Galois Theory HT25, Cubic equations
Finding a cubic formula
How can we use Galois theory to derive a formula for the cubic (in terms of radicals)?
Say $f(x) = x^3 + px + q$ is some irreducible polynomial over $\mathbb Q$ and suppose the roots are $\alpha _ 1, \alpha _ 2, \alpha _ 3$. The goal is to have expressions for these roots in terms of $p$ and $q$ which just involve $+, -, \times, \div$ and $\sqrt[n]{\cdot}$.
Let $G := \text{Gal} _ \mathbb Q(f) = \text{Gal}(K/\mathbb Q)$ be the Galois group of this polynomial, which is the Galois group of the splitting field $K = \mathbb Q(\alpha _ 1, \alpha _ 2, \alpha _ 3)$ for $f$. We assume for simplicity that $G \cong S _ 3$.
Possible in principle
First we show that solving the polynomial by radicals is always possible in principle without actually coming up with a formula. Since $G \cong S _ 3$, we have the sequence of normal subgroups
\[G \trianglerighteq A_3 \trianglerighteq \\{e\\}\]and so the Galois correspondence tells us we also have the sequence of subfields
\[\mathbb Q \le K^{A_3} \le K\]By previous results ( [[Notes - Galois Theory HT25, Determinant and discriminant]]U), we actually have $K^{A _ 3} = \mathbb Q(\delta)$ where $\delta := (\alpha _ 1 - \alpha _ 2)(\alpha _ 1 - \alpha _ 3)(\alpha _ 2 - \alpha _ 3)$, and so to keep everything very explicit we could write this sequence of field extensions as
\[\mathbb Q \le \mathbb Q(\delta) \le \mathbb Q(\alpha_1, \alpha_2, \alpha_3)\]Since $\delta$ is a symmetric function in the roots, it follows from the fundamental theorem of symmetric functions that $\delta$ can be written in terms of $p$ and $q$. So if we could somehow find a way to write the elements of $\mathbb Q(\alpha _ 1, \alpha _ 2, \alpha _ 3)$ in terms of $\delta$, we would be done.
To do this, we can use results from [[Notes - Galois Theory HT25, Kummer extensions]]U. The important result is the following:
If:
- $K/F$ is a Galois extension
- $[K : F] = p$ where $p$ prime
- For some $\varepsilon \in F$ where $\varepsilon \ne 1$, we have $\varepsilon^p = 1$
Then there exists $u \in K$ such that $u^p \in F$ and $K = F(u)$.
This would almost imply
\[\mathbb Q(\alpha_1, \alpha_2, \alpha_3) = \mathbb Q(\delta, u), \text{ where} u = \sqrt[3]{\ell} \text{ for some } \ell \in \mathbb Q(\delta)\]since $[\mathbb Q(\alpha _ 1, \alpha _ 2, \alpha _ 3) : \mathbb Q(\delta)] = 3$ (since $ \vert A _ 3 \vert = 3$) but the problem is that we can’t guarantee the last condition, namely that $\mathbb Q(\alpha _ 1, \alpha _ 2, \alpha _ 3)$ contains a non-trivial 3rd root of unity. To resolve this, we instead consider the sequence of subfields
\[\mathbb Q \le \mathbb Q(\delta, \omega) \le \mathbb Q(\alpha_1, \alpha_2, \alpha_3, \omega)\]where $\omega$ is a 3rd root of unity. You can then carefully check that $\mathbb Q(\alpha _ 1, \alpha _ 2, \alpha _ 3, \omega) / \mathbb Q$ is still a Galois extension and that $\text{Gal}(\mathbb Q(\alpha _ 1, \alpha _ 2, \alpha _ 3, \omega) / \mathbb Q(\delta, \omega)) \cong A _ 3$ so that now the requirements of the above results are satisfied with $K = \mathbb Q(\alpha _ 1, \alpha _ 2, \alpha _ 3, \omega)$, $F = \mathbb Q(\alpha _ 1, \alpha _ 2, \alpha _ 3)$ and $\varepsilon = \omega$. Thus we can conclude
\[\mathbb Q(\alpha_1, \alpha_2, \alpha_3, \omega) = \mathbb Q(\delta, \omega, u), \text{where }u = \sqrt[3]{\ell} \text{ for some } \ell \in \mathbb Q(\delta, \omega)\]and with this, we are done since we can now in principle find expressions for $\alpha _ 1, \alpha _ 2, \alpha _ 3$ in terms of $\delta, \omega, u$.
Possible in practice
Actually finding the expressions requires some work. Firstly, direct calculation shows that
\[\delta = \sqrt{-4p^3 - 27q^2}\]The proof of the Kummer extension result finds $u$ as an $\omega$-eigenvector of $\sigma$ where $\langle \sigma \rangle = \text{Gal}(\mathbb Q(\alpha _ 1, \alpha _ 2, \alpha _ 3, \omega) / \mathbb Q(\delta, \omega))$. Since the group is cyclic, one eigenvector is given by
\[u = \frac{\alpha_1 + \omega \alpha_2 + \omega^2 \alpha_3}{3}\]We could have also considered $\omega^2$ to give another eigenvector $v$ of $\sigma$ (this time with eigenvalue $\omega^2$):
\[v = \frac{\alpha_1 + \omega^2 \alpha_2 + \omega \alpha_3}{3}\]Remember that $u^3$ and $v^3$ are in $\mathbb Q(\omega, \delta)$. By direct computation, we can show
- $u^3 v^3 = \frac{-p^3}{27}$
- $u^3 + v^3 = -q$
and thus $u^3$ and $v^3$ are the roots of the quadratic
\[z^2 + qz - \frac{p^3}{27} = 0\]and so may be written in terms of $p$ and $q$, and so $u$ and $v$ can be extracted by taking $p$th roots. Finally, we may write $\alpha _ 1, \alpha _ 2, \alpha _ 3$ in terms of $u$ and $v$ like so:
- $\alpha _ 1 = u + v$
- $\alpha _ 2 = \omega^2 u + \omega v$
- $\alpha _ 3 = \omega u + \omega^2 \beta$
as required.
More general results
There is a more general result in [[Notes - Galois Theory HT25, Solvability by radicals]]U which says:
If:
- $F$ is a field of characteristic zero
- $\alpha$ is algebraic over $F$
Then:
- $\alpha$ lies in a radical extension of $F$ if and only if $\text{Gal} _ F(m _ {F, \alpha})$ is solvable.
The “Possible in principle” section above is just a special case.
Flashcards
A complete characterisation of the Galois group in terms of the roots
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 $A _ 3$:
Let $\alpha$ be a root of $f$. Since $f$ is irreducible over $F$, $m _ {F, \alpha} = f$. Hence
\[[F(\alpha) : F] = \deg m_{F, \alpha} = \deg f = 3\]Hence $[F(\alpha) : F] \mid [K : F]$ 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
\[f(x) = (x - \alpha_1)(x - \alpha_2)(x - \alpha_3) \in \mathbb Q[x]\]
Can you give necessary and sufficient conditions for when $f$ has Galois group $C _ 3$?
It is tempting to say (by appealing to other roots) that this occurs when
\[\Delta = (\alpha_1 - \alpha_2)^2 (\alpha_1 - \alpha_3)^2(\alpha_2 - \alpha_3)^2\]is square in $\mathbb Q$. This is almost correct, but neglects reducibility: the correct answer is that the Galois group is $C _ 3$ iff $\alpha _ 1, \alpha _ 2, \alpha _ 3 \notin \mathbb Q$ and $\Delta \in \mathbb Q^2$.
@exam~
Simplifying substitution
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).
Resolvent quadratic
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 $v$
Why this is relevant:
- This is basically a derivation of the cubic formula; if $\alpha _ 1$ can be written in terms of $u$ and $v$ and $u^3$ and $v^3$ are the roots of a quadratic polynomial with coefficients in $F$, then you can calculate $\alpha _ 1$ by first solving the “resolvent quadratic” to derive $u^3$ and $v^3$, then plugging $u$ and $v$ into the expression for $\alpha _ 1$.
- This seems a lot more like random algebra than Galois theory. However, Galois theory is the reason we knew what to set $u$ and $v$ to, these are eigenvectors of $K(\omega)$ where $K$ is a splitting field of $f$ and $\omega$ is a cube root of unity.
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
\[\alpha_1 = u - \frac{p}{3u}\]