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




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    up 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




    share|cite|improve this question






















      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




      share|cite|improve this question












      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





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      asked Aug 23 at 4:48









      sinbadh

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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.






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          • 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











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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.






          share|cite|improve this answer




















          • 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















          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.






          share|cite|improve this answer




















          • 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













          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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

















          • 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


















           

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