpha_1 \\\\ 3(\omega u + \omega^2 v) &= (\omega + \omega^2)\alpha_1 + (\omega^2 + \omega)\alpha_2 + 2\alpha_3 = 3\alpha_3 \\\\ 3(\omega^2 u + \omega v) &= (\omega^2 + \omega) \alpha_1 + 2\alpha_2 + 9\omega + \omega^2) \alpha_3 = 3\alpha_2 \end{aligned}\]For (d), this follows from (b, c), since they are the roots of
\[z^2 + qz - \frac{p^3}{27} = 0\]Finally, since
\[\begin{aligned} u + v &= \frac{\alpha_1 + \omega \alpha_2 + \omega^2 \alpha_3 + \alpha_1 + \omega^2 \alpha_2 + \omega \alpha_3 }{3} \\\\ &= \frac{2 \alpha_1 - \alpha_2 - \alpha_3}{3} \\\\ &= \alpha_1 \end{aligned}\]and $uv = -\frac p 3$, it follows that
\[\alpha_1 = u - \frac{p}{3u}\]