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All $mathbb{Q}$-automorphism of $mathbb{Q}(sqrt{3},sqrt{5})$
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I have a question related to count the number of all $mathbb{Q}$-automorphisms of $mathbb{Q}(sqrt{3},sqrt{5})$. Since $mathbb{Q}(sqrt{3},sqrt{5})$ is a splitting field of $(x^2-3)(x^2-5)$ over $mathbb{Q}$. There is a theorem guarantee that (such theorem require $mathbb{Q}(sqrt{3},sqrt{5})$ to be a splitting field of some polynomial):
- There is an automorphism maps $sqrt{3}$ to $sqrt{3}$ and fix $Bbb Q$.
- There is an automorphism maps $sqrt{3}$ to $-sqrt{3}$ and fix $Bbb Q$.
- There is an automorphism maps $sqrt{5}$ to $sqrt{5}$ and fix $Bbb Q$.
- There is an automorphism maps $sqrt{5}$ to $-sqrt{5}$ and fix $Bbb Q$.
So there are $2times 2=4$ possible automorphisms, $|textbf{Aut}_{Bbb Q}mathbb{Q}(sqrt{3},sqrt{5})|=4$. But my question is, is there indeed an automorphism maps $sqrt{3}$ to $sqrt{3}$ and meanwhile maps $sqrt{5}$ to $-sqrt{5}$? I know the answer is yes, but is it definitely true? Or is it so just by trivial reason? Or should we need a theorem to state this result? PS: I found that Hungerford explicit construct such four automorphisms, to check these are all indeed exist.
abstract-algebra extension-field
asked yesterday
Eric
1,737515
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up vote
1
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I have a question related to count the number of all $mathbb{Q}$-automorphisms of $mathbb{Q}(sqrt{3},sqrt{5})$. Since $mathbb{Q}(sqrt{3},sqrt{5})$ is a splitting field of $(x^2-3)(x^2-5)$ over $mathbb{Q}$. There is a theorem guarantee that (such theorem require $mathbb{Q}(sqrt{3},sqrt{5})$ to be a splitting field of some polynomial):
- There is an automorphism maps $sqrt{3}$ to $sqrt{3}$ and fix $Bbb Q$.
- There is an automorphism maps $sqrt{3}$ to $-sqrt{3}$ and fix $Bbb Q$.
- There is an automorphism maps $sqrt{5}$ to $sqrt{5}$ and fix $Bbb Q$.
- There is an automorphism maps $sqrt{5}$ to $-sqrt{5}$ and fix $Bbb Q$.
So there are $2times 2=4$ possible automorphisms, $|textbf{Aut}_{Bbb Q}mathbb{Q}(sqrt{3},sqrt{5})|=4$. But my question is, is there indeed an automorphism maps $sqrt{3}$ to $sqrt{3}$ and meanwhile maps $sqrt{5}$ to $-sqrt{5}$? I know the answer is yes, but is it definitely true? Or is it so just by trivial reason? Or should we need a theorem to state this result? PS: I found that Hungerford explicit construct such four automorphisms, to check these are all indeed exist.
abstract-algebra extension-field
asked yesterday
Eric
1,737515
Using that the elements of $mathbb{Q}(sqrt{3},sqrt{5})$ are of the form $$a_0+a_1sqrt{3}+a_2sqrt{5}+a_3sqrt{3}sqrt{5},$$ you can easily construct such automorphism.
– rldias
yesterday
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up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have a question related to count the number of all $mathbb{Q}$-automorphisms of $mathbb{Q}(sqrt{3},sqrt{5})$. Since $mathbb{Q}(sqrt{3},sqrt{5})$ is a splitting field of $(x^2-3)(x^2-5)$ over $mathbb{Q}$. There is a theorem guarantee that (such theorem require $mathbb{Q}(sqrt{3},sqrt{5})$ to be a splitting field of some polynomial):
- There is an automorphism maps $sqrt{3}$ to $sqrt{3}$ and fix $Bbb Q$.
- There is an automorphism maps $sqrt{3}$ to $-sqrt{3}$ and fix $Bbb Q$.
- There is an automorphism maps $sqrt{5}$ to $sqrt{5}$ and fix $Bbb Q$.
- There is an automorphism maps $sqrt{5}$ to $-sqrt{5}$ and fix $Bbb Q$.
So there are $2times 2=4$ possible automorphisms, $|textbf{Aut}_{Bbb Q}mathbb{Q}(sqrt{3},sqrt{5})|=4$. But my question is, is there indeed an automorphism maps $sqrt{3}$ to $sqrt{3}$ and meanwhile maps $sqrt{5}$ to $-sqrt{5}$? I know the answer is yes, but is it definitely true? Or is it so just by trivial reason? Or should we need a theorem to state this result? PS: I found that Hungerford explicit construct such four automorphisms, to check these are all indeed exist.
abstract-algebra extension-field
asked yesterday
Eric
1,737515
I have a question related to count the number of all $mathbb{Q}$-automorphisms of $mathbb{Q}(sqrt{3},sqrt{5})$. Since $mathbb{Q}(sqrt{3},sqrt{5})$ is a splitting field of $(x^2-3)(x^2-5)$ over $mathbb{Q}$. There is a theorem guarantee that (such theorem require $mathbb{Q}(sqrt{3},sqrt{5})$ to be a splitting field of some polynomial):
- There is an automorphism maps $sqrt{3}$ to $sqrt{3}$ and fix $Bbb Q$.
- There is an automorphism maps $sqrt{3}$ to $-sqrt{3}$ and fix $Bbb Q$.
- There is an automorphism maps $sqrt{5}$ to $sqrt{5}$ and fix $Bbb Q$.
- There is an automorphism maps $sqrt{5}$ to $-sqrt{5}$ and fix $Bbb Q$.
So there are $2times 2=4$ possible automorphisms, $|textbf{Aut}_{Bbb Q}mathbb{Q}(sqrt{3},sqrt{5})|=4$. But my question is, is there indeed an automorphism maps $sqrt{3}$ to $sqrt{3}$ and meanwhile maps $sqrt{5}$ to $-sqrt{5}$? I know the answer is yes, but is it definitely true? Or is it so just by trivial reason? Or should we need a theorem to state this result? PS: I found that Hungerford explicit construct such four automorphisms, to check these are all indeed exist.
abstract-algebra extension-field
abstract-algebra extension-field
asked yesterday
Eric
1,737515
asked yesterday
Eric
1,737515
asked yesterday
Eric
1,737515
asked yesterday
Eric
1,737515
asked yesterday
Eric
1,737515
1,737515
Using that the elements of $mathbb{Q}(sqrt{3},sqrt{5})$ are of the form $$a_0+a_1sqrt{3}+a_2sqrt{5}+a_3sqrt{3}sqrt{5},$$ you can easily construct such automorphism.
– rldias
yesterday
add a comment |
Using that the elements of $mathbb{Q}(sqrt{3},sqrt{5})$ are of the form $$a_0+a_1sqrt{3}+a_2sqrt{5}+a_3sqrt{3}sqrt{5},$$ you can easily construct such automorphism.
– rldias
yesterday
Using that the elements of $mathbb{Q}(sqrt{3},sqrt{5})$ are of the form $$a_0+a_1sqrt{3}+a_2sqrt{5}+a_3sqrt{3}sqrt{5},$$ you can easily construct such automorphism.
– rldias
yesterday
Using that the elements of $mathbb{Q}(sqrt{3},sqrt{5})$ are of the form $$a_0+a_1sqrt{3}+a_2sqrt{5}+a_3sqrt{3}sqrt{5},$$ you can easily construct such automorphism.
– rldias
yesterday
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3 Answers
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Your descriptions of the four automorphisms ar lacking. For 1. and 2. you describe what happens to $sqrt 3$, but not what happens to $sqrt5$, and vice versa for 3. and 4.
Given an arbitrary element $a+bsqrt3+csqrt5+dsqrt{15}$ in your field, where $a, b, c, d$ are rational, the four automorphisms are:
$$
a+bsqrt3+csqrt5+dsqrt{15}mapsto a+bsqrt3+csqrt5+dsqrt{15}\
a+bsqrt3+csqrt5+dsqrt{15}mapsto a-bsqrt3+csqrt5-dsqrt{15}\
a+bsqrt3+csqrt5+dsqrt{15}mapsto a+bsqrt3-csqrt5-dsqrt{15}\
a+bsqrt3+csqrt5+dsqrt{15}mapsto a-bsqrt3-csqrt5+dsqrt{15}
$$
Or, in words:
- One automorphism that maps $sqrt3$ to $sqrt3$ and $sqrt5$ to $sqrt5$
- One automorphism that maps $sqrt3$ to $-sqrt3$ and $sqrt 5$ to $sqrt5$
- One automorphism that maps $sqrt3$ to $sqrt3$ and $sqrt 5$ to $-sqrt5$
- One automorphism that maps $sqrt3$ to $-sqrt3$ and $sqrt5$ to $-sqrt5$
with the implicit understanding that the image of all other elements are defined by the properties of homomorphisms (as well as by fixing $Bbb Q$, although that's superfluous in this specific case).
You should, of course, check that these are indeed automorphisms by showing that they are bijective homomorphisms.
answered yesterday
Arthur
108k7103186
So must I check that, say $phi:a+bsqrt3+csqrt5+dsqrt{15}mapsto a-bsqrt3+csqrt5-dsqrt{15}$ is indeed an automorphism? (In other words, it is not automatically so by any theorem). But for the case $Bbb Q(sqrt{2})$ over $Bbb Q$. Since there is a theorem guarantee so, I don't need to check there is an automorphism maps $sqrt{2}$ to $-sqrt{2}$ this time.
– Eric
yesterday
The theorem I mentioned is this
– Eric
yesterday
1
@Eric Depends on what you mean by "automatically". Yes, Galois theory gives you that there is a unique automorphism which fixes $Bbb Q$, sends $sqrt3$ to $-sqrt3$ and $sqrt 5$ to $sqrt5$. On the other hand, your $phi$ is a function which sends $sqrt 3$ to $-sqrt3$ and $sqrt5$ to $sqrt5$. Connecting the two is not completely automatic, and should in principle be done. One way of doing that is showing that your $phi$ is an automorphism. The same is actually true for $Bbb Q(sqrt2)$ and the (quite obvious) non-trivial automorphism there.
– Arthur
yesterday
I see! Yes, I think you completely get what I ask. Actually I haven't studied Galois theorem, I encountered this exercise in the section in the Hungerford's textbook before the Galois theorem section. My teacher told me that this it "redundant" to check such $phi$ must exists. (She means it is trivial, and I think she means it does not need Galois theory either.) But I can't find a trivial reason that such automorphism which involves determining the way two "basis" map is definitely, trivially exist.
– Eric
yesterday
add a comment |
up vote
1
down vote
You don't really have to check all the computations suggested by @Arthur 's answer.
The degree of the extension $mathbb{Q}(sqrt{3},sqrt{5})$ is 4 (because $1,sqrt{3},sqrt{5}$ and $sqrt{15}$ are linearly independent, for example). Also, the field being a splitting field of a separable polynomial(over Q any poly with simple roots is separable) implies that the extension is Galois. So the group of automorphisms will be either $mathbb{Z}_2 times mathbb{Z}_2$ or $mathbb{Z}_4$. But since $sqrt{3}$ $mapsto -sqrt{3}$ (and leave Q fixed) and $sqrt{5} mapsto -sqrt{5}$(and leave Q fixed again) are distinct automorphisms of $mathbb{Q}(sqrt{3})$ and $mathbb{Q}(sqrt{5})$ respectively, that can be extended to automorphisms of $mathbb{Q}(sqrt{3},sqrt{5})$ (any automorphism of a smaller field may be extended to an automorphism of the larger field) and generate subgroups of order two in our Galois group G, it follows that the group cannot be $mathbb{Z}_4$ (since $mathbb{Z}_4$ has only one subgroup of order 2).
answered yesterday
Sorin Tirc
5039
Thanks for the comment @Arthur you are right, I have updated my answer accordingly.
– Sorin Tirc
yesterday
add a comment |
up vote
0
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Yes. You have an automorphism for every election of signs. But you are no counting your morphisms well. For example, the morphisms 1. and 3. could be the identity.
answered yesterday
Dante Grevino
1463
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Your descriptions of the four automorphisms ar lacking. For 1. and 2. you describe what happens to $sqrt 3$, but not what happens to $sqrt5$, and vice versa for 3. and 4.
Given an arbitrary element $a+bsqrt3+csqrt5+dsqrt{15}$ in your field, where $a, b, c, d$ are rational, the four automorphisms are:
$$
a+bsqrt3+csqrt5+dsqrt{15}mapsto a+bsqrt3+csqrt5+dsqrt{15}\
a+bsqrt3+csqrt5+dsqrt{15}mapsto a-bsqrt3+csqrt5-dsqrt{15}\
a+bsqrt3+csqrt5+dsqrt{15}mapsto a+bsqrt3-csqrt5-dsqrt{15}\
a+bsqrt3+csqrt5+dsqrt{15}mapsto a-bsqrt3-csqrt5+dsqrt{15}
$$
Or, in words:
- One automorphism that maps $sqrt3$ to $sqrt3$ and $sqrt5$ to $sqrt5$
- One automorphism that maps $sqrt3$ to $-sqrt3$ and $sqrt 5$ to $sqrt5$
- One automorphism that maps $sqrt3$ to $sqrt3$ and $sqrt 5$ to $-sqrt5$
- One automorphism that maps $sqrt3$ to $-sqrt3$ and $sqrt5$ to $-sqrt5$
with the implicit understanding that the image of all other elements are defined by the properties of homomorphisms (as well as by fixing $Bbb Q$, although that's superfluous in this specific case).
You should, of course, check that these are indeed automorphisms by showing that they are bijective homomorphisms.
answered yesterday
Arthur
108k7103186
So must I check that, say $phi:a+bsqrt3+csqrt5+dsqrt{15}mapsto a-bsqrt3+csqrt5-dsqrt{15}$ is indeed an automorphism? (In other words, it is not automatically so by any theorem). But for the case $Bbb Q(sqrt{2})$ over $Bbb Q$. Since there is a theorem guarantee so, I don't need to check there is an automorphism maps $sqrt{2}$ to $-sqrt{2}$ this time.
– Eric
yesterday
The theorem I mentioned is this
– Eric
yesterday
1
@Eric Depends on what you mean by "automatically". Yes, Galois theory gives you that there is a unique automorphism which fixes $Bbb Q$, sends $sqrt3$ to $-sqrt3$ and $sqrt 5$ to $sqrt5$. On the other hand, your $phi$ is a function which sends $sqrt 3$ to $-sqrt3$ and $sqrt5$ to $sqrt5$. Connecting the two is not completely automatic, and should in principle be done. One way of doing that is showing that your $phi$ is an automorphism. The same is actually true for $Bbb Q(sqrt2)$ and the (quite obvious) non-trivial automorphism there.
– Arthur
yesterday
I see! Yes, I think you completely get what I ask. Actually I haven't studied Galois theorem, I encountered this exercise in the section in the Hungerford's textbook before the Galois theorem section. My teacher told me that this it "redundant" to check such $phi$ must exists. (She means it is trivial, and I think she means it does not need Galois theory either.) But I can't find a trivial reason that such automorphism which involves determining the way two "basis" map is definitely, trivially exist.
– Eric
yesterday
add a comment |
up vote
5
down vote
accepted
Your descriptions of the four automorphisms ar lacking. For 1. and 2. you describe what happens to $sqrt 3$, but not what happens to $sqrt5$, and vice versa for 3. and 4.
Given an arbitrary element $a+bsqrt3+csqrt5+dsqrt{15}$ in your field, where $a, b, c, d$ are rational, the four automorphisms are:
$$
a+bsqrt3+csqrt5+dsqrt{15}mapsto a+bsqrt3+csqrt5+dsqrt{15}\
a+bsqrt3+csqrt5+dsqrt{15}mapsto a-bsqrt3+csqrt5-dsqrt{15}\
a+bsqrt3+csqrt5+dsqrt{15}mapsto a+bsqrt3-csqrt5-dsqrt{15}\
a+bsqrt3+csqrt5+dsqrt{15}mapsto a-bsqrt3-csqrt5+dsqrt{15}
$$
Or, in words:
- One automorphism that maps $sqrt3$ to $sqrt3$ and $sqrt5$ to $sqrt5$
- One automorphism that maps $sqrt3$ to $-sqrt3$ and $sqrt 5$ to $sqrt5$
- One automorphism that maps $sqrt3$ to $sqrt3$ and $sqrt 5$ to $-sqrt5$
- One automorphism that maps $sqrt3$ to $-sqrt3$ and $sqrt5$ to $-sqrt5$
with the implicit understanding that the image of all other elements are defined by the properties of homomorphisms (as well as by fixing $Bbb Q$, although that's superfluous in this specific case).
You should, of course, check that these are indeed automorphisms by showing that they are bijective homomorphisms.
answered yesterday
Arthur
108k7103186
So must I check that, say $phi:a+bsqrt3+csqrt5+dsqrt{15}mapsto a-bsqrt3+csqrt5-dsqrt{15}$ is indeed an automorphism? (In other words, it is not automatically so by any theorem). But for the case $Bbb Q(sqrt{2})$ over $Bbb Q$. Since there is a theorem guarantee so, I don't need to check there is an automorphism maps $sqrt{2}$ to $-sqrt{2}$ this time.
– Eric
yesterday
The theorem I mentioned is this
– Eric
yesterday
1
@Eric Depends on what you mean by "automatically". Yes, Galois theory gives you that there is a unique automorphism which fixes $Bbb Q$, sends $sqrt3$ to $-sqrt3$ and $sqrt 5$ to $sqrt5$. On the other hand, your $phi$ is a function which sends $sqrt 3$ to $-sqrt3$ and $sqrt5$ to $sqrt5$. Connecting the two is not completely automatic, and should in principle be done. One way of doing that is showing that your $phi$ is an automorphism. The same is actually true for $Bbb Q(sqrt2)$ and the (quite obvious) non-trivial automorphism there.
– Arthur
yesterday
I see! Yes, I think you completely get what I ask. Actually I haven't studied Galois theorem, I encountered this exercise in the section in the Hungerford's textbook before the Galois theorem section. My teacher told me that this it "redundant" to check such $phi$ must exists. (She means it is trivial, and I think she means it does not need Galois theory either.) But I can't find a trivial reason that such automorphism which involves determining the way two "basis" map is definitely, trivially exist.
– Eric
yesterday
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Your descriptions of the four automorphisms ar lacking. For 1. and 2. you describe what happens to $sqrt 3$, but not what happens to $sqrt5$, and vice versa for 3. and 4.
Given an arbitrary element $a+bsqrt3+csqrt5+dsqrt{15}$ in your field, where $a, b, c, d$ are rational, the four automorphisms are:
$$
a+bsqrt3+csqrt5+dsqrt{15}mapsto a+bsqrt3+csqrt5+dsqrt{15}\
a+bsqrt3+csqrt5+dsqrt{15}mapsto a-bsqrt3+csqrt5-dsqrt{15}\
a+bsqrt3+csqrt5+dsqrt{15}mapsto a+bsqrt3-csqrt5-dsqrt{15}\
a+bsqrt3+csqrt5+dsqrt{15}mapsto a-bsqrt3-csqrt5+dsqrt{15}
$$
Or, in words:
- One automorphism that maps $sqrt3$ to $sqrt3$ and $sqrt5$ to $sqrt5$
- One automorphism that maps $sqrt3$ to $-sqrt3$ and $sqrt 5$ to $sqrt5$
- One automorphism that maps $sqrt3$ to $sqrt3$ and $sqrt 5$ to $-sqrt5$
- One automorphism that maps $sqrt3$ to $-sqrt3$ and $sqrt5$ to $-sqrt5$
with the implicit understanding that the image of all other elements are defined by the properties of homomorphisms (as well as by fixing $Bbb Q$, although that's superfluous in this specific case).
You should, of course, check that these are indeed automorphisms by showing that they are bijective homomorphisms.
answered yesterday
Arthur
108k7103186
Your descriptions of the four automorphisms ar lacking. For 1. and 2. you describe what happens to $sqrt 3$, but not what happens to $sqrt5$, and vice versa for 3. and 4.
Given an arbitrary element $a+bsqrt3+csqrt5+dsqrt{15}$ in your field, where $a, b, c, d$ are rational, the four automorphisms are:
$$
a+bsqrt3+csqrt5+dsqrt{15}mapsto a+bsqrt3+csqrt5+dsqrt{15}\
a+bsqrt3+csqrt5+dsqrt{15}mapsto a-bsqrt3+csqrt5-dsqrt{15}\
a+bsqrt3+csqrt5+dsqrt{15}mapsto a+bsqrt3-csqrt5-dsqrt{15}\
a+bsqrt3+csqrt5+dsqrt{15}mapsto a-bsqrt3-csqrt5+dsqrt{15}
$$
Or, in words:
- One automorphism that maps $sqrt3$ to $sqrt3$ and $sqrt5$ to $sqrt5$
- One automorphism that maps $sqrt3$ to $-sqrt3$ and $sqrt 5$ to $sqrt5$
- One automorphism that maps $sqrt3$ to $sqrt3$ and $sqrt 5$ to $-sqrt5$
- One automorphism that maps $sqrt3$ to $-sqrt3$ and $sqrt5$ to $-sqrt5$
with the implicit understanding that the image of all other elements are defined by the properties of homomorphisms (as well as by fixing $Bbb Q$, although that's superfluous in this specific case).
You should, of course, check that these are indeed automorphisms by showing that they are bijective homomorphisms.
answered yesterday
Arthur
108k7103186
edited yesterday
answered yesterday
Arthur
108k7103186
answered yesterday
Arthur
108k7103186
answered yesterday
Arthur
108k7103186
108k7103186
So must I check that, say $phi:a+bsqrt3+csqrt5+dsqrt{15}mapsto a-bsqrt3+csqrt5-dsqrt{15}$ is indeed an automorphism? (In other words, it is not automatically so by any theorem). But for the case $Bbb Q(sqrt{2})$ over $Bbb Q$. Since there is a theorem guarantee so, I don't need to check there is an automorphism maps $sqrt{2}$ to $-sqrt{2}$ this time.
– Eric
yesterday
The theorem I mentioned is this
– Eric
yesterday
1
@Eric Depends on what you mean by "automatically". Yes, Galois theory gives you that there is a unique automorphism which fixes $Bbb Q$, sends $sqrt3$ to $-sqrt3$ and $sqrt 5$ to $sqrt5$. On the other hand, your $phi$ is a function which sends $sqrt 3$ to $-sqrt3$ and $sqrt5$ to $sqrt5$. Connecting the two is not completely automatic, and should in principle be done. One way of doing that is showing that your $phi$ is an automorphism. The same is actually true for $Bbb Q(sqrt2)$ and the (quite obvious) non-trivial automorphism there.
– Arthur
yesterday
I see! Yes, I think you completely get what I ask. Actually I haven't studied Galois theorem, I encountered this exercise in the section in the Hungerford's textbook before the Galois theorem section. My teacher told me that this it "redundant" to check such $phi$ must exists. (She means it is trivial, and I think she means it does not need Galois theory either.) But I can't find a trivial reason that such automorphism which involves determining the way two "basis" map is definitely, trivially exist.
– Eric
yesterday
add a comment |
So must I check that, say $phi:a+bsqrt3+csqrt5+dsqrt{15}mapsto a-bsqrt3+csqrt5-dsqrt{15}$ is indeed an automorphism? (In other words, it is not automatically so by any theorem). But for the case $Bbb Q(sqrt{2})$ over $Bbb Q$. Since there is a theorem guarantee so, I don't need to check there is an automorphism maps $sqrt{2}$ to $-sqrt{2}$ this time.
– Eric
yesterday
The theorem I mentioned is this
– Eric
yesterday
1
@Eric Depends on what you mean by "automatically". Yes, Galois theory gives you that there is a unique automorphism which fixes $Bbb Q$, sends $sqrt3$ to $-sqrt3$ and $sqrt 5$ to $sqrt5$. On the other hand, your $phi$ is a function which sends $sqrt 3$ to $-sqrt3$ and $sqrt5$ to $sqrt5$. Connecting the two is not completely automatic, and should in principle be done. One way of doing that is showing that your $phi$ is an automorphism. The same is actually true for $Bbb Q(sqrt2)$ and the (quite obvious) non-trivial automorphism there.
– Arthur
yesterday
I see! Yes, I think you completely get what I ask. Actually I haven't studied Galois theorem, I encountered this exercise in the section in the Hungerford's textbook before the Galois theorem section. My teacher told me that this it "redundant" to check such $phi$ must exists. (She means it is trivial, and I think she means it does not need Galois theory either.) But I can't find a trivial reason that such automorphism which involves determining the way two "basis" map is definitely, trivially exist.
– Eric
yesterday
So must I check that, say $phi:a+bsqrt3+csqrt5+dsqrt{15}mapsto a-bsqrt3+csqrt5-dsqrt{15}$ is indeed an automorphism? (In other words, it is not automatically so by any theorem). But for the case $Bbb Q(sqrt{2})$ over $Bbb Q$. Since there is a theorem guarantee so, I don't need to check there is an automorphism maps $sqrt{2}$ to $-sqrt{2}$ this time.
– Eric
yesterday
So must I check that, say $phi:a+bsqrt3+csqrt5+dsqrt{15}mapsto a-bsqrt3+csqrt5-dsqrt{15}$ is indeed an automorphism? (In other words, it is not automatically so by any theorem). But for the case $Bbb Q(sqrt{2})$ over $Bbb Q$. Since there is a theorem guarantee so, I don't need to check there is an automorphism maps $sqrt{2}$ to $-sqrt{2}$ this time.
– Eric
yesterday
The theorem I mentioned is this
– Eric
yesterday
The theorem I mentioned is this
– Eric
yesterday
1
1
@Eric Depends on what you mean by "automatically". Yes, Galois theory gives you that there is a unique automorphism which fixes $Bbb Q$, sends $sqrt3$ to $-sqrt3$ and $sqrt 5$ to $sqrt5$. On the other hand, your $phi$ is a function which sends $sqrt 3$ to $-sqrt3$ and $sqrt5$ to $sqrt5$. Connecting the two is not completely automatic, and should in principle be done. One way of doing that is showing that your $phi$ is an automorphism. The same is actually true for $Bbb Q(sqrt2)$ and the (quite obvious) non-trivial automorphism there.
– Arthur
yesterday
@Eric Depends on what you mean by "automatically". Yes, Galois theory gives you that there is a unique automorphism which fixes $Bbb Q$, sends $sqrt3$ to $-sqrt3$ and $sqrt 5$ to $sqrt5$. On the other hand, your $phi$ is a function which sends $sqrt 3$ to $-sqrt3$ and $sqrt5$ to $sqrt5$. Connecting the two is not completely automatic, and should in principle be done. One way of doing that is showing that your $phi$ is an automorphism. The same is actually true for $Bbb Q(sqrt2)$ and the (quite obvious) non-trivial automorphism there.
– Arthur
yesterday
I see! Yes, I think you completely get what I ask. Actually I haven't studied Galois theorem, I encountered this exercise in the section in the Hungerford's textbook before the Galois theorem section. My teacher told me that this it "redundant" to check such $phi$ must exists. (She means it is trivial, and I think she means it does not need Galois theory either.) But I can't find a trivial reason that such automorphism which involves determining the way two "basis" map is definitely, trivially exist.
– Eric
yesterday
I see! Yes, I think you completely get what I ask. Actually I haven't studied Galois theorem, I encountered this exercise in the section in the Hungerford's textbook before the Galois theorem section. My teacher told me that this it "redundant" to check such $phi$ must exists. (She means it is trivial, and I think she means it does not need Galois theory either.) But I can't find a trivial reason that such automorphism which involves determining the way two "basis" map is definitely, trivially exist.
– Eric
yesterday
add a comment |
up vote
1
down vote
You don't really have to check all the computations suggested by @Arthur 's answer.
The degree of the extension $mathbb{Q}(sqrt{3},sqrt{5})$ is 4 (because $1,sqrt{3},sqrt{5}$ and $sqrt{15}$ are linearly independent, for example). Also, the field being a splitting field of a separable polynomial(over Q any poly with simple roots is separable) implies that the extension is Galois. So the group of automorphisms will be either $mathbb{Z}_2 times mathbb{Z}_2$ or $mathbb{Z}_4$. But since $sqrt{3}$ $mapsto -sqrt{3}$ (and leave Q fixed) and $sqrt{5} mapsto -sqrt{5}$(and leave Q fixed again) are distinct automorphisms of $mathbb{Q}(sqrt{3})$ and $mathbb{Q}(sqrt{5})$ respectively, that can be extended to automorphisms of $mathbb{Q}(sqrt{3},sqrt{5})$ (any automorphism of a smaller field may be extended to an automorphism of the larger field) and generate subgroups of order two in our Galois group G, it follows that the group cannot be $mathbb{Z}_4$ (since $mathbb{Z}_4$ has only one subgroup of order 2).
answered yesterday
Sorin Tirc
5039
Thanks for the comment @Arthur you are right, I have updated my answer accordingly.
– Sorin Tirc
yesterday
add a comment |
up vote
1
down vote
You don't really have to check all the computations suggested by @Arthur 's answer.
The degree of the extension $mathbb{Q}(sqrt{3},sqrt{5})$ is 4 (because $1,sqrt{3},sqrt{5}$ and $sqrt{15}$ are linearly independent, for example). Also, the field being a splitting field of a separable polynomial(over Q any poly with simple roots is separable) implies that the extension is Galois. So the group of automorphisms will be either $mathbb{Z}_2 times mathbb{Z}_2$ or $mathbb{Z}_4$. But since $sqrt{3}$ $mapsto -sqrt{3}$ (and leave Q fixed) and $sqrt{5} mapsto -sqrt{5}$(and leave Q fixed again) are distinct automorphisms of $mathbb{Q}(sqrt{3})$ and $mathbb{Q}(sqrt{5})$ respectively, that can be extended to automorphisms of $mathbb{Q}(sqrt{3},sqrt{5})$ (any automorphism of a smaller field may be extended to an automorphism of the larger field) and generate subgroups of order two in our Galois group G, it follows that the group cannot be $mathbb{Z}_4$ (since $mathbb{Z}_4$ has only one subgroup of order 2).
answered yesterday
Sorin Tirc
5039
Thanks for the comment @Arthur you are right, I have updated my answer accordingly.
– Sorin Tirc
yesterday
add a comment |
up vote
1
down vote
up vote
1
down vote
You don't really have to check all the computations suggested by @Arthur 's answer.
The degree of the extension $mathbb{Q}(sqrt{3},sqrt{5})$ is 4 (because $1,sqrt{3},sqrt{5}$ and $sqrt{15}$ are linearly independent, for example). Also, the field being a splitting field of a separable polynomial(over Q any poly with simple roots is separable) implies that the extension is Galois. So the group of automorphisms will be either $mathbb{Z}_2 times mathbb{Z}_2$ or $mathbb{Z}_4$. But since $sqrt{3}$ $mapsto -sqrt{3}$ (and leave Q fixed) and $sqrt{5} mapsto -sqrt{5}$(and leave Q fixed again) are distinct automorphisms of $mathbb{Q}(sqrt{3})$ and $mathbb{Q}(sqrt{5})$ respectively, that can be extended to automorphisms of $mathbb{Q}(sqrt{3},sqrt{5})$ (any automorphism of a smaller field may be extended to an automorphism of the larger field) and generate subgroups of order two in our Galois group G, it follows that the group cannot be $mathbb{Z}_4$ (since $mathbb{Z}_4$ has only one subgroup of order 2).
answered yesterday
Sorin Tirc
5039
You don't really have to check all the computations suggested by @Arthur 's answer.
The degree of the extension $mathbb{Q}(sqrt{3},sqrt{5})$ is 4 (because $1,sqrt{3},sqrt{5}$ and $sqrt{15}$ are linearly independent, for example). Also, the field being a splitting field of a separable polynomial(over Q any poly with simple roots is separable) implies that the extension is Galois. So the group of automorphisms will be either $mathbb{Z}_2 times mathbb{Z}_2$ or $mathbb{Z}_4$. But since $sqrt{3}$ $mapsto -sqrt{3}$ (and leave Q fixed) and $sqrt{5} mapsto -sqrt{5}$(and leave Q fixed again) are distinct automorphisms of $mathbb{Q}(sqrt{3})$ and $mathbb{Q}(sqrt{5})$ respectively, that can be extended to automorphisms of $mathbb{Q}(sqrt{3},sqrt{5})$ (any automorphism of a smaller field may be extended to an automorphism of the larger field) and generate subgroups of order two in our Galois group G, it follows that the group cannot be $mathbb{Z}_4$ (since $mathbb{Z}_4$ has only one subgroup of order 2).
answered yesterday
Sorin Tirc
5039
edited yesterday
answered yesterday
Sorin Tirc
5039
answered yesterday
Sorin Tirc
5039
answered yesterday
Sorin Tirc
5039
5039
Thanks for the comment @Arthur you are right, I have updated my answer accordingly.
– Sorin Tirc
yesterday
add a comment |
Thanks for the comment @Arthur you are right, I have updated my answer accordingly.
– Sorin Tirc
yesterday
Thanks for the comment @Arthur you are right, I have updated my answer accordingly.
– Sorin Tirc
yesterday
Thanks for the comment @Arthur you are right, I have updated my answer accordingly.
– Sorin Tirc
yesterday
add a comment |
up vote
0
down vote
Yes. You have an automorphism for every election of signs. But you are no counting your morphisms well. For example, the morphisms 1. and 3. could be the identity.
answered yesterday
Dante Grevino
1463
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up vote
0
down vote
Yes. You have an automorphism for every election of signs. But you are no counting your morphisms well. For example, the morphisms 1. and 3. could be the identity.
answered yesterday
Dante Grevino
1463
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
up vote
0
down vote
Yes. You have an automorphism for every election of signs. But you are no counting your morphisms well. For example, the morphisms 1. and 3. could be the identity.
answered yesterday
Dante Grevino
1463
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Yes. You have an automorphism for every election of signs. But you are no counting your morphisms well. For example, the morphisms 1. and 3. could be the identity.
answered yesterday
Dante Grevino
1463
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Dante Grevino
1463
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Dante Grevino
1463
answered yesterday
Dante Grevino
1463
1463
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Using that the elements of $mathbb{Q}(sqrt{3},sqrt{5})$ are of the form $$a_0+a_1sqrt{3}+a_2sqrt{5}+a_3sqrt{3}sqrt{5},$$ you can easily construct such automorphism.
– rldias
yesterday