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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?

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    $\begingroup$ $\Bbb Q(i\sqrt[4]2)$ is not the splitting field of $t^4-2$; indeed it is not even Galois over $\Bbb Q$. $\endgroup$ – Lord Shark the Unknown Feb 12 at 20:24
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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 $\Bbb Q$.

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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.

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