# Example of a Galois extension $L/\mathbb{Q}$ with $\text{Gal}(L/\mathbb{Q})\cong\mathbb{Z}_{4}$

I propose that $$L=\mathbb{Q}(i\sqrt[4]{2})$$.

Obviously $$\mathbb{Q}(i\sqrt[4]{2})$$ is the splitting field of $$f=t^{4}-2\in\mathbb{Q}[t]$$, since

$$N:=\{\text{zeros of f}\}=\{\sqrt[4]{2},-\sqrt[4]{2},i\sqrt[4]{2},-i\sqrt[4]{2}\}\subseteq\mathbb{Q}(i\sqrt[4]{2})$$

Hence, since $$f$$ is irreducible by Eisenstein's criterion, we get that

$$[\mathbb{Q}(i\sqrt[4]{2}):\mathbb{Q}]=[f:\mathbb{Q}]:=\text{deg}(m_{f})=4$$

Because $$\mathbb{Q}(i\sqrt[4]{2})$$ is the splitting field of $$f$$ over $$\mathbb{Q}$$ we know that the extension $$\mathbb{Q}(i\sqrt[4]{2})/\mathbb{Q}$$ is galois and therefore $$[\mathbb{Q}(i\sqrt[4]{2}):\mathbb{Q}]=|\text{Gal}(\mathbb{Q}(i\sqrt[4]{2})/\mathbb{Q})|=4$$

It is know, that there are only 2 groups of order $$4$$ up to isomorphisms: $$\mathbb{Z}_{4}$$ and $$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$$

No I'm not quite sure whether this part of my "proof" is correct

$$\text{Gal}(\mathbb{Q}(i\sqrt[4]{2})/\mathbb{Q}):=\langle\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\rangle$$

Let $$\sigma_{i}$$ be defined as followed

$$\sigma_{i|\mathbb{Q}}=$$ id for $$i=1,2,3,4$$ and

$$\sigma_{1}(i\sqrt[4]{2})=\sqrt[4]{2}$$

$$\sigma_{2}(i\sqrt[4]{2})=-\sqrt[4]{2}$$

$$\sigma_{3}(i\sqrt[4]{2})=i\sqrt[4]{2}$$

$$\sigma_{4}(i\sqrt[4]{2})=-i\sqrt[4]{2}$$

We now see that $$\text{Gal}(\mathbb{Q}(i\sqrt[4]{2})/\mathbb{Q})=\langle\sigma_{3}\rangle$$

thus, it is a cyclic group and since $$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$$ is not cyclic we get

$$\text{Gal}(\mathbb{Q}(i\sqrt[4]{2})/\mathbb{Q})\cong\mathbb{Z}_{4}$$

Now is this approach correct, and are there ways to improve it?

• $\Bbb Q(i\sqrt[4]2)$ is not the splitting field of $t^4-2$; indeed it is not even Galois over $\Bbb Q$. Feb 12 '19 at 20:24

If you want a field with Galois group $$\Bbb Z_4$$ over $$\Bbb Q$$, take a prime $$p$$ with $$p\equiv1\pmod4$$. Then $$\Bbb Q(\zeta_p)$$ has Galois group $$G \cong\Bbb Z_{p-1}$$ over $$\Bbb Q$$ (where $$\zeta_p=\exp(2\pi i/p)$$). But $$G$$ has a subgroup $$H$$ with $$G/H\cong \Bbb Z_4$$. Then the fixed field of $$H$$ has Galois group $$\Bbb Z_4$$ over $$\Bbb Q$$.
• Why is $G\cong\mathbb{Z}_{p-1}$? Feb 12 '19 at 20:31
• (+1) - and just to say the obvious: a fine example, of course, is with $p=5$. Feb 12 '19 at 20:32
All cyclic extensions $$L/\mathbf Q$$ of degree $$4$$ can be obtained in two successive "kummerian steps" in the following way. Take first a quadratic field $$K=\mathbf Q (\sqrt d)$$, with Galois group $$G=$$ over $$\mathbf Q$$, then a quadratic extension $$L/K$$, which can alxays be written as $$L=K(\sqrt a)$$, with $$a \in K^*/{K^*}^2$$ (slightly abusing langauage). It is easy to prove that $$L/K$$ is Galois iff $$a \in {(K^*/{K^*}^2)}^G$$ (fixed points by $$G$$) [Hint : try to extend to action of $$G$$ to $$\sqrt a$$ ], i.e. $$s(a)/a =x^2$$, with $$x\in K^*$$. By Hilbert's thm. 90, this hypothesis is equivalent to $$N(x)=\pm 1$$, where $$N$$ is the norm map of $$K/\mathbf Q$$, and it is again easy to show that $$Gal(L/\mathbf Q)$$ is cyclic iff $$N(x)=-1$$. By quadratic reciprocity, the existence of such an $$x$$ is equivalent to $$d$$ being a norm in $$\mathbf Q (i)/\mathbf Q$$, i.e. $$d$$ being the sum of two squares in $${\mathbf Q}^*$$. This "explains" in particular the condition $$p\equiv 1$$ mod $$4$$ in the cyclotomic example given by @Lord Shark the Unknown.