Fields isomorphic to $mathbbQ$
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We know that certain fields has an isomorphic copy of $mathbbQ$ (for example, every ordered field). But, is there an no trivial explicit example of a field that is isomorphic to $mathbbQ$?
field-theory rational-numbers
asked Aug 23 at 4:48
sinbadh
6,283724
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up vote
2
down vote
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We know that certain fields has an isomorphic copy of $mathbbQ$ (for example, every ordered field). But, is there an no trivial explicit example of a field that is isomorphic to $mathbbQ$?
field-theory rational-numbers
asked Aug 23 at 4:48
sinbadh
6,283724
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
We know that certain fields has an isomorphic copy of $mathbbQ$ (for example, every ordered field). But, is there an no trivial explicit example of a field that is isomorphic to $mathbbQ$?
field-theory rational-numbers
asked Aug 23 at 4:48
sinbadh
6,283724
We know that certain fields has an isomorphic copy of $mathbbQ$ (for example, every ordered field). But, is there an no trivial explicit example of a field that is isomorphic to $mathbbQ$?
field-theory rational-numbers
asked Aug 23 at 4:48
sinbadh
6,283724
asked Aug 23 at 4:48
sinbadh
6,283724
asked Aug 23 at 4:48
sinbadh
6,283724
asked Aug 23 at 4:48
sinbadh
6,283724
6,283724
add a comment |Â
add a comment |Â
1 Answer
1
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5
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You're not gonna find a field that is isomorphic to $Bbb Q$ but doesn't "look just like" it: if $K$ is a field isomorphic to $Bbb Q$ and $1_K$ is the identity element, then $Bbb Z$ uniquely embeds in $K$ by identifying $aleftrightarrow acdot 1_K$. Then the only option for the isomorphism $Bbb Qcong K$ is $a/bleftrightarrow (acdot 1_K)/(bcdot 1_K)$.
Basically the point is that you'll never see a field isomorphic to $Bbb Q$ which the author doesn't just call $Bbb Q$ because the identification is so "explicit". If you want some trivial answer, you can take any countable set $X$ and use a bijection $Bbb Qto X$ to put a field structure on $X$ that will make $X$ isomorphic to $Bbb Q$ as a field.
answered Aug 23 at 4:57
Alex Mathers
10.2k21343
Ok. I understood it. is not last statement false? For example, $mathbbQ[sqrt2]$ is biyectable with $mathbbQ$ but they are not isomorphic fields
– sinbadh
Aug 23 at 5:10
2
Just because you can take a bijection $f:Bbb QtoBbb Q[sqrt2]$ and use this to put some (emphasize some) field structure on $Bbb Q[sqrt 2]$ doesn't mean that this must be the same as the "standard" field structure on $Bbb Q[sqrt2]$.
– Alex Mathers
Aug 23 at 5:12
Yes. I thought it after writting my last comment
– sinbadh
Aug 23 at 5:15
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
You're not gonna find a field that is isomorphic to $Bbb Q$ but doesn't "look just like" it: if $K$ is a field isomorphic to $Bbb Q$ and $1_K$ is the identity element, then $Bbb Z$ uniquely embeds in $K$ by identifying $aleftrightarrow acdot 1_K$. Then the only option for the isomorphism $Bbb Qcong K$ is $a/bleftrightarrow (acdot 1_K)/(bcdot 1_K)$.
Basically the point is that you'll never see a field isomorphic to $Bbb Q$ which the author doesn't just call $Bbb Q$ because the identification is so "explicit". If you want some trivial answer, you can take any countable set $X$ and use a bijection $Bbb Qto X$ to put a field structure on $X$ that will make $X$ isomorphic to $Bbb Q$ as a field.
answered Aug 23 at 4:57
Alex Mathers
10.2k21343
Ok. I understood it. is not last statement false? For example, $mathbbQ[sqrt2]$ is biyectable with $mathbbQ$ but they are not isomorphic fields
– sinbadh
Aug 23 at 5:10
2
Just because you can take a bijection $f:Bbb QtoBbb Q[sqrt2]$ and use this to put some (emphasize some) field structure on $Bbb Q[sqrt 2]$ doesn't mean that this must be the same as the "standard" field structure on $Bbb Q[sqrt2]$.
– Alex Mathers
Aug 23 at 5:12
Yes. I thought it after writting my last comment
– sinbadh
Aug 23 at 5:15
add a comment |Â
up vote
5
down vote
You're not gonna find a field that is isomorphic to $Bbb Q$ but doesn't "look just like" it: if $K$ is a field isomorphic to $Bbb Q$ and $1_K$ is the identity element, then $Bbb Z$ uniquely embeds in $K$ by identifying $aleftrightarrow acdot 1_K$. Then the only option for the isomorphism $Bbb Qcong K$ is $a/bleftrightarrow (acdot 1_K)/(bcdot 1_K)$.
Basically the point is that you'll never see a field isomorphic to $Bbb Q$ which the author doesn't just call $Bbb Q$ because the identification is so "explicit". If you want some trivial answer, you can take any countable set $X$ and use a bijection $Bbb Qto X$ to put a field structure on $X$ that will make $X$ isomorphic to $Bbb Q$ as a field.
answered Aug 23 at 4:57
Alex Mathers
10.2k21343
Ok. I understood it. is not last statement false? For example, $mathbbQ[sqrt2]$ is biyectable with $mathbbQ$ but they are not isomorphic fields
– sinbadh
Aug 23 at 5:10
2
Just because you can take a bijection $f:Bbb QtoBbb Q[sqrt2]$ and use this to put some (emphasize some) field structure on $Bbb Q[sqrt 2]$ doesn't mean that this must be the same as the "standard" field structure on $Bbb Q[sqrt2]$.
– Alex Mathers
Aug 23 at 5:12
Yes. I thought it after writting my last comment
– sinbadh
Aug 23 at 5:15
add a comment |Â
up vote
5
down vote
up vote
5
down vote
You're not gonna find a field that is isomorphic to $Bbb Q$ but doesn't "look just like" it: if $K$ is a field isomorphic to $Bbb Q$ and $1_K$ is the identity element, then $Bbb Z$ uniquely embeds in $K$ by identifying $aleftrightarrow acdot 1_K$. Then the only option for the isomorphism $Bbb Qcong K$ is $a/bleftrightarrow (acdot 1_K)/(bcdot 1_K)$.
Basically the point is that you'll never see a field isomorphic to $Bbb Q$ which the author doesn't just call $Bbb Q$ because the identification is so "explicit". If you want some trivial answer, you can take any countable set $X$ and use a bijection $Bbb Qto X$ to put a field structure on $X$ that will make $X$ isomorphic to $Bbb Q$ as a field.
answered Aug 23 at 4:57
Alex Mathers
10.2k21343
You're not gonna find a field that is isomorphic to $Bbb Q$ but doesn't "look just like" it: if $K$ is a field isomorphic to $Bbb Q$ and $1_K$ is the identity element, then $Bbb Z$ uniquely embeds in $K$ by identifying $aleftrightarrow acdot 1_K$. Then the only option for the isomorphism $Bbb Qcong K$ is $a/bleftrightarrow (acdot 1_K)/(bcdot 1_K)$.
Basically the point is that you'll never see a field isomorphic to $Bbb Q$ which the author doesn't just call $Bbb Q$ because the identification is so "explicit". If you want some trivial answer, you can take any countable set $X$ and use a bijection $Bbb Qto X$ to put a field structure on $X$ that will make $X$ isomorphic to $Bbb Q$ as a field.
answered Aug 23 at 4:57
Alex Mathers
10.2k21343
answered Aug 23 at 4:57
Alex Mathers
10.2k21343
answered Aug 23 at 4:57
Alex Mathers
10.2k21343
answered Aug 23 at 4:57
Alex Mathers
10.2k21343
10.2k21343
Ok. I understood it. is not last statement false? For example, $mathbbQ[sqrt2]$ is biyectable with $mathbbQ$ but they are not isomorphic fields
– sinbadh
Aug 23 at 5:10
2
Just because you can take a bijection $f:Bbb QtoBbb Q[sqrt2]$ and use this to put some (emphasize some) field structure on $Bbb Q[sqrt 2]$ doesn't mean that this must be the same as the "standard" field structure on $Bbb Q[sqrt2]$.
– Alex Mathers
Aug 23 at 5:12
Yes. I thought it after writting my last comment
– sinbadh
Aug 23 at 5:15
add a comment |Â
Ok. I understood it. is not last statement false? For example, $mathbbQ[sqrt2]$ is biyectable with $mathbbQ$ but they are not isomorphic fields
– sinbadh
Aug 23 at 5:10
2
Just because you can take a bijection $f:Bbb QtoBbb Q[sqrt2]$ and use this to put some (emphasize some) field structure on $Bbb Q[sqrt 2]$ doesn't mean that this must be the same as the "standard" field structure on $Bbb Q[sqrt2]$.
– Alex Mathers
Aug 23 at 5:12
Yes. I thought it after writting my last comment
– sinbadh
Aug 23 at 5:15
Ok. I understood it. is not last statement false? For example, $mathbbQ[sqrt2]$ is biyectable with $mathbbQ$ but they are not isomorphic fields
– sinbadh
Aug 23 at 5:10
Ok. I understood it. is not last statement false? For example, $mathbbQ[sqrt2]$ is biyectable with $mathbbQ$ but they are not isomorphic fields
– sinbadh
Aug 23 at 5:10
2
2
Just because you can take a bijection $f:Bbb QtoBbb Q[sqrt2]$ and use this to put some (emphasize some) field structure on $Bbb Q[sqrt 2]$ doesn't mean that this must be the same as the "standard" field structure on $Bbb Q[sqrt2]$.
– Alex Mathers
Aug 23 at 5:12
Just because you can take a bijection $f:Bbb QtoBbb Q[sqrt2]$ and use this to put some (emphasize some) field structure on $Bbb Q[sqrt 2]$ doesn't mean that this must be the same as the "standard" field structure on $Bbb Q[sqrt2]$.
– Alex Mathers
Aug 23 at 5:12
Yes. I thought it after writting my last comment
– sinbadh
Aug 23 at 5:15
Yes. I thought it after writting my last comment
– sinbadh
Aug 23 at 5:15
add a comment |Â
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