A particular field generated from a set
$begingroup$
I am trying to understand a problem from my book regarding field extensions and fields generated from sets.
I have shown the set $B_0 = lbrace (0,0),(1,0) rbrace$ to generate a field $Q$ which consists of all elements of the form $(a,0)$, where $a in mathbb{Q}$. This field is then isomorphic to $mathbb{Q}$ with the map
$sigma : Q to mathbb{Q} quad $ , $quad sigma [(a,0)] = a.$
Suppose now we add an element $(1,sqrt{3}) $ to $B_0$ such that we get $B_1 = lbrace (0,0),(1,0),(1,sqrt{3}) rbrace $. What field would $B_1$ generate? My book claims that this field is $mathbb{Q}(sqrt{3})$, but i have trouble seeing this. In similar way to how we found $sigma$ for $B_0$, what would an isomorphism look like for the field generated by $B_1$?
Thank you very much!
abstract-algebra galois-theory extension-field
asked Nov 30 '18 at 23:10
kasp9201kasp9201
526
$endgroup$
add a comment |
$begingroup$
I am trying to understand a problem from my book regarding field extensions and fields generated from sets.
I have shown the set $B_0 = lbrace (0,0),(1,0) rbrace$ to generate a field $Q$ which consists of all elements of the form $(a,0)$, where $a in mathbb{Q}$. This field is then isomorphic to $mathbb{Q}$ with the map
$sigma : Q to mathbb{Q} quad $ , $quad sigma [(a,0)] = a.$
Suppose now we add an element $(1,sqrt{3}) $ to $B_0$ such that we get $B_1 = lbrace (0,0),(1,0),(1,sqrt{3}) rbrace $. What field would $B_1$ generate? My book claims that this field is $mathbb{Q}(sqrt{3})$, but i have trouble seeing this. In similar way to how we found $sigma$ for $B_0$, what would an isomorphism look like for the field generated by $B_1$?
Thank you very much!
abstract-algebra galois-theory extension-field
asked Nov 30 '18 at 23:10
kasp9201kasp9201
526
$endgroup$
add a comment |
$begingroup$
I am trying to understand a problem from my book regarding field extensions and fields generated from sets.
I have shown the set $B_0 = lbrace (0,0),(1,0) rbrace$ to generate a field $Q$ which consists of all elements of the form $(a,0)$, where $a in mathbb{Q}$. This field is then isomorphic to $mathbb{Q}$ with the map
$sigma : Q to mathbb{Q} quad $ , $quad sigma [(a,0)] = a.$
Suppose now we add an element $(1,sqrt{3}) $ to $B_0$ such that we get $B_1 = lbrace (0,0),(1,0),(1,sqrt{3}) rbrace $. What field would $B_1$ generate? My book claims that this field is $mathbb{Q}(sqrt{3})$, but i have trouble seeing this. In similar way to how we found $sigma$ for $B_0$, what would an isomorphism look like for the field generated by $B_1$?
Thank you very much!
abstract-algebra galois-theory extension-field
asked Nov 30 '18 at 23:10
kasp9201kasp9201
526
$endgroup$
I am trying to understand a problem from my book regarding field extensions and fields generated from sets.
I have shown the set $B_0 = lbrace (0,0),(1,0) rbrace$ to generate a field $Q$ which consists of all elements of the form $(a,0)$, where $a in mathbb{Q}$. This field is then isomorphic to $mathbb{Q}$ with the map
$sigma : Q to mathbb{Q} quad $ , $quad sigma [(a,0)] = a.$
Suppose now we add an element $(1,sqrt{3}) $ to $B_0$ such that we get $B_1 = lbrace (0,0),(1,0),(1,sqrt{3}) rbrace $. What field would $B_1$ generate? My book claims that this field is $mathbb{Q}(sqrt{3})$, but i have trouble seeing this. In similar way to how we found $sigma$ for $B_0$, what would an isomorphism look like for the field generated by $B_1$?
Thank you very much!
abstract-algebra galois-theory extension-field
abstract-algebra galois-theory extension-field
asked Nov 30 '18 at 23:10
kasp9201kasp9201
526
asked Nov 30 '18 at 23:10
kasp9201kasp9201
526
asked Nov 30 '18 at 23:10
kasp9201kasp9201
526
asked Nov 30 '18 at 23:10
kasp9201kasp9201
526
asked Nov 30 '18 at 23:10
kasp9201kasp9201
526
526
add a comment |
add a comment |
1 Answer
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oldest
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$begingroup$
If we call the field generated by $B_1$ something, say $F$, then the map begin{align*}F &to mathbb{Q}(sqrt{3})\(x,ysqrt{3})&mapsto x+ysqrt{3} end{align*}
is an isomorphism.
To prove that:
Firstly, it is well-defined, as $x$ and $y$ are rational for all elements of $F$, so $x+ysqrt{3}$ is, indeed, in $mathbb{Q}(sqrt{3})$.
Secondly, it's clearly a field homomorphism (check everything if you're so inclined).
Thirdly, it's injective, since if $x + y sqrt{3} = 0$ with $x$ and $y$ rational, then $x = y = 0$.
Finally, it's surjective, since we can get all $x + ysqrt{3}$ in this way, as the image of $(x,ysqrt{3}) = y(1,sqrt{3})+(x-y)(1,0)$.
answered Nov 30 '18 at 23:16
user3482749user3482749
4,266919
$endgroup$
$begingroup$
Perfect, thank you very much!
$endgroup$
– kasp9201
Nov 30 '18 at 23:28
$begingroup$
Sorry to bother you again @user3482749, but I have been thinking about your answer and something still does not make sense to me. Is (x,ysqrt{3}) really the form every number in F has? Is it not true that we also have elements (a,b) in F, where b is an integer? If so, I do not see how we can write this as (x,ysqrt{3}). Thank you very much!
$endgroup$
– kasp9201
Dec 1 '18 at 5:41
1
$begingroup$
This is where we need slightly more detail: what, exactly, are your field operations? My answer is correct for the ones that I was thinking of, but it occurs to me that these aren't the only option, or even necessarily the most obvious ones.
$endgroup$
– user3482749
Dec 1 '18 at 10:47
$begingroup$
Aha, that was exactly what I needed to hear. I proved the isomorphism you provided me with, and it makes sense to me now. Again, thank you very much!
$endgroup$
– kasp9201
Dec 1 '18 at 16:51
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If we call the field generated by $B_1$ something, say $F$, then the map begin{align*}F &to mathbb{Q}(sqrt{3})\(x,ysqrt{3})&mapsto x+ysqrt{3} end{align*}
is an isomorphism.
To prove that:
Firstly, it is well-defined, as $x$ and $y$ are rational for all elements of $F$, so $x+ysqrt{3}$ is, indeed, in $mathbb{Q}(sqrt{3})$.
Secondly, it's clearly a field homomorphism (check everything if you're so inclined).
Thirdly, it's injective, since if $x + y sqrt{3} = 0$ with $x$ and $y$ rational, then $x = y = 0$.
Finally, it's surjective, since we can get all $x + ysqrt{3}$ in this way, as the image of $(x,ysqrt{3}) = y(1,sqrt{3})+(x-y)(1,0)$.
answered Nov 30 '18 at 23:16
user3482749user3482749
4,266919
$endgroup$
$begingroup$
Perfect, thank you very much!
$endgroup$
– kasp9201
Nov 30 '18 at 23:28
$begingroup$
Sorry to bother you again @user3482749, but I have been thinking about your answer and something still does not make sense to me. Is (x,ysqrt{3}) really the form every number in F has? Is it not true that we also have elements (a,b) in F, where b is an integer? If so, I do not see how we can write this as (x,ysqrt{3}). Thank you very much!
$endgroup$
– kasp9201
Dec 1 '18 at 5:41
1
$begingroup$
This is where we need slightly more detail: what, exactly, are your field operations? My answer is correct for the ones that I was thinking of, but it occurs to me that these aren't the only option, or even necessarily the most obvious ones.
$endgroup$
– user3482749
Dec 1 '18 at 10:47
$begingroup$
Aha, that was exactly what I needed to hear. I proved the isomorphism you provided me with, and it makes sense to me now. Again, thank you very much!
$endgroup$
– kasp9201
Dec 1 '18 at 16:51
add a comment |
$begingroup$
If we call the field generated by $B_1$ something, say $F$, then the map begin{align*}F &to mathbb{Q}(sqrt{3})\(x,ysqrt{3})&mapsto x+ysqrt{3} end{align*}
is an isomorphism.
To prove that:
Firstly, it is well-defined, as $x$ and $y$ are rational for all elements of $F$, so $x+ysqrt{3}$ is, indeed, in $mathbb{Q}(sqrt{3})$.
Secondly, it's clearly a field homomorphism (check everything if you're so inclined).
Thirdly, it's injective, since if $x + y sqrt{3} = 0$ with $x$ and $y$ rational, then $x = y = 0$.
Finally, it's surjective, since we can get all $x + ysqrt{3}$ in this way, as the image of $(x,ysqrt{3}) = y(1,sqrt{3})+(x-y)(1,0)$.
answered Nov 30 '18 at 23:16
user3482749user3482749
4,266919
$endgroup$
$begingroup$
Perfect, thank you very much!
$endgroup$
– kasp9201
Nov 30 '18 at 23:28
$begingroup$
Sorry to bother you again @user3482749, but I have been thinking about your answer and something still does not make sense to me. Is (x,ysqrt{3}) really the form every number in F has? Is it not true that we also have elements (a,b) in F, where b is an integer? If so, I do not see how we can write this as (x,ysqrt{3}). Thank you very much!
$endgroup$
– kasp9201
Dec 1 '18 at 5:41
1
$begingroup$
This is where we need slightly more detail: what, exactly, are your field operations? My answer is correct for the ones that I was thinking of, but it occurs to me that these aren't the only option, or even necessarily the most obvious ones.
$endgroup$
– user3482749
Dec 1 '18 at 10:47
$begingroup$
Aha, that was exactly what I needed to hear. I proved the isomorphism you provided me with, and it makes sense to me now. Again, thank you very much!
$endgroup$
– kasp9201
Dec 1 '18 at 16:51
add a comment |
$begingroup$
If we call the field generated by $B_1$ something, say $F$, then the map begin{align*}F &to mathbb{Q}(sqrt{3})\(x,ysqrt{3})&mapsto x+ysqrt{3} end{align*}
is an isomorphism.
To prove that:
Firstly, it is well-defined, as $x$ and $y$ are rational for all elements of $F$, so $x+ysqrt{3}$ is, indeed, in $mathbb{Q}(sqrt{3})$.
Secondly, it's clearly a field homomorphism (check everything if you're so inclined).
Thirdly, it's injective, since if $x + y sqrt{3} = 0$ with $x$ and $y$ rational, then $x = y = 0$.
Finally, it's surjective, since we can get all $x + ysqrt{3}$ in this way, as the image of $(x,ysqrt{3}) = y(1,sqrt{3})+(x-y)(1,0)$.
answered Nov 30 '18 at 23:16
user3482749user3482749
4,266919
$endgroup$
If we call the field generated by $B_1$ something, say $F$, then the map begin{align*}F &to mathbb{Q}(sqrt{3})\(x,ysqrt{3})&mapsto x+ysqrt{3} end{align*}
is an isomorphism.
To prove that:
Firstly, it is well-defined, as $x$ and $y$ are rational for all elements of $F$, so $x+ysqrt{3}$ is, indeed, in $mathbb{Q}(sqrt{3})$.
Secondly, it's clearly a field homomorphism (check everything if you're so inclined).
Thirdly, it's injective, since if $x + y sqrt{3} = 0$ with $x$ and $y$ rational, then $x = y = 0$.
Finally, it's surjective, since we can get all $x + ysqrt{3}$ in this way, as the image of $(x,ysqrt{3}) = y(1,sqrt{3})+(x-y)(1,0)$.
answered Nov 30 '18 at 23:16
user3482749user3482749
4,266919
answered Nov 30 '18 at 23:16
user3482749user3482749
4,266919
answered Nov 30 '18 at 23:16
user3482749user3482749
4,266919
answered Nov 30 '18 at 23:16
user3482749user3482749
4,266919
4,266919
$begingroup$
Perfect, thank you very much!
$endgroup$
– kasp9201
Nov 30 '18 at 23:28
$begingroup$
Sorry to bother you again @user3482749, but I have been thinking about your answer and something still does not make sense to me. Is (x,ysqrt{3}) really the form every number in F has? Is it not true that we also have elements (a,b) in F, where b is an integer? If so, I do not see how we can write this as (x,ysqrt{3}). Thank you very much!
$endgroup$
– kasp9201
Dec 1 '18 at 5:41
1
$begingroup$
This is where we need slightly more detail: what, exactly, are your field operations? My answer is correct for the ones that I was thinking of, but it occurs to me that these aren't the only option, or even necessarily the most obvious ones.
$endgroup$
– user3482749
Dec 1 '18 at 10:47
$begingroup$
Aha, that was exactly what I needed to hear. I proved the isomorphism you provided me with, and it makes sense to me now. Again, thank you very much!
$endgroup$
– kasp9201
Dec 1 '18 at 16:51
add a comment |
$begingroup$
Perfect, thank you very much!
$endgroup$
– kasp9201
Nov 30 '18 at 23:28
$begingroup$
Sorry to bother you again @user3482749, but I have been thinking about your answer and something still does not make sense to me. Is (x,ysqrt{3}) really the form every number in F has? Is it not true that we also have elements (a,b) in F, where b is an integer? If so, I do not see how we can write this as (x,ysqrt{3}). Thank you very much!
$endgroup$
– kasp9201
Dec 1 '18 at 5:41
1
$begingroup$
This is where we need slightly more detail: what, exactly, are your field operations? My answer is correct for the ones that I was thinking of, but it occurs to me that these aren't the only option, or even necessarily the most obvious ones.
$endgroup$
– user3482749
Dec 1 '18 at 10:47
$begingroup$
Aha, that was exactly what I needed to hear. I proved the isomorphism you provided me with, and it makes sense to me now. Again, thank you very much!
$endgroup$
– kasp9201
Dec 1 '18 at 16:51
$begingroup$
Perfect, thank you very much!
$endgroup$
– kasp9201
Nov 30 '18 at 23:28
$begingroup$
Perfect, thank you very much!
$endgroup$
– kasp9201
Nov 30 '18 at 23:28
$begingroup$
Sorry to bother you again @user3482749, but I have been thinking about your answer and something still does not make sense to me. Is (x,ysqrt{3}) really the form every number in F has? Is it not true that we also have elements (a,b) in F, where b is an integer? If so, I do not see how we can write this as (x,ysqrt{3}). Thank you very much!
$endgroup$
– kasp9201
Dec 1 '18 at 5:41
$begingroup$
Sorry to bother you again @user3482749, but I have been thinking about your answer and something still does not make sense to me. Is (x,ysqrt{3}) really the form every number in F has? Is it not true that we also have elements (a,b) in F, where b is an integer? If so, I do not see how we can write this as (x,ysqrt{3}). Thank you very much!
$endgroup$
– kasp9201
Dec 1 '18 at 5:41
1
1
$begingroup$
This is where we need slightly more detail: what, exactly, are your field operations? My answer is correct for the ones that I was thinking of, but it occurs to me that these aren't the only option, or even necessarily the most obvious ones.
$endgroup$
– user3482749
Dec 1 '18 at 10:47
$begingroup$
This is where we need slightly more detail: what, exactly, are your field operations? My answer is correct for the ones that I was thinking of, but it occurs to me that these aren't the only option, or even necessarily the most obvious ones.
$endgroup$
– user3482749
Dec 1 '18 at 10:47
$begingroup$
Aha, that was exactly what I needed to hear. I proved the isomorphism you provided me with, and it makes sense to me now. Again, thank you very much!
$endgroup$
– kasp9201
Dec 1 '18 at 16:51
$begingroup$
Aha, that was exactly what I needed to hear. I proved the isomorphism you provided me with, and it makes sense to me now. Again, thank you very much!
$endgroup$
– kasp9201
Dec 1 '18 at 16:51
add a comment |
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