how to find the Galois group?

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 \(\displaystyle (x^2 - 2)\)
 
logistic_guy said:
does this mean, the number of roots doesn't matter, the galois group is always Z/(2Z)?
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No, it doesn't. If you have only one non-trivial automorphism [imath]\sigma[/imath] then what are the values of [imath]\sigma(\sqrt{3})[/imath] and [imath]\sigma(\sqrt{5})[/imath]?
 
blamocur said:
No, it doesn't. If you have only one non-trivial automorphism [imath]\sigma[/imath] then what are the values of [imath]\sigma(\sqrt{3})[/imath] and [imath]\sigma(\sqrt{5})[/imath]?
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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?
 
logistic_guy said:
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?
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Of course: you've listed at least four of them in your last post.
 
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 \(\displaystyle \mathbb{Z}/2\mathbb{Z}\)
 
logistic_guy said:
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 \(\displaystyle \mathbb{Z}/2\mathbb{Z}\)
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There are infinitely many finite groups of order 2, not just [imath]\mathbb Z_2[/imath].
P.S. Not sure about terminology, but by 'order' I mean integer [imath]d[/imath] such that [imath]x^d = 1[/imath] for all [imath]x[/imath] in the group.
 
blamocur said:
There are infinitely many finite groups of order 2, not just [imath]\mathbb Z_2[/imath].
P.S. Not sure about terminology, but by 'order' I mean integer [imath]d[/imath] such that [imath]x^d = 1[/imath] for all [imath]x[/imath] in the group.
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do i have to say the galois group is \(\displaystyle \mathbb{Z_2}..........\)

the dots emphsizes there are many of them. is that correct?
 
logistic_guy said:
do i have to say the galois group is \(\displaystyle \mathbb{Z_2}..........\)

the dots emphsizes there are many of them. is that correct?
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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?
 
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?
 
logistic_guy said:
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?
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Very good! Can you find a minimal set of non-identity elements which generate the whole group?
 
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
 
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 said:
\(\displaystyle 5 - 3 + 2 = 4\)
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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 [imath]a_2, a_3, a_5[/imath] -- can you combine them to get [imath]a_4[/imath]?
 
by \(\displaystyle \sigma_2\), do you means \(\displaystyle \sigma_2 = \sqrt{2}\)?

by \(\displaystyle \sigma_4\), do you means \(\displaystyle \sigma_4 = \sqrt{4} = 2\)?
 
logistic_guy said:
by \(\displaystyle \sigma_2\), do you means \(\displaystyle \sigma_2 = \sqrt{2}\)?

by \(\displaystyle \sigma_4\), do you means \(\displaystyle \sigma_4 = \sqrt{4} = 2\)?
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No. Here is what I said earlier:
blamocur said:
In your list in post #29 number 1 is an identity, but numbers 3, 4 and 6 each flip a single root.
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So, for example, by [imath]a_4[/imath] I mean automorphism number 4. in your list from your post # 29.
 
if i combine \(\displaystyle \sigma_2, \sigma_3, \sigma_5\) i get \(\displaystyle \sigma_2\) not \(\displaystyle \sigma_4\)
 
logistic_guy said:
if i combine \(\displaystyle \sigma_2, \sigma_3, \sigma_5\) i get \(\displaystyle \sigma_2\) not \(\displaystyle \sigma_4\)
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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 [imath](a_3,a_4,a_6)[/imath] can be used to get any other element of the group -- can you see that?
P.S. I prefer [imath]a_k[/imath] instead of [imath]\sigma_k[/imath] to reduce typing.
 
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