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How do I show that $\mathbb{Q}(\sqrt[4]{7},\sqrt{-1})$ is Galois?

At first I thought it was the splitting field of $x^4-7$, but I was only able to prove that it was a subfield of the splitting field. Any ideas?

I'm trying to find all the intermediate fields in terms of their generators, but I don't understand how. I am trying to imitate Dummit and Foote on this. I am looking at the subgroups of the Galois group in terms of $\tau$ where $\tau$ is the automorphism that takes $\sqrt[4]{7}$ to itself and $i$ to $-i$, and $\sigma$ that takes $\sqrt[4]{7}$ to $i\sqrt[4]{7}$ and $i$ to itself. How do I, for example, find the subfield corresponding to $\{1, \tau\sigma^3\}$? I know I am supposed to find four elements of the galois group that $\tau\sigma^3$ fixes, but so far i can only find $-\sqrt[4]{7}^3$.

MathTeacher
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  • If you can show that $\mathbb{Q}(\sqrt[4]{7},i)$ is a subfield of the splitting field, then show that every root of $x^4-7$ is in $\mathbb{Q}(\sqrt[4]{7})$ and you'll be done. – Arturo Magidin Dec 04 '11 at 06:55
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    @MathMastersStudent: You are on the right track with only a small step left, what went wrong? – Eric Naslund Dec 04 '11 at 06:58
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    You have asked 9 questions, and you haven't accepted any answers. Do you know how to accept answers? Do you understand the importance to this website of accepting answers? – Gerry Myerson Dec 04 '11 at 07:00
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    Gerry, so sorry! I didn't know there was such a thing. I will check the meta-section and accept ASAP. This ite is wonderful. – MathTeacher Dec 04 '11 at 07:24

2 Answers2

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The splitting field of your polynomial is generated by the $a=\sqrt 7$, $-a$, $ia$ and $-ia$, its roots. It therefore contains $a$ and $i=\frac{ia}{a}$. Obviously, the field generated by $a$ and $i$ also contains the four roots.

Mariano Suárez-Álvarez
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The polynomial $f(x)=x^4-7$ factors as $(x-7^{1/4})(x+7^{1/4})(x-i7^{1/4})(x+i7^{1/4})$, and all these irreducible factors are distinct. Hence, $x^4-7$ is separable. Moreover, the field $L= \mathbb{Q}(7^{1/4},i)$ contains all its roots and is the minimal field where $x^4-7$ factors completely in. Hence, it is the minimal splitting field of $f(x)$. So the extension $K \subset L$ is both normal and separable and hence, is Galois.

Fourier_T
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