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Let $K = \mathbb{Q}(i2^{1/3}, 3^{1/4})$. Is this a Galois extension?

Text: Essentials of Modern Algebra, Cheryl Chute Miller

user26857
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user148884
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  • That doesn't seem like a reason why it would not be a Galois extension as abelian Galois extensions are cyclotomic by the Kronecker-Weber theorem. – Samuel Reid Jun 10 '14 at 22:37
  • If you know all the roots of the splitting field are in $K$, then it is by definition a normal extension of $\mathbb Q$, and so, since separability isn't a problem over $\mathbb Q$, is Galois. No? – jdc Jun 11 '14 at 01:09
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    What are you getting for the minimal polynomials? The minimal polynomial of $i2^{1/3}$ should have degree 6, while the minimal polynomial of $3^{1/4}$ is of degree 4. In particular, their product is not of degree 7. – RghtHndSd Jun 11 '14 at 01:18
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    @lhf What about using $\alpha^3 = -2i$ to show $i \in K$? Then you get $\omega \in K$ – John M Jun 11 '14 at 02:01
  • @JohnM what would the monic polynomial for this be if it was a root field? I think it has degree 12 but K has degree 24 as an extension. How can this be? – user148884 Jun 12 '14 at 22:28
  • This follow up question regarding degree is answered here. – John M Jun 13 '14 at 00:12

1 Answers1

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The conjugates of $\alpha=i2^{1/3}$ are $2^{1/3}\omega^k$, for $k=1,3,5,7,9,11$ and $\omega$ is a primitive $12$-root of unit: $\omega = (\sqrt 3 + i)/2$.

$\alpha^3 = -2i$ and so $i \in K$ and $2^{1/3} \in K$.

The conjugates of $\beta=3^{1/4}$ are $3^{1/4}i^k$ for $k=0,1,2,3$, and so are all in $K$.

$\beta^2=\sqrt 3$ implies that $\omega \in K$ and so all conjugates of $\alpha$ are in $K$.

Hence, $K=\mathbb Q(\alpha,\beta)$ contains all conjugates of $\alpha$ and $\beta$ and so $K$ is normal.

lhf
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