how to find the Galois group?

[logistic_guy]

thank you blamocur for this clarfication. things become more clear.

i know now why $\displaystyle \sigma(\sqrt(3)) \neq \sqrt{2}$

more importantly, i know now, the automorphisms are of order two which make the galois group of integers modulo $\displaystyle 2$, $\displaystyle \mathbb{Z}/2\mathbb{Z}$.

does this mean, the number of roots doesn't matter, the galois group is always $\displaystyle \mathbb{Z}/2\mathbb{Z}$?

this is the same answer we got if you had only one root $\displaystyle (x^2 - 2)$

 

[blamocur]

does this mean, the number of roots doesn't matter, the galois group is always Z/(2Z)?

No, it doesn't. If you have only one non-trivial automorphism $\sigma$ then what are the values of $\sigma(\sqrt{3})$ and $\sigma(\sqrt{5})$?

 

[logistic_guy]

No, it doesn't. If you have only one non-trivial automorphism $\sigma$ then what are the values of $\sigma(\sqrt{3})$ and $\sigma(\sqrt{5})$?

i din't understand what non-trivial autophormisim mean?

we already said what are the values of $\displaystyle \sigma(\sqrt{3})$ and $\displaystyle \sigma(\sqrt{5})$

$\displaystyle \sigma(\sqrt{3}) = +\sqrt{3}$ or $\displaystyle \sigma(\sqrt{3}) = -\sqrt{3}$

$\displaystyle \sigma(\sqrt{5}) = +\sqrt{5}$ or $\displaystyle \sigma(\sqrt{5}) = -\sqrt{5}$

do you mean since we have three roots, we must have more than one non-trivial autophorimisim? what are they?

 

[blamocur]

i din't understand what non-trivial autophormisim mean?

we already said what are the values of $\displaystyle \sigma(\sqrt{3})$ and $\displaystyle \sigma(\sqrt{5})$

$\displaystyle \sigma(\sqrt{3}) = +\sqrt{3}$ or $\displaystyle \sigma(\sqrt{3}) = -\sqrt{3}$

$\displaystyle \sigma(\sqrt{5}) = +\sqrt{5}$ or $\displaystyle \sigma(\sqrt{5}) = -\sqrt{5}$

do you mean since we have three roots, we must have more than one non-trivial autophorimisim? what are they?

Of course: you've listed at least four of them in your last post.

 

[logistic_guy]

i'm glad i'm understanding non-trivial automorphisism, but even if i've $\displaystyle 6$ non-trivial automorphisims, the automorphisims are still of order $\displaystyle 2$ and the galois group is $\displaystyle \mathbb{Z}/2\mathbb{Z}$

 

[blamocur]

i'm glad i'm understanding non-trivial automorphisism, but even if i've $\displaystyle 6$ non-trivial automorphisims, the automorphisims are still of order $\displaystyle 2$ and the galois group is $\displaystyle \mathbb{Z}/2\mathbb{Z}$

There are infinitely many finite groups of order 2, not just $\mathbb Z_2$.
P.S. Not sure about terminology, but by 'order' I mean integer $d$ such that $x^d = 1$ for all $x$ in the group.

 

[logistic_guy]

There are infinitely many finite groups of order 2, not just $\mathbb Z_2$.
P.S. Not sure about terminology, but by 'order' I mean integer $d$ such that $x^d = 1$ for all $x$ in the group.

do i have to say the galois group is $\displaystyle \mathbb{Z_2}..........$

the dots emphsizes there are many of them. is that correct?

 

[blamocur]

do i have to say the galois group is $\displaystyle \mathbb{Z_2}..........$

the dots emphsizes there are many of them. is that correct?

I am not sure about this notation and its meaning.

Lets answer this concrete question: how many different automorphisms are in the group? What are they? I.e., is there a simple way to describe each of them?

 

[logistic_guy]

the automorphisms are

$\displaystyle 1. \ \sigma : a + b\sqrt{2} + c\sqrt{3} + d\sqrt{5}$

$\displaystyle 2. \ \sigma : a - b\sqrt{2} - c\sqrt{3} - d\sqrt{5}$

$\displaystyle 3. \ \sigma : a + b\sqrt{2} + c\sqrt{3} - d\sqrt{5}$

$\displaystyle 4. \ \sigma : a + b\sqrt{2} - c\sqrt{3} + d\sqrt{5}$

$\displaystyle 5. \ \sigma : a - b\sqrt{2} - c\sqrt{3} + d\sqrt{5}$

$\displaystyle 6. \ \sigma : a - b\sqrt{2} + c\sqrt{3} + d\sqrt{5}$

$\displaystyle 7. \ \sigma : a - b\sqrt{2} + c\sqrt{3} - d\sqrt{5}$

$\displaystyle 8. \ \sigma : a + b\sqrt{2} - c\sqrt{3} - d\sqrt{5}$

there are $\displaystyle 8$ automorphisims if i don't forget any combination

what is the simple way to describe each one of them?

 

[blamocur]

the automorphisms are

$\displaystyle 1. \ \sigma : a + b\sqrt{2} + c\sqrt{3} + d\sqrt{5}$

$\displaystyle 2. \ \sigma : a - b\sqrt{2} - c\sqrt{3} - d\sqrt{5}$

$\displaystyle 3. \ \sigma : a + b\sqrt{2} + c\sqrt{3} - d\sqrt{5}$

$\displaystyle 4. \ \sigma : a + b\sqrt{2} - c\sqrt{3} + d\sqrt{5}$

$\displaystyle 5. \ \sigma : a - b\sqrt{2} - c\sqrt{3} + d\sqrt{5}$

$\displaystyle 6. \ \sigma : a - b\sqrt{2} + c\sqrt{3} + d\sqrt{5}$

$\displaystyle 7. \ \sigma : a - b\sqrt{2} + c\sqrt{3} - d\sqrt{5}$

$\displaystyle 8. \ \sigma : a + b\sqrt{2} - c\sqrt{3} - d\sqrt{5}$

there are $\displaystyle 8$ automorphisims if i don't forget any combination

what is the simple way to describe each one of them?

Very good! Can you find a minimal set of non-identity elements which generate the whole group?

 

[logistic_guy]

i know the identity elements are $\displaystyle 0$ and $\displaystyle 1$ but i don't know how to think of a set that will generate all the roots

 

[blamocur]

I would not call 0 an identity -- it is not even an automorphism.

In you list in post #29 number 1 is an identity, but numbers 3, 4 and 6 each flip a single root. Do you agree that all other automorphism can be expressed through products of those three?

 

[logistic_guy]

$\displaystyle 3, 4, 6$ or $\displaystyle 2, 3, 5$???

 

[blamocur]

$\displaystyle 3, 4, 6$ or $\displaystyle 2, 3, 5$???

Can you express 4 as a composition of 2,3 and 5?

 

[logistic_guy]

$\displaystyle 5 - 3 + 2 = 4$

 

[blamocur]

$\displaystyle 5 - 3 + 2 = 4$

You seem to be talking about natural numbers, but I was talking about automorphisms from your list in post #29. To avoid confusion we can call them $a_2, a_3, a_5$ -- can you combine them to get $a_4$?

 

[logistic_guy]

by $\displaystyle \sigma_2$, do you means $\displaystyle \sigma_2 = \sqrt{2}$?

by $\displaystyle \sigma_4$, do you means $\displaystyle \sigma_4 = \sqrt{4} = 2$?

 

[blamocur]

by $\displaystyle \sigma_2$, do you means $\displaystyle \sigma_2 = \sqrt{2}$?

by $\displaystyle \sigma_4$, do you means $\displaystyle \sigma_4 = \sqrt{4} = 2$?

No. Here is what I said earlier:

In your list in post #29 number 1 is an identity, but numbers 3, 4 and 6 each flip a single root.

So, for example, by $a_4$ I mean automorphism number 4. in your list from your post # 29.

 

[logistic_guy]

if i combine $\displaystyle \sigma_2, \sigma_3, \sigma_5$ i get $\displaystyle \sigma_2$ not $\displaystyle \sigma_4$

 

[blamocur]

if i combine $\displaystyle \sigma_2, \sigma_3, \sigma_5$ i get $\displaystyle \sigma_2$ not $\displaystyle \sigma_4$

How did you get that? BTW, you don't need to combine all three to get one of them. My claim is that all or some elements from the set $(a_3,a_4,a_6)$ can be used to get any other element of the group -- can you see that?
P.S. I prefer $a_k$ instead of $\sigma_k$ to reduce typing.