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galois theory thread - learning basics

Basically I struggle IMMENSELY with Galois Theory, and while I have numerous notes with abstract definitions etc, I am someone who needs to see concrete examples to see what is going on. As a result I'm working through various questions and solutions, and would appreciate if anyone can help explain things as simply as possible.

Consider the polynomial

f = t^4 - 3

over Q. Find the splitting field L of f over Q (in C) and state the degree [L:Q].

f is irreducible of the rationals bu Eisenstein at p = 3. Its roots in C are

\alpha, i \alpha, - \alpha, - i \alpha

where

\alpha = 3^{1/4} > 0

and the splitting field is

L = Q(\pm \alpha, \pm i \alpha) = Q(\alpha, i)

which has degree 8 over Q.

OK firstly, I understand how to use Eisenstein's criteria, but how do you work out what the roots in C are?

Thanks in advance!

Reply 1

Zhen Lin

Solve the given equation directly:

t^4 - 3 = 0 \iff t^4 = 3

, so there are 4 roots, related by factors given by the fourth roots of unity.

Reply 2

Jake22

So as Zhen Lin has mentioned, the four roots are the real fourth root of three and then its products with the four fourth roots of unity. This generalises to calculating the complex nth roots of any real number i.e. if

\omega

is an nth root of unity and

\beta

is the real nth root of say,

b\in \mathbb{R}

then

(\omega\beta)^n=\omega^n\beta^n =b

.

So back to the example. Let

\alpha

be the real fourth root of three, then the four fourth roots of unity are

1,-1,i,-i

and so you have that the roots of

t^4-3

are

\alpha,-\alpha,i\alpha,-i\alpha

. So

L=\mathbb{Q}(\alpha,i)\geq \mathbb{Q}(\alpha)\geq \mathbb{Q}

.

Now

\alpha

has minimal polynomial

t^4-3

over

\mathbb{Q}

and so

[\mathbb{Q}(\alpha):\mathbb{Q}]=4

. Also,

i

has minimal polynomial

t^2+1

over

\mathbb{Q}

and this remains irreducible over

\mathbb{Q}(\alpha)

(as

\mathbb{Q}(\alpha)\subseteq \mathbb{R}

) and so

[L:\mathbb{Q}(\alpha)]=2

and then by the tower law

[L:\mathbb{Q}]=2\times 4=8

.

Now, if we consider automorphisms of

L

that fix

\mathbb{Q}

, we see that the only possible images for

\alpha

are

\pm \alpha, \pm i \alpha

and the only possible images for

i

are

\pm i

(We only need to work out the action on basis elements) but by the fundamental theorem of Galois Theory, the order of the Galois group of

L:\mathbb{Q}

|\Gamma(L:\mathbb{Q})|=8

so each of the 8 possible combinations is an element. So you have enough information to know everything about the Galois group (which in this case can be seen fairly easily to be the Dihedral group of order 8 i.e. the symmetry group of the square). Just calculate the images of products of the automorphisms on

\alpha,i

. To explicitly work out the Galois correspondance takes a little more work but essentially, you know the entire Galois Theory of the extension.

Reply 3

sexyzebra

WOW thank you for such a detailed explanation Jake.

Show that the polynomial

t^5 -6t + 3

is not solvable by radicals over Q. Your state clearly any necessary theorems.

Looking through the solution to this, and it uses the theorem:

If p is prime and f is irreducible over Q of degree p, having precisely 2 nonreal zeros in C, then the Galois group of f is isomorphic to S_p.

It then goes on to test 4 values of t, so that by the IVT f has at least 3 real zeros, and then by looking at f' we see that this has only real solutions, and hence there are exactly 3 real zeros.

I don't understand why this theorem is being used when it talks about 2 nonreal zeros, and in this example it has 3 real zeros? Or does 3 real zeros imply there are 2 nonreal zeros??

Reply 4

DFranklin

How many roots does a 5th degree polynomial have? So if 3 of them are real, how many are non-real?

Reply 5

sexyzebra

If an angle

\theta

is trisectable by ruler and compass then the polynomial

g = t^3 - 3t - 2 cos \theta

is reducible over the field

K = Q(cos \theta)

Show that the acute angle

cos \theta = \frac{37}{125}

can be trisected by ruler and compass.

We have

g = t^3 - 3t - \frac{74}{125}

. Multiplying by 125 and putting x = 5t we may consider instead the integral polynomial

h = x^3 - 75x - 74

etc etc...

I'm just trying a similar question for

cos \theta = \frac{13}{27}

and was wondering if there is a quick method/way of spotting what to set x equal to? I tried x = 3t here and it didn't get me anywhere...

Thanks in advance!

Reply 6

DFranklin

Post your working. At first glance (i.e. doing it in my head) it looks like it should work exactly the same way as the 37/125 case.

Reply 7

sexyzebra

Thanks David I sorted out that question in the end.

Could someone please check if this answer is OK, and worthy of 8 marks in an exam?

If an angle

\theta

is trisectable by ruler and compass then the polynomial

g = t^3 - 3t - 2 cos \theta

is reducible over the field

K = Q(cos \theta)

Show that the acute angle cos

\theta = \frac{1}{4}

cannot be trisected by ruler and compass.

We have

g = t^3 - 3t -\frac{1}{2}

and K = Q.

It is a question of whether or not g has a rational root.

Putting x = 4t we get:

h = \frac{x^3}{64} - \frac{3x}{4} - \frac{1}{2}

Multiplying by 64:

h = x^3 - 48x - 32

Any rational root would in fact be an integer x with 4 | x

Then

4^3 | 32

which is a contradiction.

Since h is irreducible, so is g, and hence

\theta

is not trisectable.

Thanks in advance!

Reply 8

DFranklin

Comments and additions in bold italic

sexyzebra

If an angle

\theta

is trisectable by ruler and compass then the polynomial

g = t^3 - 3t - 2 cos \theta

is reducible over the field

K = Q(cos \theta)

Show that the acute angle cos

\theta = \frac{1}{4}

cannot be trisected by ruler and compass.

We have

g = t^3 - 3t -\frac{1}{2}

and K = Q.

It is a question of whether or not g has a rational root.

Putting x = 4t we get: I'm not sure why you've done this

h = \frac{x^3}{64} - \frac{3x}{4} - \frac{1}{2}

Multiplying by 64:

h = x^3 - 48x - 32

It doesn't really matter, but I don't like to see someone write h = {stuff} and then say h = 64 {stuff} on the next line. Introduce another variable!

By Gauss's Lemma, Any rational root would in fact be an integer x.

with 4 | x I don't see where this comes from.

Then

4^3 | 32

which is a contradiction. If you're right that 4|x this is fine. But I don't see why it has to be true.

Since h is irreducible, so is g, and hence

\theta

is not trisectable.

Reply 9

sexyzebra

Thanks or the comments :smile:

Reply 10

DFranklin

sexyzebra

Thanks or the comments :smile:

I wasn't sure about the "putting x = 4t" bit myself...it goes back to what I said in post 6, I don't understand how you're supposed to work out what to put x equal to. I thought it might not matter and it was just a way of getting the equation into a nicer form but seems it does indeed matter...

Could you possibly please explain this bit?Looking at post 6:

You have a cubic g=t^3-3t-74/125. We want to get rid of any fractions, so we multiply by 125 to get 125g = 125t^3-375t-74. But most of our results about reducability only really work with monic polynomials. So we put x = 5t to get the coefficient of x^3 = 1.

Now in the later question you have a cubic t^3 - 3t - 1/2. Multiply by 2 to clear the fractions and you get 2t^3-6t-1. To get a monic polynomial, we really need the leading term to be a cube, so multiply again by 4 to get 8t^3-24t-4.
Now put x=2t to get x^3 - 12x - 4.

But (unless I'm missing something), there's nothing actually wrong with putting x = 4t, it does work here, but it just seems picked at random.

Where I think you have gone wrong is thinking that because x = 4t, x has to be a multiple of 4. But we are looking for rational roots, not integer ones. So it might be that, for example, t = 1/4, and then x = 1.

But it is true that if x^3-48x-32 = 0 (and x is rational) then x has to be an integer. Moreover, x has to be an integer dividing 32. There aren't many possibilities, and it's easy to show that none of them work. (Note that there are even fewer possibilities if you'd set x=2t to get x^3-12x-4).

Reply 11

hohoho11

Hi, sorry to take over your thread, but I am struggling with a similar question the ones you posted recently.

My example is:

If an angle

\theta

is trisectable by ruler and compass then the polynomial

g = t^3 - 3t - 2 cos \theta

is reducible over the field

K = Q(cos \theta)

Show that the acute angle

cos \theta = \frac{47}{128}

can be trisected by ruler and compass.

However my solution is different to yours and I don't understand it.

Solution:

We ask whether or not

f(t) = t^3 - 3t - \frac{47}{64}

is irreducible/Q.

Suppose

\frac{m}{n} \in Q

(reduced) is a zero. Then

64m^3 - 192mn^2 - 47n^3 = 0. n^2|64

, so

n|8

. If

n = \pm 8

then

8^3

divides the second and third terms of the equation, hence

8^3|64m^3

and

8|m

which is a contradiction. So

n|4

. But from the equation

64|47n^3

, so that

4|n

. Hence

n = \pm 4

. Trying n = 4 we get the solution m = -1.

Hence it is trisectable.

I can't follow this answer at all. Where is it getting all the knowledge that this divides this etc etc, is it do with Eisensteins criteria? Please help me I have been pondering over this for some time :mad:

Thanks in advance!

Reply 12

DFranklin

Which is the first "a | b" that you don't understand? Is it

n^2 | 64

Reply 13

hohoho11

Hi DFranklin, sorry for the bad latex. Yes its all of them really...I don't understand where they come from or why they are true?

Reply 14

DFranklin

64m^3 - 192mn^2 -47n^3, so (192m+47n)n^2 = 64 m^3. So n^2 | 64m^3. But m/n is in reduced form, so...

Reply 15

hohoho11

Thank you thank you, okay that makes sense, but then why if

n = \pm 8

does 8^3 have to divide the second and third terms of the equation??

Reply 16

DFranklin

If n = +/-8, then n^3 = +/-64, so 47 n^3 is a multiple of 64. Similarly for 192mn^2.

Reply 17

sexyzebra

Thanks for the helpful post David. Just one thing I'm not sure about...

DFranklin

But it is true that if x^3-48x-32 = 0 (and x is rational) then x has to be an integer. Moreover, x has to be an integer dividing 32.

How do you know this, is it from a theorem or just "common knowledge"?

Also could possibly give me another angle that it definitely not trisectable so I can make sure I'm OK with this question?

Reply 18

DFranklin

It's a (fairly) trivial corollary of Gauss's Lemma.

I don't know how to come up with angles that are provably non-trisectable. I'd just use google to search for some examples.