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I would appreciate it if you could verify my reasoning is correct or inform me of where the flaws in my reasoning are for the following problem:

"Find the Galois Group of $x^3-3x+3$ over

(a)$\mathbb{Q}$

(b)$\mathbb{Q}(i\sqrt{15})$

(a) Since the discriminant of $x^3-3x+3$ is negative, I know this polynomial has one real root and two complex conjugate roots. Since this polynomial is irreducible by eisenstein's citerion the degree of each root is $3$ over $\mathbb{Q}$. With a little application of the tower rule I determined that the splitting field of the polynomial is of degree $6$ over $\mathbb{Q}$. Since this extension is the splitting field of a polynomial it is normal, and so the Galois group for this polynomial has exactly 6 $\mathbb{Q}$-automorphisms which consist of permuting the roots of $x^3-3x+3$.

(b) My question comes in when I am asked to find the Galois group of this polynomial over $\mathbb{Q}(i\sqrt{15})$, From my understanding this question is asking me to find the $\mathbb{Q}(i\sqrt{15})$-automorphisms of the splitting field of $x^3-3x+3$. Since I am now looking for $\mathbb{Q}(i\sqrt{15})$-automorphisms, I know they must fix $\sqrt{15}$ and $i$. Since every $\mathbb{Q}(i\sqrt{15})$-atomorphism must fix $i$, any automomorphism must act as the identity on the complex conjugate roots, and the real root cannot be sent to a complex conjugate root because that would imply that the inverse of that automorphism does not fix $i$. Therefore the Galois group of $x^3-3x+3$ is trivial over $\mathbb{Q}(i\sqrt{15})$.

Is this reasoning correct?

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    $\begingroup$ It’s not correct that if something fixes $i\sqrt{5}=\sqrt{-15}$ then it fixes both $i$ and$\sqrt{15}$. $\endgroup$ – Lubin Nov 16 '15 at 5:26
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No, sorry, your reasoning in the second part is not correct (and your conclusion isn’t right either). It’s not true, as I said in a comment, that if something leaves $\sqrt{-15}$ fixed, it fixes both $i$ and $\sqrt{15}$. Much more importantly, however, you expressed the notion that if an automorphism leaves $i$ fixed, it leaves all complex numbers fixed. For a good example, consider the field gotten by adjoining the twelfth roots of unity. This is acutally $\Bbb Q(i,\omega)$ where $\omega$ is a primitive cube root of unity. But you get the field as a simple extension, $\Bbb Q(\zeta)$, where $\zeta$ is a primitive twelfth root such as $e^{2\pi i/12}=\cos30^\circ+i\sin30^\circ=\sqrt3/2+i/2$. Any element of the Galois group must send $\zeta$ to another primitive twelfth root of $1$, and these are just $\zeta^{\pm1}$ and $\zeta^{\pm5}$. The automorphism that sends $\zeta$ to $\zeta^{-1}$ does indeed coincide with complex conjugation on our field, but since $i=\zeta^3$ and $\omega=\zeta^4$, the auromorphism that sends $\zeta$ to $\zeta^5$ also sends $i=\zeta^3$ to $\zeta^{15}=\zeta^3=i$ but sends $\omega=\zeta^4$ to $\zeta^{20}=\zeta^8=\omega^2$.

You should expect in this case that adjoining the square root of the discriminant, $3\sqrt{-15}$ here, will get you to a field over which the big field has cyclic Galois group of order $3$. Maybe you can prove this now.

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