Show that $\mathbb{Q}(\sqrt[3]{2})$ does not have non trivial automorphisms

Question

Since I started the abstract algebra course, I haven't had exercises associated with the existence or non-existence of automorphisms, and I haven’t found exercises similar to this either so I don't know how to proceed. I would greatly appreciate your help.

Answer 1

The other answer use a bit too much machinery than what is required, I'll propose an answer that doesnt require any background knowledge.

Suppose that $\sigma : \mathbb{Q}(\sqrt[3]{2}) \to \mathbb{Q}(\sqrt[3]{2})$ is an automorphism. Since we must have that $\sigma(1) = 1$, its also true that $\sigma\left(\frac{a}{b}\right) = \frac{a\cdot\sigma(1)}{b\cdot\sigma(1)} = \frac{a}{b}$ for any $a,b \in \mathbb{Z}$, so $\sigma$ fixes every element in $\mathbb{Q}$.

Since $\sigma$ acts like the identity on $\mathbb{Q}$, the image of $\sqrt[3]{2}$ now uniquely determines all the values of $\sigma$ (since $\mathbb{Q}(\sqrt[3]{2})$ is generated by this element, by definition), so if we have that $\sigma(\sqrt[3]{2}) = \sqrt[3]{2}$, it must be the case that $\sigma = \text{id}$.

And we indeed have that, since $$ (\sqrt[3]{2})^3-2 = 0 \implies \sigma((\sqrt[3]{2})^3-2) = \sigma(0) \implies \left[\sigma(\sqrt[3]{2})\right]^3 - 2 = 0 $$

solving the last equation for $\sigma(\sqrt[3]{2})$, we must either have that, $\sigma(\sqrt[3]{2}) = \sqrt[3]{2}$, or $\sigma(\sqrt[3]{2}) = \omega\sqrt[3]{2}$, or $\sigma(\sqrt[3]{2}) = \omega^2\sqrt[3]{2}$, where $\omega = e^{2\pi i /3}$, the primitive $3^{\text{rd}}$ root of unity. The only one possible out of these is $\sigma(\sqrt[3]{2}) = \sqrt[3]{2}$, since $\sigma(\sqrt[3]{2})$ is necessarily a real value (its in $\mathbb{Q}(\sqrt[3]{2})$).

Answer 2

This problem is a special case of a general phenomenon. If $T$ is an automorphism on a field $F$, if $p(x)$ is a polynomial with coefficients in $F$, and if $\alpha \in F$ is a root of $p(x)$, then $T(\alpha) \in F$ is also a root of $p(x)$. Try showing this and then applying it here.

Answer 3

The Galois group of a field extension consists in the automorphisms that fix the base field and move the elements that generates the extension. If the extension is generated by a finite number of elements then the automorphisms characterized by the images of these elements, that are also roots of the minimal polynomial of these elements. Since the minimal polynomial of $\sqrt[3]{2}$ is $t^3-2$, which roots are $\sqrt[3]{2}, \sqrt[3]{2}w^2, \sqrt[3]{2}w^2$ where $w=e^{2\pi i/3}$, the unique roots in the extension is the first, so the only movement between roots is $\sqrt[3]{2}\to \sqrt[3]{2}$, that is the identity

Answer 4

It suffices to show that any automorphism $a$ of $\Bbb Q(\alpha)$ will restrict to an automorphism of $\Bbb Q$. This can be done, and is actually true for any field extension of $\Bbb Q$.

Then it must be the identity on $\Bbb Q$.

Then, $a(\alpha)=\alpha $. That's because $a$ has to permute the roots $x^3-2$. But the other two roots are not real. And clearly $\Bbb Q(\alpha)\subset\Bbb R$.

Then it's the identity on $\Bbb Q(\alpha)$.

Answer 5

The field extension $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ is not normal (and therefore not galois) as the minimal polynom $X^3-2$ of $\sqrt[3]{2}$ has other solutions ($\zeta_3\sqrt[3]{2}$ and $\zeta_3^2\sqrt[3]{2}$ with $\zeta_3=e^{\frac{2\pi i}{3}}$) that are not in $\mathbb{Q}(\sqrt[3]{2})$. Let $H=\operatorname{Aut}(\mathbb{Q}(\sqrt[3]{2}))=\operatorname{Aut}(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q})$, then according to Artin's theorem, we have: $$|H| =[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}(\sqrt[3]{2})^H] =\frac{[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]}{[\mathbb{Q}(\sqrt[3]{2})^H:\mathbb{Q}]} \Rightarrow \frac{3}{|H|} =[\mathbb{Q}(\sqrt[3]{2})^H:\mathbb{Q}]\in\mathbb{N},$$ which implies $|H|=1$ or $|H|=3$. The latter case is not possible as $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ is not galois.