Why isn’t $\mathbb{Q}[x]/\langle x^3-2\rangle$ a normal extension over $\mathbb{Q}$?

Question

The extension $\mathbb{Q}[x]/\langle x^3-2\rangle$ is given on my lecture notes as an example of a non normal extension over $\mathbb{Q}$. I understand why $\mathbb{Q}[\sqrt[3]{2}]$ is not: because it contains one of the roots of the irreducible polynomial $x^3-2$ (namely, $\sqrt[3]{2}$) but not all of them (the two non-real ones).

I know $\mathbb{Q}[x]/\langle x^3-2\rangle \cong \mathbb{Q}[\sqrt[3]{2}]$ and I guess that’s enough to conclude, but is there a way of checking the definition of normal extension doesn’t apply directly to $\mathbb{Q}[x]/\langle x^3-2\rangle$?

Answer 1

Consider a field $\mathbf Q(\alpha)$ where $\alpha^3 = 2$, which is a cubic extension. If this field has a second root $\alpha'$ of $x^3 - 2$, then $\omega = \alpha'/\alpha$ is a cube root of unity that is not $1$: a nontrivial cube root of unity. That makes $\omega$ a root of $(x^3-1)/(x-1) = x^2+x+1$, but this polynomial is irreducible over $\mathbf Q$, so $\mathbf Q(\omega)$ is a quadratic extension of $\mathbf Q$, and a quadratic extension can't lie inside a cubic extension.

The same argument shows for an arbitrary rational number $r$ that is not a cube (replace $2$ above with $r$), the cubic field $\mathbf Q(\sqrt[3]{r})$ is not a normal extension of $\mathbf Q$: if it were normal then it would contain $\mathbf Q(\omega)$, which is a quadratc extension.

More generally, if $n > 1$ is odd and $r$ is a rational number such that $x^n - r$ is irreducble over $\mathbf Q$, then we can show the field $K := \mathbf Q(\sqrt[n]{r})$ is not a normal extension of $\mathbf Q$ by contradiction. Assuming it is normal, $x^n - r$ splits completely over $K$ since it has a root in $K$: $$ x^n - r = \prod_{j=1}^n (x - \alpha_i), $$ where $\alpha_1, \ldots, \alpha_n$ are in $K$. Then the ratios $\alpha_i/\alpha_1$ (dividing all $n$th roots of $r$ by a fixed choice of $n$th root of $r$) are a full set of $n$th roots of unity in $K$: $$ \prod_{j=1}^n (x - \alpha_i/\alpha_1) = \frac{1}{\alpha_1^n}\prod_{j=1}^n ((\alpha_1x) - \alpha_i) = \frac{1}{r}((\alpha_1x)^n - r) = \frac{rx^n-r}{r} = x^n-1, $$ so the $n$th cyclotomic field is contained in $\mathbf Q(\sqrt[n]{r})$. Counting field degrees, we get $\varphi(n) \mid n$, but $\varphi(n)$ is even and $n$ is odd, so we have a contradiction.

Answer 2

Let $K=\Bbb{Q}[a]/(a^3-2)$. To show that $K/\Bbb{Q}$ isn't normal without appealing to $\Bbb{R}$:

$(ax)^3-2= 2(x^3-1)$ would split completely over $K$ if $K/\Bbb{Q}$ was normal.

Check that $x^2+x+1$ is irreducible over $\Bbb{Q}$.

So it stays irreducible over the cubic extension $K$ (if $K/E/\Bbb{Q}$ then $[E:\Bbb{Q}]|[K:\Bbb{Q}]$)

Whence $x^3-1=(x-1)(x^2+x+1)$ doesn't split completely over $K$.