I want to show that $\mathbb{Q}(i,\sqrt[4]{1+i} )$ is normal and thus nonabelian extension of $\mathbb{Q}$, but got stuck.
The main problem is that for the extension to be normal $\root4\of{1-i}$ should also be an element. I failed to show that.
Thank you for your helps.
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My question arose from this document, page 5, Remark 2.7. It is given as a counterexample of Kronecker and Weber theorem:
When $K=\mathbb{Q}(i)$, $K(\root4\of{1-i})$ can never be any intermediate field of $K({ \zeta }_{ n })/K$, where ${ \zeta }_{ n }$ is a primitive $n$-th root of unity.
I first expected that above statement could be proved by showing that $K(\root4\of{1-i})$ is not abelian over $\mathbb{Q}$ since any cyclotomic extension of K is a cyclotomic extension of $\mathbb{Q}$ thus abelian.
However from your advices, it seems that it's not even normal. But I can't still understand the method you showed me..
Is there any more easy explanation for this?
When $K=\mathbb{Q}(i)$, $K(\root4\of{1-i})$ can never be any intermediate field of $K({ \zeta }_{ n })/K$, where ${ \zeta }_{ n }$ is a primitive $n$-th root of unity.
I only know about elementary field and Galois theory.
Best Answer
Introduce the extensions $k=\mathbf Q(i)=\mathbf Q(1+i), K=k(\sqrt {1+i})=\mathbf Q(i,\sqrt {1+i}), L=\mathbf Q(i,\sqrt [4] {1+i})$. By construction, the extension $K/k$ is quadratic, and from the definition of normality, $K/\mathbf Q$ is normal iff $K$ is stabilized by all the prolongations of the generator $\gamma$ of $\mathrm{Gal}(k/\mathbf Q)$ ("complex conjugation") to embeddings of $K$ into $\bar {\mathbf Q}$. Obviously, this happens iff $\gamma ({1+i})/(1+i)\in {k^*}^{2}$. But the quotient is equal to $i$, so $K/\mathbf Q$ is not normal.
Let us look now at $L/\mathbf Q$. Since $k=\mathbf Q(1+i)$ contains the $4$-th roots of unity, Kummer theory tells us that $L/k$ is cyclic of degree $4$. It follows that the Kummer dual $R=\left<1+i\right>\bmod {k^*}^4\subset {k^*/k^*}^4$ of $\mathrm{Gal}(L/k)$ is also cyclic of order $4$, and $L/\mathbf Q$ is normal iff $\gamma (R)=R$, which means that $ (1-i)/(1+i)^h\in {k^*}^{4}$, for a certain $h$ which is invertible mod $4$, i.e. $h \equiv \pm 1$ mod $4$. All calculations done, I find that this cannot happen. So there must be an error, either in my calculations or in the question.