Why is group of automorphism of $\mathbb{Q}[\sqrt[3]{2}]/\mathbb{Q}$ is of order 1

Question

the question is as stated in the title. Why is this that $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$ is Galois (as the Galois group has two elements), but the corresponding one for third root of two is not Galois because the Automorphism group in $\mathbb{Q}(\sqrt[3]{2})$ that fixes $\mathbb{Q}$ only has one element?

Thank you.

Hi guys again,

I want to say thank to everyone. The answers are all very helpful. And I think I totally understand now. (Sorry that I can only accept one answer and can not vote up yet, I'm new here). Here would be my answer based on your answers (I prefer to not assume the existence of real number and complex number).

1. We start with $f(x) = x^3 - 2 \in \mathbb{Q}[x],$ which is irreducible. Thus $Q_1 = Q[x]/f(x)$ is a field. Let $\bar{X} \in Q_1,$ then $\bar{X}$ is a root for f(x) in $Q_1$. Now we need to prove that the other roots are not in $Q_1.$
2. Assume that $\exists a+b\bar{X};a,b\in \mathbb{Q}$ is a root of f(X). Then equating $(a+b\bar{X})^3 = \bar{X}^3$ in $\mathbb{Q}[x]/f(x),$ we will have that other root to be $\bar{X},$ which is a contradiction as polynomial in $\mathbb{Q}$ is separable. So $Q_1 = \mathbb{Q}(\bar{X})$ does not contain any of the other two roots.
3. Thus there can be only one automorphism in $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$, as any automorphism is completely characterized by how it acts on the roots (of the minimal polynomial?).

Please tell me if I get anything incorrect. Thank you again I appreciate the helps.

Answer 1

Every automorphism of $\mathbb{Q}(\sqrt[3]{2})$ has to send $\sqrt[3]{2}$ to a conjugate of $\sqrt[3]{2}$ in $\mathbb{Q}(\sqrt[3]{2})$.

Since $\sqrt[3]{2}$ is real and its other conjugates are not, every automorphism of $\mathbb{Q}(\sqrt[3]{2})$ has to send $\sqrt[3]{2}$ to itself and hence is the identity.

Answer 2

$\mathbb{Q}[\sqrt[3]{2}]$ is not a splitting field of any polynomial in $\mathbb{Q}[x]$. Therefore, it is not a Galois extension, and we conclude that the order of $Aut(\mathbb{Q}[\sqrt[3]{2}]/\mathbb{Q})$ is strictly less than $[\mathbb{Q}[\sqrt[3]{2}]:\mathbb{Q}] = 3$.

It is a theorem that the order of $Aut(\mathbb{Q}[\sqrt[3]{2}]/\mathbb{Q})$ must divide the degree of the field extension. Since $|Aut(\mathbb{Q}[\sqrt[3]{2}]/\mathbb{Q})| < 3$ by the above, then the only proper divisor of $3$ is $1$. Hence, the automorphism group has only $1$ element, namely the identity.

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On the other hand, $\mathbb{Q}[\sqrt{2}]$ is a splitting field of a polynomial in $\mathbb{Q}[x]$. Therefore, it is a Galois extension, so the order of $Aut(\mathbb{Q}[\sqrt{2}]/\mathbb{Q})$ is equal to $[\mathbb{Q}[\sqrt{2}]:\mathbb{Q}] = 2$.

Answer 3

A field extension $E/F$ is Galois if (and only if) it is the splitting field of a separable polynomial. (Over $\mathbb{Q}$, every irreducible polynomial is separable, so I won't mention this condition again.) A field $E$ is the splitting field for a polynomial $f(x)$ if it is the smallest field over which $f$ splits into linear factors, i.e., such that $f(x) = a(x - \alpha_1)(x - \alpha_2) \cdots (x - \alpha_n)$ where $\alpha_i \in E$ are all the roots of $f(x)$. That is, the splitting field is the smallest field containing all of the roots of $f(x)$.

Now let's consider the examples you mentioned. Note that $\sqrt{2}$ has minimal polynomial $f(x) = x^2 - 2$ over $\mathbb{Q}$. To see that $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$ is Galois, we must show that all the roots of $f(x)$ lie in $\mathbb{Q}(\sqrt{2})$. But $f$ has only one other root, namely $-\sqrt{2} \in \mathbb{Q}(\sqrt{2})$ and $f$ splits as $f(x) = (x - \sqrt{2})(x + \sqrt{2})$. Thus $\mathbb{Q}(\sqrt{2})$ is the splitting field for $f$.

Now let's consider your other example. Note that $\sqrt[3]{2}$ has minimal polynomial $g(x) = x^3 - 2$ over $\mathbb{Q}$. What are the other roots of $g(x)$? It's not too hard to see that they are simply $\zeta \sqrt[3]{2}$ and $\zeta^2 \sqrt[3]{2}$ where $\zeta$ is a primitive third root of unity. (Since we're working inside $\mathbb{C}$, we can just take $\zeta = e^{2\pi i /3}$.) Are these contained in $\mathbb{Q}(\sqrt[3]{2})$? The easiest way to see that the answer is no is to notice that we can embed $\mathbb{Q}(\sqrt[3]{2})$ inside $\mathbb{R}$ by taking $\sqrt[3]{2}$ to be the usual real cubed root of $2$. The other two roots of $g$, $\zeta \sqrt[3]{2}$ and $\zeta^2 \sqrt[3]{2}$, are complex, hence not contained in $\mathbb{R}$, hence not contained in $\mathbb{Q}(\sqrt[3]{2})$. This shows that $\mathbb{Q}(\sqrt[3]{2})$ is not the splitting field for $g$.