how to find the Galois group?

[logistic_guy]

this is the question

Determine the Galois group of \(\displaystyle (x^2 - 2)(x^2 - 3)(x^2 - 5)\). Determine all the subfields of the splitting field of this polynomial.

 

[blamocur]

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[logistic_guy]

i'm very bad in Galois theory. if it ask about only \(\displaystyle (x^2 - 2)\), i know partially to solve

the roots are \(\displaystyle \sqrt{2}\), \(\displaystyle -\sqrt{2}\), so the splitting field is \(\displaystyle \mathbb{Q(\sqrt{2})}\) and i need to find Gal\(\displaystyle (\mathbb{Q(\sqrt{2})}\text{/} \mathbb{Q})\)

since the roots are \(\displaystyle \sqrt{2}\), \(\displaystyle -\sqrt{2}\), i've 2 automorphisms

for any automorphism \(\displaystyle \sigma \), i've \(\displaystyle \sigma^2\sqrt{2} = \sqrt{2}\)

is it correct to conclude from that the automorphisms are of order \(\displaystyle 2\) and the Galois group is \(\displaystyle \mathbb{Z}/2\mathbb{Z}\)?

 

[blamocur]

i'm very bad in Galois theory. if it ask about only \(\displaystyle (x^2 - 2)\), i know partially to solve

the roots are \(\displaystyle \sqrt{2}\), \(\displaystyle -\sqrt{2}\), so the splitting field is \(\displaystyle \mathbb{Q(\sqrt{2})}\) and i need to find Gal\(\displaystyle (\mathbb{Q(\sqrt{2})}\text{/} \mathbb{Q})\)

since the roots are \(\displaystyle \sqrt{2}\), \(\displaystyle -\sqrt{2}\), i've 2 automorphisms

for any automorphism \(\displaystyle \sigma \), i've \(\displaystyle \sigma^2\sqrt{2} = \sqrt{2}\)

is it correct to conclude from that the automorphisms are of order \(\displaystyle 2\) and the Galois group is \(\displaystyle \mathbb{Z}/2\mathbb{Z}\)?

I agree with your analysis for [imath]x^2-2[/imath]. What is the splitting field for [imath]p(x) = (x^2-2)(x^2-3)(x^3-5)[/imath] ? I.e., what are its elements?

 

[logistic_guy]

thank you blamocur. i didn't expect to get it correct

the field of \(\displaystyle p(x)\) is \(\displaystyle \mathbb{Q(\sqrt{2},\sqrt{3},\sqrt{5})}\)

the elements are the roots and combination of the roots

\(\displaystyle \sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{6}, \sqrt{10}, \sqrt{15}\)

can i say the field is \(\displaystyle \mathbb{Q(\sqrt{2})}, \mathbb{Q(\sqrt{3})}, \mathbb{Q(\sqrt{5})}\) and find each one sepeartly as i do in post 3?

 

[blamocur]

thank you blamocur. i didn't expect to get it correct

the field of \(\displaystyle p(x)\) is \(\displaystyle \mathbb{Q(\sqrt{2},\sqrt{3},\sqrt{5})}\)

the elements are the roots and combination of the roots

\(\displaystyle \sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{6}, \sqrt{10}, \sqrt{15}\)

can i say the field is \(\displaystyle \mathbb{Q(\sqrt{2})}, \mathbb{Q(\sqrt{3})}, \mathbb{Q(\sqrt{5})}\) and find each one sepeartly as i do in post 3?

Not sure what the last statement would mean, but your second line looks good to me.

What can you say about automorphisms of that field?

 

[logistic_guy]

since we've \(\displaystyle 6\) roots, there are \(\displaystyle 6\) automorphisms or more. how to find the exact number of automorphisims?

 

[blamocur]

how to find the exact number of automorphisims?

You extension field is generated by a small number of elements, which means it is enough to define your automorphisms only on those elements. And it helps to remember that the automorphisms aren't allowed to change anything in [imath]\mathbb Q[/imath].

 

[logistic_guy]

the extension field and the subfield is the same field?

 

[blamocur]

the extension field and the subfield is the same field?

What do you mean? I am guessing that subfield is [imath]\mathbb Q[/imath] and the extension field is [imath]\mathbb Q(\sqrt{2},\sqrt{3},\sqrt{5}[/imath] -- how can they be the same?

 

[logistic_guy]

i'm sorry. still not familiar with all words

how to determine the order of automorphisims? is it correct to say that each root has two signs, then the automorphisimis are of order two?

two signs, i mean like this \(\displaystyle +\sqrt{2}\), \(\displaystyle -\sqrt{2}\)

 

[blamocur]

i'm sorry. still not familiar with all words

I am not good on the terms myself, but which ones do you have troubles with?

is it correct to say that each root has two signs, then the automorphisimis are of order two?

It happens to be correct, but can you show/prove this? Here is a question to help you out: for an arbitrary automorphism [imath]\sigma[/imath] what are possible values of [imath]\sigma (\sqrt{3})[/imath]? And why?

 

[logistic_guy]

I am not good on the terms myself, but which ones do you have troubles with?

i mixed between extension field and subfield, but so far so good

for an arbitrary automorphism [imath]\sigma[/imath] what are possible values of [imath]\sigma (\sqrt{3})[/imath]? And why?

i'm not sure how to answer this but possible ways to map \(\displaystyle \sigma\) is

\(\displaystyle \sigma : a + b\sqrt{2} + c\sqrt{3} + d\sqrt{5} \mapsto a + b\sqrt{2} + c\sqrt{3} + d\sqrt{5}\)

another way

\(\displaystyle \sigma : a + b\sqrt{2} + c\sqrt{3} + d\sqrt{5} \mapsto a + b\sqrt{2} - c\sqrt{3} + d\sqrt{5}\)

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

And why?

i think because there is two way to map the automorphism of \(\displaystyle \sqrt{3}\)

 

[blamocur]

Can you show why, for example, one cannot have [imath]\sigma(\sqrt{3}) = \sqrt(2)[/imath], or, for that matter why can't [imath]\sigma(\sqrt{3})[/imath] be anything but [imath]\pm \sqrt{3}[/imath] ?

 

[logistic_guy]

i'll analyze the reason in this way. tell me if there something wrong

if \(\displaystyle \sigma(\sqrt{3}) = \sqrt{2}\), it mean

\(\displaystyle \sqrt{3} = a + b\sqrt{2}\), where \(\displaystyle a\) and \(\displaystyle b\) rational numbers

squaring

\(\displaystyle 3 = a^2 + 2ab\sqrt{2} + 2b^2\)

i delete the second term because \(\displaystyle \sqrt{2}\) is irrational. this give \(\displaystyle a = 0\) or \(\displaystyle b = 0\)

if \(\displaystyle a = 0\), \(\displaystyle \frac{3}{2} = b^2\), this is wrong because \(\displaystyle \frac{3}{2}\) is not squared rational number in \(\displaystyle \mathbb{Q}\)

if \(\displaystyle b = 0\), \(\displaystyle 3 = a^2\), this is also wrong because \(\displaystyle 3\) is not squared rational number in \(\displaystyle \mathbb{Q}\)

 

[blamocur]

How do you deduce [imath]\sqrt{3} = a + b\sqrt{2}[/imath] from [imath]\sigma(\sqrt{3}) = \sqrt{2}[/imath]?

A simpler approach: if [imath]\sigma(\sqrt{3}) = \sqrt{2}[/imath] then what can we say about [imath]\sigma\left(\left(\sqrt{3}\right)^2\right)[/imath] ?

 

[logistic_guy]

How do you deduce [imath]\sqrt{3} = a + b\sqrt{2}[/imath] from [imath]\sigma(\sqrt{3}) = \sqrt{2}[/imath]?

i thought that the only way to get a relationship between automorphism and the root. was it wrong?

A simpler approach: if [imath]\sigma(\sqrt{3}) = \sqrt{2}[/imath] then what can we say about [imath]\sigma\left(\left(\sqrt{3}\right)^2\right)[/imath] ?

not sure but i think, \(\displaystyle \sigma\left(\left(\sqrt{3}\right)^2\right) = (\sqrt{2})^2 = 2\)

maybe the reason this is wrong because automorphism can't map to a lower value, or to be more precise automorphism map to its self or its inverse.

 

[blamocur]

i thought that the only way to get a relationship between automorphism and the root. was it wrong?


not sure but i think, \(\displaystyle \sigma\left(\left(\sqrt{3}\right)^2\right) = (\sqrt{2})^2 = 2\)

maybe the reason this is wrong because automorphism can't map to a lower value, or to be more precise automorphism map to its self or its inverse.

Why not sure? If [imath]\sigma[/imath] is automorphism of a field then [imath]\sigma(x\cdot y) = \sigma(x)\cdot\sigma(y)[/imath]. And in you case you also have [imath]\forall x\in \mathbb Q: \sigma (x) = x[/imath].

 

[logistic_guy]

nice definition, but do you mean

\(\displaystyle \forall x\in \mathbb Q: \sigma^2 (x) = x\).

because \(\displaystyle \sigma (x)\) could also equal \(\displaystyle -x\)

 

[blamocur]

nice definition, but do you mean

\(\displaystyle \forall x\in \mathbb Q: \sigma^2 (x) = x\).

because \(\displaystyle \sigma (x)\) could also equal \(\displaystyle -x\)

Not for [imath]x\in \mathbb Q[/imath]. Galois groups only consider automorphisms which fix the subfield, which is [imath]\mathbb Q[/imath] in your case.