A particular field generated from a set












0












$begingroup$


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







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    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







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      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







      share|cite|improve this question









      $endgroup$




      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






      share|cite|improve this question













      share|cite|improve this question











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      asked Nov 30 '18 at 23:10









      kasp9201kasp9201

      526




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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 $(x,ysqrt{3}) = y(1,sqrt{3})+(x-y)(1,0)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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











          Your 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 $(x,ysqrt{3}) = y(1,sqrt{3})+(x-y)(1,0)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          1












          $begingroup$

          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 $(x,ysqrt{3}) = y(1,sqrt{3})+(x-y)(1,0)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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














          1












          1








          1





          $begingroup$

          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 $(x,ysqrt{3}) = y(1,sqrt{3})+(x-y)(1,0)$.






          share|cite|improve this answer









          $endgroup$



          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 $(x,ysqrt{3}) = y(1,sqrt{3})+(x-y)(1,0)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • $begingroup$
            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


















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