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


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


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

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





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


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

  • 2
    $\begingroup$ $\Bbb Q(i\sqrt[4]2)$ is not the splitting field of $t^4-2$; indeed it is not even Galois over $\Bbb Q$. $\endgroup$ 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 $\Bbb Q$.


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 $G=<s>$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.