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I am trying to do the first exercise of Chapter VI of the book Algebra, by Serge Lang (page 320, whoever is using the revised third edition). The exercise is:

What is the Galois group of the following polynomials?

(c) $X^3-10$ over $\mathbb{Q}(\sqrt{2})$.

(d) $X^3-10$ over $\mathbb{Q}(\sqrt{-3})$.

(e) $X^3-X-1$ over $\mathbb{Q}(\sqrt{-23})$.

(f) $X^4-5$ over $\mathbb{Q}(\sqrt{5})$, $\mathbb{Q}(\sqrt{-5})$, $\mathbb{Q}(i)$.

Any help would be welcome, if there is any general strategy, at least to start. I am only used to determine the Galois group over $\mathbb{Q}$, using the examples from Dummit and Foote as a reference (they have plenty).

EDIT: I figure it out how to do (c), (d) and (e) using the discriminant method for cubic polynomials described in chapter VI, §2:

Let $f(X)$ be a cubic polynomial in $k[X]$, and assume char $k \neq 2$, $3$. Then:

(a) $f$ is irreducible over $k$ if and only if $f$ has no root in $k$,

(b) Assume $f$ irreducible . Then the Galois group of $f$ is $S_3$ if and only if the discriminant of $f$ is not a square in $k$. If the discriminant is a square, then the Galois group is cyclic of order 3, equal to the alternating group $A_3$ as a permutation of the roots of $f$.

The discriminant of $X^3-10$ is $-2700=-2^2\times 3^3\times 5^2$ so $\sqrt{-2700}=30\sqrt{-3}$ which belongs to $\mathbb{Q}(\sqrt{-3})$ but does not belong to $\mathbb{Q}(\sqrt{2})$. So (c) is $S_3$ and (d) is $A_3$.

For (e) the discriminant is $-23$, which is a square in $\mathbb{Q}(\sqrt{-23})$, so the Galois group is $A_3$.

As for (f), the only thing I know is that the roots are $\sqrt[4]{5}$, $i\sqrt[4]{5}$, $-\sqrt[4]{5}$ and $-i\sqrt[4]{5}$. Since $i\sqrt[4]{5}=\sqrt[4]{-5}$ so I'm assuming I have to compare $\sqrt[4]{\cdot}$ with $\sqrt{\cdot}$

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    For the first two, think of the automorphisms in the Galois group over the rationals that also fix the adjoined element. There's a difference between the two that becomes quite clear if you think of the splitting field embedded into the complex numbers. – 2017-02-09
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    Well, the maps I used over $\mathbb{Q}$ were $\sqrt[3]{10}\mapsto\sqrt[3]{10}$ or $\zeta_3\sqrt[3]{10}$ or $\zeta_3^2\sqrt[3]{10}$ and $\zeta_3\mapsto\zeta_3$ or $\zeta_3^2$. Any of those cases has nothing to do with $\sqrt{2}$. In fact, $\sqrt{2}$ don't affect anything on my previous solution over $\mathbb{Q}$. So I'm concerned what would be the difference on the previous proof. – 2017-02-09
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    You're right that they have no effect on $\sqrt{2}$. Now look at (d). – 2017-02-10
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    I am afraid there is no "magic theorem" which will do the job in all cases. In general, IF the given polynomial is simple enough that you can see what its roots are (all cases above except perhaps (e)), the natural procedure is to introduce intermediary fields by adding roots step by step. In case (c) for example, you must visibly introduce $ K = Q(\sqrt 2, \zeta_3)$, which is quadratic over $Q(\sqrt 2)$, and then show that the splitting field $L$ of $X^3 - 10$ is $K( \sqrt [3] {10})$ , with $Gal(L/K) = C_3$. – 2017-02-11
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    ... The final step is to show that $Gal(L/K) = S_3$ (not $C_6$). Note that the result is the same as over $Q$. This is because $Q(\sqrt 2)$ is real. The comparison with case (d) will be instructive. – 2017-02-11
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    Thanks for all the help guys, I put a solution for (c), (d) and (e) over there. But item (f) is still halfway through. – 2017-02-11

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