# Galois Theory questions

1. Dec 4, 2016

### PsychonautQQ

1. The problem statement, all variables and given/known data
Let K be the splitting field in C of the polynomial g(X) = x^3 + 4 over Q
a) determine the degree of the extension K:Q,
b) determine the structure of the galois group Gal(K:Q)
c) Using the Galois correspondence, determine all subfields of K
d) Let r be an element of K and let m(x) be the minimal polynomial of r over Q. is m solvable by radicals? explain

2. Relevant equations

3. The attempt at a solution
a) Q splits in Q(b,c) where b the 3rd real root of 4 and c is a primitive 6th root of unity. Then all 3 roots are in Q(b,c), and they take the form -b, cb and (c^5)b. The degree of the extension will be 6 because the minimal polynomial of b over Q has degree 3 and the minimal polynomial of c over Q(b) has degree 2.

b) The Galois group is $D_3=S_3$ and so there are 6 elements in it.

c) The subgroups of $S_3$ are <(12)> <(13)> <(23)> and <(123)>. Lets say 1=-b, 2=cb and 3=bc^5.

The fixed field of <(12)> will be Q(b(c^5)), whilst it swaps the other two roots -b and cb. I am a bit confused here because I think (cb - b) should be included in the fixed field because (12) would swap cb an b around and thus the term (cb - b) would stay the same. Is Q(bc^5,cb-b) the same as Q(bc^5) somehow?

Anyway, following my same logic, the fixed field of (13) will be Q(cb) or Q(cb, bc^5-b) or Q(bc^5-b). I hope the thing that is confusing me here makes sense.

Then (23) would fix cb + cb^5 and b so would the fixed field simply be Q(b,cb + bc^5)? My gut is telling me it's just Q(b) but I'm not sure why cb + bc^5 wouldn't be included, is this element in Q(b)? I mean c+c^5 would equal a real number I believe if you look at them as points on the unit circle, but I highly doubt this real number is linearly independent from the basis {1,b} over Q and thus cb+cb^5 would not be in Q(b).

And finally we have (123). I believe the fixed field of this would be Q(b+bc+bc^5).

d). I believe the answer is yes it would be a radical extension, since b^3 is an element of Q and so is c^6.

Thanks PF!

2. Dec 4, 2016

### Staff: Mentor

I assume you meant $K=\mathbb{Q}(b,c)$. Anyway. Why did you choose $c$ to be a sixth root of unity? Shouldn't it be $\sqrt{-3}$ instead and you get directly the degree $2$ extension? But maybe they work both, I simply saw no reason for it and thus didn't really follow your constructions. I would have expected $<(1,2,3)>=\{1,b,b^2\}$ and $<(1,2)>=<\sqrt{-3}>$.

An argument, why $Gal(K,\mathbb{Q}) \ncong \mathbb{Z}_6$ would be nice.

Regarding the solvability, you could simply calculate all roots and show (and prove) a basis of $K$ over $\mathbb{Q}$, or otherwise argue with the corresponding property of $S_3$.

3. Dec 5, 2016

### PsychonautQQ

Well, a 6th root of unity has a minimal polynomial of degree 2 and if we let c be the sixth root of unity then c^3 = -1 and (c^5)^3 = -1, and these are the only two powers of c that when raised to the third are equal to -1,. Furthermore, I believe that Q(b,c) is a splitting field of x^3+4, because the roots are -b, -bc, and -bc^5 and all of these roots are in Q(b,c), and also Q(b,c) is contained in Q(b,bc,bc^5) as (bc)^2 * (bc^5) = b^3 * c^7 = 3*c so c is an element of Q(b,bc,bc^5). I'm pretty sure what I wrote works, and I'm only saying all this so that you might find a flaw in my logic, because believe it or not I've been wrong before.

$Gal(K,Q) does not equal Z_6$ because the minimal polynomial of b is x^3-4 and the minimal polynomial of c is x^2-x+1, where we will have one automorphism in the Galois group that takes b to the other roots of it's minimal polynomial. So there will exist an autmorphism such that b --> bc and an automorphism that takes c-->c^2, and these automorphisms are not commutative so the Galois group will not be $Z_6$

______________________________

I'm not understanding your idea that b=the real cube root of 3 and c = (-3)^1/2. I understand that c and -c will both be roots of x^3+4 but how will we get the third root?

4. Dec 5, 2016

### Staff: Mentor

I simply calculated $x^3+4=(x-b)(x- \frac{b}{2}(-1+\sqrt{-3}))(x- \frac{b}{2}(-1-\sqrt{-3}))$ and $\{1,b,b^2,\sqrt{-3},b\sqrt{-3},b^2\sqrt{-3}\}$ is a nice basis (to be proven) for $K \supseteq \mathbb{Q}$ which almost already displays the group structure of $S_3 = D_3 = \mathbb{Z}_3 \rtimes \mathbb{Z}_2$.

What has puzzled me with the sixth root of unity $c$ is, that $c \notin K$, at least I couldn't immediately see it and you provided no proof. This way you get an isomorphic field $K'$ of $K$, which doesn't really affect the solution, but I think $K' \cong K$ should be explained.
This has been the reason for me to chose $\sqrt{-3}$ instead.

5. Dec 5, 2016

### PsychonautQQ

Ahh cool so my solution does work also? If so, awesome. What will the fixed fields look like?

6. Dec 5, 2016

### Staff: Mentor

If you either prove that $K$ contains the sixth roots of unity (which I doubt is true), or establish an isomorphism $K \rightarrow \mathbb{Q}(b,c)$. Simply assuming one of them is true, is a bit weak in my opinion.
It'll look like $\mathbb{Q}$. What do you mean? Did you mean the intermediate fields? In my solution they will be $\mathbb{Q}(b)$ and $\mathbb{Q}(\sqrt{-3})$. I have difficulties to clearly understand your solution, e.g. the connection between permutations to automorphisms should explicitly be defined in terms of basis elements. Also $c^5$ appears on the scene without any motivation. I know why, but shouldn't you explain it? And why are $\{-b,cb,c^5b\}$ all the roots? I think you leave a bit too much to the reader. (Or maybe I'm not that smart.)

7. Dec 6, 2016

### PsychonautQQ

well the polynomial has 3 roots and {-b,bc,bc^5} are all roots, so therefore this set must contain ALL the roots. (-b)^3 = -4, (bc)^3 = 4*-1 = -4, (bc^5)^3 = 4*c^15=4*(c^12)*c^3=4*1*-1=-4. Therefore Q(b,c) contains all the roots. However, to be a splitting field you have to be the smallest field to contain all the roots, which my field is not, because your field is contained in mine but mine is not contained in yours.

8. Dec 6, 2016

### Staff: Mentor

Nope. $K = \mathbb{Q}(b,c)$ - no isomorphisms or alike needed.
$c=\exp(2\pi i \cdot \frac{1}{6}) = \cos (\frac{\pi}{3}) + i\cdot \sin (\frac{\pi}{3}) = \frac{1}{2}(1+i\cdot \sqrt{3})$ and similar for $c^5$ which shows $\mathbb{Q}(b,c)=\mathbb{Q}(b,\sqrt{-3})=K$. (I only needed this line to see we're talking about the same numbers.)

9. Dec 7, 2016

### PsychonautQQ

So what you're saying is we're both genius's? Lol.

But actually, can you help me find the intermediate subfields using Galois subgroups and the Galois correspondence?

The Galois group is $S_3$ the subgroups are generated by (12), (13), (23), and (123). Let's consider 1 to be the real root, 2 to be bc and 3 to be bc^5.

That means that (23) is complex conjugation, so the fixed field of (23) is Q(b), because b is the only real root and the other 2 roots are flipped by complex conjugation. That being said, the element bc+bc^5 will also be fixed, because they will both go to each other, so is the fixed field of (23) actually Q(b,bc+bc^5)?

I am a bit confused here, because I feel like the fixed field should be just Q(b). I guess (bc+bc^5)=b(c+c^5) where c+c^5 is a real number because their imaginary parts cancel out (imagine where they lie on the unit circle to see this). But still, I doubt that c+c^5 is Q-linearly dependent from b.

. I believe the fixed field of (12) will be Q(b+bc), the fixed field of (13) will be Q(b+bc^2), and the fixed field of (123) will be Q(b+bc+bc^5).

But then again won't (12) also fix Q(bc^5)? So would the fixed field be Q(b+bc,bc^5)?

Do you see why I am confused? Thanks in advance.

10. Dec 7, 2016

### Staff: Mentor

It helps to work with explicit numbers in parallel.

Let us actually split $x^3+4=(x+b)(x-bc)(x-bc^5)$ with $b=\sqrt[3]{4}\in \mathbb{R}\, , \, c=\frac{1}{2}(1+i\sqrt{3})\, , \,c^5=\frac{1}{2}(1-i\sqrt{3})$. Thus the identification with permutations should be $1 \triangleq -b\, , \,2 \triangleq bc\, , \, 3 \triangleq bc^5$. (The minus sign has to be somewhere. I found it easier here, than to bother with it in any multiplications.)

Let us now consider $bc+bc^5$. This is $bc+bc^5=b(c+c^5)=b\frac{1}{2}(1+i\sqrt{3}+1-i\sqrt{3})=b \frac{1}{2}\cdot 2=b$, so $\mathbb{Q}(b,bc+bc^5)=\mathbb{Q}(b)$. I haven't calculated the whole multiplication table in $\mathbb{Q}(b,c)$ which would certainly help, since, e.g. $(1,3,2)$ is actually an automorphism $\sigma : \mathbb{Q}(b,c) \longrightarrow \mathbb{Q}(b,c)$ with $\sigma(-b)=bc^5\, , \,\sigma(bc^5)=bc\, , \,\sigma(bc)=-b$.

11. Dec 7, 2016

### PsychonautQQ

Yes, all these automorphisms are permutations. Would the fixed field of (1,3,2) be Q(-b+bc+bc^5)?

12. Dec 7, 2016

### Staff: Mentor

$-b+bc+bc^5=b(-1+c+c^5)=b(-1+1)=0$ and therefore $\mathbb{Q}(-b+bc+bc^5)=\mathbb{Q}$.

On the other hand $(1,3,2)$ belongs to the automorphism $\varphi : \mathbb{Q}(b,c) \longrightarrow \mathbb{Q}(b,c)$ with
$\varphi(-b)=bc^5\, , \,\varphi(bc^5)=bc\, , \,\varphi(bc)=-b$. Let us calculate $\varphi(c)$.
$$\varphi(c)=\varphi(b^{-1}bc)=(\varphi(b))^{-1}\varphi(bc)=-(bc^5)^{-1}(-b)=c^{-5}=1\cdot c^{-5}=c^6 \cdot c^{-5}=c$$
Here $\varphi$ is an automorphism of order three, and $(1,3,2)^2=(1,2,3)$ the other $3-$cycle. I would expect one subfield of degree $3$ and three isomorphic subfileds of degree $2$.

Last edited: Dec 7, 2016
13. Dec 8, 2016

### PsychonautQQ

Do you know how to calculate these subfields explicitly? Or even what the final result would be? What elements do we have to adjoin to Q to get the intermediate field that corresponds to the fixed field of <(123)>=<(132)>?

14. Dec 8, 2016

### Staff: Mentor

The only explicit method which comes to my mind is the brute force way to do it:
Chose a basis to express an invariant element, apply the automorphisms and solve the equation.
However, this doesn't seem to be very pleasant. I would take the subfields I suspect to exist instead, and verify their invariance.

As to $(1,2,3)$. Here we know that $\varphi(c)=c$ by the calculation above (post #12), so we are left with $\mathbb{Q}(b,c) \supseteq \mathbb{Q}(c)$ which is of degree $3$ and $Gal(\mathbb{Q}(b,c),\mathbb{Q}(c)) = <\varphi> \cong \mathbb{Z}_3 = \{1,(1,2,3),(1,3,2)\}$ as expected.