Show that $\mathbb{Q}(a)/\mathbb{Q}$ is normal, where $a$ is a root of the irreducible polynomial $x^3-3x-1$

Question

We have that $E=\mathbb{Q}(a)$, where $a\in \mathbb{C}$ is a root of the irreducible polynomial $x^3-3x-1\in \mathbb{Q}[x]$.

I want to show that $E/\mathbb{Q}$ is normal.

I have done the following:

Let $b\in E$. A basis of the extension is $1, a, a^2$. So, $b$ can be written as $$b=q_0+q_1a+q_2a^2$$

We have to find the minimal irreducible polynomial of $b$ over $\mathbb{Q}$ and compute the other roots to check if they are in $E$, or not?

Since $[E:\mathbb{Q}]=3$, we have that $\deg m(b,\mathbb{Q})\leq 3$.

So, the general form of that polynomial is $Ax^3+Bx^2+Cx+D$.

So, do we have to replace $x$ with $b=q_0+q_1a+q_2a^2$, compute that polynomial, knowing that $a^3=3a+1$, and find the other roots?

Or is there an other way to show that?

You only have to show that the other two roots of $x^3 - 3x - 1$ are in $E$. That makes $E$ the splitting field of $x^3 - 3x - 1$ over $\Bbb Q$, which by definition makes $E/\Bbb Q$ a normal extension. — Arthur, Dec 31 '16 at 14:25
Are the other two roots $\omega a$ and $\omega^2 a$, where $\omega$ is the cubic root of $1$ ? @Arthur — Mary Star, Dec 31 '16 at 14:28
Do we use the Euclidean division to find the other roots? @Arthur — Mary Star, Dec 31 '16 at 14:30
Using the Euclidean division we get $x^3-3x-1=(x-1)(x^2+ax+(a^2-3))$. Using the discriminant we have to find the other two roots, right? @Arthur — Mary Star, Dec 31 '16 at 14:35
@MaryStar: $(x-1)$ is a factor of the polynomial $p(x)$ if and only if $1$ is a root of $p(x)$. However, $1$ is clearly not a root of $x^3-3x-1$, since $1^3-3-1=-3$, not $0$. — Zev Chonoles, Dec 31 '16 at 14:36
Sorry, I meant $x^3-3x-1=(x-a)(x^2+ax+(a^2-3))$. @ZevChonoles — Mary Star, Dec 31 '16 at 14:37
Using the disciminant, the other roots are $$x=\frac{-a+\sqrt{12-3a^2}}{2}, \ \ x=\frac{-a-\sqrt{12-3a^2}}{2} $$ right? But is this square root an element of $E$ ? — Mary Star, Dec 31 '16 at 14:39
@MaryStar Great job so far! Whether $12 - 3a^2$ has a square root in $E$ is the last thing you need to figure out. Therefore, I suggest you try to solve $$(k + la + ma^2)^2 = 12-3a^2$$for rational $k, l, m$. — Arthur, Dec 31 '16 at 14:40
We get the following system: $$\left{\begin{matrix}k^2+2kl+6lm=0 \ l^2+3m^2+2km=-3 \ k^2+2lm=12\end{matrix}\right.$$ right? How could we solve it? @Arthur — Mary Star, Dec 31 '16 at 14:59
FYI, the roots are $2\cos\frac{2\pi}{9}$, $2\cos\frac{4\pi}{9}$ and $2\cos\frac{8\pi}{9}$. Clearly the second and third root are rational expressions of the first one. — egreg, Dec 31 '16 at 15:00
@MaryStar Set $x=2\cos\theta$, so the equation becomes $4\cos^3\theta-3\cos\theta=-\frac{1}{2}$, so $\cos3\theta=-\frac{1}{2}$. Old trick by Viète. — egreg, Dec 31 '16 at 15:15
I get the equation set $$\left{\begin{align}k^2 +2lm &= 12\m^2 + 2kl + 6lm &= 0\l^2 + 2km + 3m^2 &= -3\end{align}\right.$$(I think your first $k^2$ should've been an $m^2$) but you're right, that looks like a mess to solve. — Arthur, Dec 31 '16 at 15:22
That was the straight-forward way, and apparently it failed to yield a result in any nice way. That means it's time for tricks, and I think the trigonometric substitution is the best one so far. — Arthur, Dec 31 '16 at 16:25
And see there for showing the discriminant is a square $\implies$ the over roots lie in $\mathbb{Q}(a)$ @Arthur — reuns, Jan 01 '17 at 09:20
@user1952009 We are already aware that the other roots are given by $x=\frac{-a\pm\sqrt{12-3a^2}}{2}$, which is clearly in the field if only the discriminant $12 - 3a^2$ is a square. — Arthur, Jan 01 '17 at 09:53
@Arthur How did you factor the polynomial ? That's what my link is about, from the discriminant property $\Delta = \prod_{i,j} (a_i-a_j) = -4b^3-27c^2$ — reuns, Jan 01 '17 at 09:56
@user1952009 I factored the polynomial by noting that $(x-a)$ must be a factor, and then doing long division. — Arthur, Jan 01 '17 at 09:57
I see, but in general you might have some troubles for showing the discriminant of the quadratic is a square ? — reuns, Jan 01 '17 at 10:01

score 3 · Answer 1 · edited Jun 12 '20 at 10:38

The discriminant (Wikipedia) of the polynomial $p(x)=x^3-3x+1\in\mathbb{Q}[x]$ is equal to $81$, which is a square in the field $\mathbb{Q}$. Letting $L$ be the splitting field of $p(x)$ over $\mathbb{Q}$, there is a theorem that since the discriminant is a square, we have $[L:\mathbb{Q}]=3$. Since $\mathbb{Q}(a)\subseteq L$ and $[\mathbb{Q}(a):\mathbb{Q}]=3$ also, we must have $L=\mathbb{Q}(a)$, i.e. the field $\mathbb{Q}(a)$ is the splitting field of the polynomial over $\mathbb{Q}$. Therefore it is normal over $\mathbb{Q}$.

(from Galois Theory by Jean-Pierre Escofier)

score 3 · Answer 2 · edited Apr 13 '17 at 12:20

Hint: If $a$ is one of the roots, show that $a^2-2$ is another.

As Zev (+1) explained, observing that the discriminant is a square (of a rational number) implies that the Galois group of the splitting field is cyclic of order three. Therefore:

$E=\Bbb{Q}(a)$, and
the other zeros are at most quadratic polynomials in $a$.

This is one of the two cubic polynomials, where $a\mapsto a^2-2$ permutes the three zeros cyclically. The other is discussed here.

A few words about figuring out that the zeros $x_k=2\cos(2^k\pi/9), k=1,2,3,$ come up here, or rather, why the splitting field more or less "must" be the real subfield of the ninth cyclotomic field. This is not necessarily conclusive but pointing heavily in that direction:

By the famous Kronecker-Weber theorem any abelian extension of $\Bbb{Q}$ is a subfield of $\Bbb{Q}(e^{2\pi i/m})$ for some $m$.
For any prime $p$, the discriminant of $\Bbb{Q}(e^{2\pi i/p})$ is a power of $p$ (up to sign?).
Because the prime factors of the discriminant of a subfield are also factors of the discriminant of the bigger, and Zev calculated that the discriminant is $81$, we can conclude that in our case $m$ must be divisible by $p=3$, but not necessarily divisible by any other prime.
$m=9=3^2$ is the smallest power of three such that $Q(e^{2\pi i/m})$ has a cubic subfield.

See here. Spoiler ALERT. – Jyrki Lahtonen Dec 31 '16 at 14:51 — Jyrki Lahtonen, Dec 31 '16 at 14:51

Show that $\mathbb{Q}(a)/\mathbb{Q}$ is normal, where $a$ is a root of the irreducible polynomial $x^3-3x-1$

2 Answers2