Let \(K=\mathbb{Q}(\sqrt{2})\). The element \(\sqrt{2}\) satisfies
\[
x^2-2=0.
\]
The polynomial \(x^2-2\) is irreducible over \(\mathbb{Q}\) by the rational root test, so it is the minimal
polynomial of \(\sqrt{2}\). Hence
\[
[K:\mathbb{Q}] = 2.
\]

If \(\sigma\in \operatorname{Aut}_{\mathbb{Q}}(K)\), then \(\sigma(\sqrt{2})\) must be a root of \(x^2-2\). Therefore
\[
\sigma(\sqrt{2})=\sqrt{2}\quad \text{or}\quad \sigma(\sqrt{2})=-\sqrt{2}.
\]
Both possibilities occur.

Therefore
\[
\operatorname{Aut}_{\mathbb{Q}}(\mathbb{Q}(\sqrt{2}))\cong C_2.
\]
Moreover, \(x^2-2\) splits completely in \(\mathbb{Q}(\sqrt{2})\), so this extension is Galois.[1], [2]

Let \(K=\mathbb{Q}(\alpha)\) with \(\alpha=\sqrt[3]{2}\). Then \(\alpha\) satisfies
\[
x^3-2=0.
\]
The polynomial \(x^3-2\) is irreducible over \(\mathbb{Q}\) by Eisenstein’s criterion at \(p=2\).[5]
Therefore
\[
[K:\mathbb{Q}] = 3.
\]

The roots of \(x^3-2\) in \(\mathbb{C}\) are
\[
\alpha,\qquad \alpha\zeta_3,\qquad \alpha\zeta_3^2.
\]
But \(K=\mathbb{Q}(\alpha)\subseteq \mathbb{R}\), because every element of \(K\) has the form
\[
a+b\alpha+c\alpha^2 \qquad (a,b,c\in \mathbb{Q}),
\]
and this is real. The other two roots are nonreal, so they do not lie in \(K\).

Therefore any \(\mathbb{Q}\)-automorphism \(\sigma\) must satisfy
\[
\sigma(\alpha)=\alpha.
\]
Since \(\alpha\) generates the field, \(\sigma\) must be the identity.

Hence
\[
\operatorname{Aut}_{\mathbb{Q}}(\mathbb{Q}(\sqrt[3]{2}))=\{\operatorname{id}\}.
\]

This is a crucial example: the field has degree \(3\), but its automorphism group has only one element.
The reason is that the extension is not normal; the polynomial \(x^3-2\) does not split in the field.
In characteristic \(0\), finite Galois extensions can also be characterized by the condition
\[
|\operatorname{Aut}(L/K)|=[L:K].
\]
That fails here.[1], [2]

Let
\[
K=\mathbb{Q}(\sqrt{2},i).
\]
We know
\[
[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2,\qquad [\mathbb{Q}(i):\mathbb{Q}]=2.
\]

To compute the degree of the composite field, note that
\[
\mathbb{Q}(\sqrt{2})\subseteq \mathbb{R},
\]
while the only real elements of \(\mathbb{Q}(i)\) are rational numbers. Therefore
\[
\mathbb{Q}(\sqrt{2})\cap \mathbb{Q}(i)=\mathbb{Q},
\]
and hence
\[
[K:\mathbb{Q}] = 2\cdot 2=4.
\]

Now let \(\sigma\in \operatorname{Aut}_{\mathbb{Q}}(K)\). Since \(\sqrt{2}\) satisfies \(x^2-2=0\),
\[
\sigma(\sqrt{2})=\pm \sqrt{2}.
\]
Since \(i\) satisfies \(x^2+1=0\),
\[
\sigma(i)=\pm i.
\]

These choices are independent, so there are four automorphisms:

Each nontrivial element has order \(2\), and the automorphisms commute. Thus
\[
\operatorname{Gal}(\mathbb{Q}(\sqrt{2},i)/\mathbb{Q})\cong C_2\times C_2.
\]

This field is the splitting field of \((x^2-2)(x^2+1)\), so it is Galois.[1], [2]

Let
\[
K=\mathbb{Q}(\alpha,i),\qquad \alpha=\sqrt[3]{2}.
\]
As above,
\[
[\mathbb{Q}(\alpha):\mathbb{Q}]=3,\qquad [\mathbb{Q}(i):\mathbb{Q}]=2.
\]
Since \(\mathbb{Q}(\alpha)\subseteq \mathbb{R}\) and the only real elements of \(\mathbb{Q}(i)\) are rational,
\[
\mathbb{Q}(\alpha)\cap \mathbb{Q}(i)=\mathbb{Q}.
\]
Hence
\[
[K:\mathbb{Q}]=6.
\]

The subtle question is the image of \(\alpha\). Since \(\alpha\) satisfies \(x^3-2=0\), we might first think
that \(\sigma(\alpha)\) could be any of
\[
\alpha,\qquad \alpha\zeta_3,\qquad \alpha\zeta_3^2.
\]
However, this only works if the target element lies inside \(K\).

Does \(K\) contain \(\zeta_3\)? Recall that
\[
\zeta_3=\frac{-1+i\sqrt{3}}{2}.
\]
If \(\zeta_3\in K=\mathbb{Q}(\alpha,i)\), then \(\sqrt{3}\) would belong to \(\mathbb{Q}(\alpha)\), because
every element of \(K\) can be written as \(u+vi\) with \(u,v\in \mathbb{Q}(\alpha)\). But that is impossible:
\(\mathbb{Q}(\alpha)\) has degree \(3\), so it cannot contain a quadratic subfield.[1], [2]

Thus every \(\mathbb{Q}\)-automorphism must fix \(\alpha\):
\[
\sigma(\alpha)=\alpha.
\]
On the other hand, \(i\) can still go to \(\pm i\), because it is a root of \(x^2+1\).

So there are exactly two automorphisms:
\[
\operatorname{id}:\ \alpha\mapsto \alpha,\ i\mapsto i,
\]
and
\[
\kappa:\ \alpha\mapsto \alpha,\ i\mapsto -i.
\]
Therefore
\[
\operatorname{Aut}_{\mathbb{Q}}(\mathbb{Q}(\sqrt[3]{2},i))\cong C_2.
\]

This extension is still not Galois, because adjoining \(i\) does not force the polynomial \(x^3-2\) to split.
The missing ingredient is \(\zeta_3\), not \(i\).[1], [2]

Let \(K=\mathbb{Q}(\zeta_5)\), where \(\zeta_5=e^{2\pi i/5}\) is a primitive fifth root of unity.
The minimal polynomial of \(\zeta_5\) over \(\mathbb{Q}\) is the fifth cyclotomic polynomial
\[
\Phi_5(x)=x^4+x^3+x^2+x+1.
\]
It is irreducible over \(\mathbb{Q}\). One standard proof is to substitute \(x+1\) and then apply
Eisenstein’s criterion.[4], [5]

Therefore
\[
[K:\mathbb{Q}] = \deg \Phi_5 = 4.
\]

The conjugates of \(\zeta_5\) are the other primitive fifth roots of unity:
\[
\zeta_5,\ \zeta_5^2,\ \zeta_5^3,\ \zeta_5^4.
\]
So any \(\mathbb{Q}\)-automorphism must send
\[
\zeta_5 \longmapsto \zeta_5^a
\]
for some \(a\in\{1,2,3,4\}\). Conversely, each such choice defines an automorphism.[3], [4]

Since \((\mathbb{Z}/5\mathbb{Z})^\times\) is cyclic of order \(4\), we obtain
\[
\operatorname{Gal}(\mathbb{Q}(\zeta_5)/\mathbb{Q})\cong C_4.
\]
A generator is the automorphism \(\sigma_2\) given by
\[
\sigma_2(\zeta_5)=\zeta_5^2.
\]

Cyclotomic fields are splitting fields of \(x^n-1\), so \(\mathbb{Q}(\zeta_5)/\mathbb{Q}\) is Galois.[3], [4]

Let
\[
K=\mathbb{Q}(\zeta_5,\zeta_3).
\]
We already know
\[
[\mathbb{Q}(\zeta_5):\mathbb{Q}]=4,\qquad [\mathbb{Q}(\zeta_3):\mathbb{Q}]=2,
\]
because \(\zeta_3\) satisfies \(x^2+x+1=0\).[3], [4]

To compute the degree of the composite, we determine the intersection. Since
\[
\mathbb{Q}(\zeta_3)=\mathbb{Q}(\sqrt{-3}),
\]
any nontrivial intersection would give a quadratic subfield of \(\mathbb{Q}(\zeta_5)\).

But \(\operatorname{Gal}(\mathbb{Q}(\zeta_5)/\mathbb{Q})\cong C_4\) has exactly one subgroup of order \(2\), so
\(\mathbb{Q}(\zeta_5)\) has exactly one quadratic subfield. One can identify it by setting
\[
s=\zeta_5+\zeta_5^{-1}.
\]
Then \(s\) satisfies
\[
s^2+s-1=0,
\]
so that quadratic subfield is \(\mathbb{Q}(\sqrt{5})\), not \(\mathbb{Q}(\sqrt{-3})\). Hence
\[
\mathbb{Q}(\zeta_5)\cap \mathbb{Q}(\zeta_3)=\mathbb{Q}.
\]

Therefore
\[
[K:\mathbb{Q}] = 4\cdot 2 = 8.
\]

A \(\mathbb{Q}\)-automorphism of \(K\) must send
\[
\zeta_5\mapsto \zeta_5^a \quad (a\in (\mathbb{Z}/5\mathbb{Z})^\times),
\]
and
\[
\zeta_3\mapsto \zeta_3^b \quad (b\in (\mathbb{Z}/3\mathbb{Z})^\times=\{1,2\}).
\]
Thus there are \(4\cdot 2=8\) automorphisms.

The two most useful generators are
\[
\sigma:\ \zeta_5\mapsto \zeta_5^2,\ \zeta_3\mapsto \zeta_3,
\]
and
\[
\tau:\ \zeta_5\mapsto \zeta_5,\ \zeta_3\mapsto \zeta_3^2.
\]
Here \(\sigma\) has order \(4\), \(\tau\) has order \(2\), and they commute. Hence
\[
\operatorname{Gal}(\mathbb{Q}(\zeta_5,\zeta_3)/\mathbb{Q})\cong C_4\times C_2.
\]

1. Show directly that every \(\mathbb{Q}\)-automorphism of \(\mathbb{Q}(\sqrt{3})\) is determined by the image of \(\sqrt{3}\), and compute the automorphism group.
2. Explain why \(\mathbb{Q}(\sqrt[3]{2})\) is contained in \(\mathbb{R}\), but \(\mathbb{Q}(\sqrt[3]{2},\zeta_3)\) is not.
3. Prove that \(\mathbb{Q}(\zeta_3)=\mathbb{Q}(\sqrt{-3})\).
4. Verify carefully that
   \[
   \zeta_5+\zeta_5^{-1}
   \]
   satisfies \(x^2+x-1=0\).
5. Find all \(\mathbb{Q}\)-automorphisms of \(\mathbb{Q}(i,\sqrt{3})\) and identify the group.

1. J. S. Milne, Fields and Galois Theory, version 5.10, 2022. Available online through James Milne’s course notes.
2. Keith Conrad, The Galois Correspondence. Introductory discussion of \(K\)-automorphisms and the basic philosophy of Galois theory.
3. Keith Conrad, Cyclotomic Extensions. Background on roots of unity and cyclotomic extensions.
4. Keith Conrad, Galois Group of Cyclotomic Fields over \(\mathbb{Q}\). A concise proof-oriented note on the structure of \(\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})\).
5. Margherita Barile, “Eisenstein’s Irreducibility Criterion,” MathWorld.