Proof that $mathbb Z[sqrt{3}]$ is a Euclidean Domain












10












$begingroup$


Let $R_d$ be the ring defined as $R_d=left { x+yomega : x,yin mathbb{Z} right}$, where
$$omega =
begin{cases}
sqrt{d}, & text{if } quad d not equiv 1mod 4 \
frac{1+sqrt{d}}{2}, & text{if } quad dequiv 1mod 4.
end{cases}$$



It has been proven that $R_d$ is Euclidean for several positive values of $d$.




Does anyone know where I can find a proof that $R_d$ is Euclidean for $d=3$?




Thank you.






abstract-algebra reference-request ring-theory







share|cite|improve this question











$endgroup$

















    10












    $begingroup$


    Let $R_d$ be the ring defined as $R_d=left { x+yomega : x,yin mathbb{Z} right}$, where
    $$omega =
    begin{cases}
    sqrt{d}, & text{if } quad d not equiv 1mod 4 \
    frac{1+sqrt{d}}{2}, & text{if } quad dequiv 1mod 4.
    end{cases}$$



    It has been proven that $R_d$ is Euclidean for several positive values of $d$.




    Does anyone know where I can find a proof that $R_d$ is Euclidean for $d=3$?




    Thank you.






    abstract-algebra reference-request ring-theory







    share|cite|improve this question











    $endgroup$















      10












      10








      10


      4



      $begingroup$


      Let $R_d$ be the ring defined as $R_d=left { x+yomega : x,yin mathbb{Z} right}$, where
      $$omega =
      begin{cases}
      sqrt{d}, & text{if } quad d not equiv 1mod 4 \
      frac{1+sqrt{d}}{2}, & text{if } quad dequiv 1mod 4.
      end{cases}$$



      It has been proven that $R_d$ is Euclidean for several positive values of $d$.




      Does anyone know where I can find a proof that $R_d$ is Euclidean for $d=3$?




      Thank you.






      abstract-algebra reference-request ring-theory







      share|cite|improve this question











      $endgroup$




      Let $R_d$ be the ring defined as $R_d=left { x+yomega : x,yin mathbb{Z} right}$, where
      $$omega =
      begin{cases}
      sqrt{d}, & text{if } quad d not equiv 1mod 4 \
      frac{1+sqrt{d}}{2}, & text{if } quad dequiv 1mod 4.
      end{cases}$$



      It has been proven that $R_d$ is Euclidean for several positive values of $d$.




      Does anyone know where I can find a proof that $R_d$ is Euclidean for $d=3$?




      Thank you.






      abstract-algebra reference-request ring-theory




      abstract-algebra reference-request ring-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 18 '15 at 4:45









      Robert Cardona

      5,29023499




      5,29023499










      asked Oct 7 '13 at 3:57









      SarahSarah

      10219




      10219






















          1 Answer
          1






          active

          oldest

          votes


















          9












          $begingroup$

          Define the norm on $mathbb Z[sqrt 3]$ to be $N(a + b sqrt 3) = vert a^2 - 3 b^2 vert$.



          Let $alpha, beta in mathbb Z[sqrt 3]$ with $beta neq 0$.



          Say $alpha = a + b sqrt 3$ and $beta = c + d sqrt 3$.



          Notice that
          begin{align*}
          fracalphabeta &= frac{a + b sqrt 3}{c + d sqrt 3} cdot frac{c - d sqrt 3}{c - d sqrt 3} \
          &= frac{ac - 3bd}{c^2 - 3d^2} + frac{-ad + bc}{c^2 - 3d^2} sqrt 3 \
          &= r + ssqrt 3
          end{align*}



          where $r = displaystyle frac{ac - 3bd}{c^2 - 3d^2}$ and $s = displaystyle frac{-ad + bc}{c^2 - 3d^2}$.



          Let $p$ be the closest integer to $r$ and let $q$ be the closest integer to $s$. Notice that $vert r - p vert leq 1/2$ and $vert s - q vert leq 1/2$.



          We want to show that $alpha = (p + qsqrt 3) beta + gamma$ for some $gamma in mathbb Z[sqrt 3]$ such that either $gamma = 0$ or $N(gamma) < N(beta)$. (We'll show the latter holds always.)



          Define $theta := (r - p) + (s - q)sqrt 3$ and define $gamma = beta cdot theta in mathbb Z[sqrt 3]$ and observe that
          begin{align*}
          gamma &= beta cdot theta\
          &= beta ( (r - p) + (s - q)sqrt 3)\
          &= beta (r + ssqrt 3) - beta(p + qsqrt 3) \
          &= beta cdotfracalphabeta - beta (p + qsqrt 3) \
          &= alpha - beta (p + qsqrt 3)
          end{align*}



          Hence we have $alpha = beta(p + qsqrt 3) + gamma$.



          Finally notice that
          begin{align*}
          N(gamma) &= N(beta cdot theta) \
          &= N(beta) cdot N(theta) \
          &= N(beta) cdot vert (r - p)^2 - 3 (s - q)^2 vert \
          &leq N(beta) cdot max{ (r - p)^2, 3(s - q)^2} \
          & leqfrac34 N(beta)\
          &< N(beta)
          end{align*}



          The key here was that $vert (r - p)^2 - 3 (s - q)^2 vert leq max{ (r - p)^2, 3(s - q)^2}$ since $(r - p)^2, 3(s - q)^2 geq 0$ and then we use that $(r - p)^2 leq 1/4$ and $3(s - q)^2 leq 3/4$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            When you're proving that $N(gamma) < N(beta)$, you say that $N(theta) = |(p-r)^2 - 3(q-s)^2|$. However, $N$ was only defined for elements in $Bbb{Z}[sqrt{3}]$. You showed that $(beta)(theta) in Bbb{Z}[sqrt{3}]$. But how does this imply that $theta in Bbb{Z}[sqrt{3}]$?
            $endgroup$
            – Artus
            Mar 12 '15 at 14:05








          • 2




            $begingroup$
            That's a very good question! We can just think of the norm $N$ on $mathbb Z[sqrt 3]$ as being the norm on $mathbb Q[sqrt 3]$, restricted to $mathbb Z[sqrt 3]$.
            $endgroup$
            – Robert Cardona
            Mar 12 '15 at 15:13










          • $begingroup$
            why is |p - r |< 1/2 and |q- s| < 1/2
            $endgroup$
            – user10024395
            Apr 23 '15 at 9:33










          • $begingroup$
            @user136266, because we define $p, q$ to be the closest integers to $r, s$, respectively. Say $p = frac14$, the closest integer is $0$ and the difference is less than $frac12$, if $p = frac34$, then the closest integer is $1$. If $p = frac12$, then pick either $0$ or $1$ and the claim holds (less than or equal to).
            $endgroup$
            – Robert Cardona
            Apr 23 '15 at 14:21










          • $begingroup$
            @RobertCardona how is 1/4 and 1/2 integers? these are not integers. i think you meant r but even so why must r be only these 3 possibilities?If i have a, c = 1 and b,d = 0. Then r = 0 and so u can't find an integer p such that p-r < 1/2.
            $endgroup$
            – user10024395
            Apr 24 '15 at 1:45













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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          9












          $begingroup$

          Define the norm on $mathbb Z[sqrt 3]$ to be $N(a + b sqrt 3) = vert a^2 - 3 b^2 vert$.



          Let $alpha, beta in mathbb Z[sqrt 3]$ with $beta neq 0$.



          Say $alpha = a + b sqrt 3$ and $beta = c + d sqrt 3$.



          Notice that
          begin{align*}
          fracalphabeta &= frac{a + b sqrt 3}{c + d sqrt 3} cdot frac{c - d sqrt 3}{c - d sqrt 3} \
          &= frac{ac - 3bd}{c^2 - 3d^2} + frac{-ad + bc}{c^2 - 3d^2} sqrt 3 \
          &= r + ssqrt 3
          end{align*}



          where $r = displaystyle frac{ac - 3bd}{c^2 - 3d^2}$ and $s = displaystyle frac{-ad + bc}{c^2 - 3d^2}$.



          Let $p$ be the closest integer to $r$ and let $q$ be the closest integer to $s$. Notice that $vert r - p vert leq 1/2$ and $vert s - q vert leq 1/2$.



          We want to show that $alpha = (p + qsqrt 3) beta + gamma$ for some $gamma in mathbb Z[sqrt 3]$ such that either $gamma = 0$ or $N(gamma) < N(beta)$. (We'll show the latter holds always.)



          Define $theta := (r - p) + (s - q)sqrt 3$ and define $gamma = beta cdot theta in mathbb Z[sqrt 3]$ and observe that
          begin{align*}
          gamma &= beta cdot theta\
          &= beta ( (r - p) + (s - q)sqrt 3)\
          &= beta (r + ssqrt 3) - beta(p + qsqrt 3) \
          &= beta cdotfracalphabeta - beta (p + qsqrt 3) \
          &= alpha - beta (p + qsqrt 3)
          end{align*}



          Hence we have $alpha = beta(p + qsqrt 3) + gamma$.



          Finally notice that
          begin{align*}
          N(gamma) &= N(beta cdot theta) \
          &= N(beta) cdot N(theta) \
          &= N(beta) cdot vert (r - p)^2 - 3 (s - q)^2 vert \
          &leq N(beta) cdot max{ (r - p)^2, 3(s - q)^2} \
          & leqfrac34 N(beta)\
          &< N(beta)
          end{align*}



          The key here was that $vert (r - p)^2 - 3 (s - q)^2 vert leq max{ (r - p)^2, 3(s - q)^2}$ since $(r - p)^2, 3(s - q)^2 geq 0$ and then we use that $(r - p)^2 leq 1/4$ and $3(s - q)^2 leq 3/4$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            When you're proving that $N(gamma) < N(beta)$, you say that $N(theta) = |(p-r)^2 - 3(q-s)^2|$. However, $N$ was only defined for elements in $Bbb{Z}[sqrt{3}]$. You showed that $(beta)(theta) in Bbb{Z}[sqrt{3}]$. But how does this imply that $theta in Bbb{Z}[sqrt{3}]$?
            $endgroup$
            – Artus
            Mar 12 '15 at 14:05








          • 2




            $begingroup$
            That's a very good question! We can just think of the norm $N$ on $mathbb Z[sqrt 3]$ as being the norm on $mathbb Q[sqrt 3]$, restricted to $mathbb Z[sqrt 3]$.
            $endgroup$
            – Robert Cardona
            Mar 12 '15 at 15:13










          • $begingroup$
            why is |p - r |< 1/2 and |q- s| < 1/2
            $endgroup$
            – user10024395
            Apr 23 '15 at 9:33










          • $begingroup$
            @user136266, because we define $p, q$ to be the closest integers to $r, s$, respectively. Say $p = frac14$, the closest integer is $0$ and the difference is less than $frac12$, if $p = frac34$, then the closest integer is $1$. If $p = frac12$, then pick either $0$ or $1$ and the claim holds (less than or equal to).
            $endgroup$
            – Robert Cardona
            Apr 23 '15 at 14:21










          • $begingroup$
            @RobertCardona how is 1/4 and 1/2 integers? these are not integers. i think you meant r but even so why must r be only these 3 possibilities?If i have a, c = 1 and b,d = 0. Then r = 0 and so u can't find an integer p such that p-r < 1/2.
            $endgroup$
            – user10024395
            Apr 24 '15 at 1:45


















          9












          $begingroup$

          Define the norm on $mathbb Z[sqrt 3]$ to be $N(a + b sqrt 3) = vert a^2 - 3 b^2 vert$.



          Let $alpha, beta in mathbb Z[sqrt 3]$ with $beta neq 0$.



          Say $alpha = a + b sqrt 3$ and $beta = c + d sqrt 3$.



          Notice that
          begin{align*}
          fracalphabeta &= frac{a + b sqrt 3}{c + d sqrt 3} cdot frac{c - d sqrt 3}{c - d sqrt 3} \
          &= frac{ac - 3bd}{c^2 - 3d^2} + frac{-ad + bc}{c^2 - 3d^2} sqrt 3 \
          &= r + ssqrt 3
          end{align*}



          where $r = displaystyle frac{ac - 3bd}{c^2 - 3d^2}$ and $s = displaystyle frac{-ad + bc}{c^2 - 3d^2}$.



          Let $p$ be the closest integer to $r$ and let $q$ be the closest integer to $s$. Notice that $vert r - p vert leq 1/2$ and $vert s - q vert leq 1/2$.



          We want to show that $alpha = (p + qsqrt 3) beta + gamma$ for some $gamma in mathbb Z[sqrt 3]$ such that either $gamma = 0$ or $N(gamma) < N(beta)$. (We'll show the latter holds always.)



          Define $theta := (r - p) + (s - q)sqrt 3$ and define $gamma = beta cdot theta in mathbb Z[sqrt 3]$ and observe that
          begin{align*}
          gamma &= beta cdot theta\
          &= beta ( (r - p) + (s - q)sqrt 3)\
          &= beta (r + ssqrt 3) - beta(p + qsqrt 3) \
          &= beta cdotfracalphabeta - beta (p + qsqrt 3) \
          &= alpha - beta (p + qsqrt 3)
          end{align*}



          Hence we have $alpha = beta(p + qsqrt 3) + gamma$.



          Finally notice that
          begin{align*}
          N(gamma) &= N(beta cdot theta) \
          &= N(beta) cdot N(theta) \
          &= N(beta) cdot vert (r - p)^2 - 3 (s - q)^2 vert \
          &leq N(beta) cdot max{ (r - p)^2, 3(s - q)^2} \
          & leqfrac34 N(beta)\
          &< N(beta)
          end{align*}



          The key here was that $vert (r - p)^2 - 3 (s - q)^2 vert leq max{ (r - p)^2, 3(s - q)^2}$ since $(r - p)^2, 3(s - q)^2 geq 0$ and then we use that $(r - p)^2 leq 1/4$ and $3(s - q)^2 leq 3/4$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            When you're proving that $N(gamma) < N(beta)$, you say that $N(theta) = |(p-r)^2 - 3(q-s)^2|$. However, $N$ was only defined for elements in $Bbb{Z}[sqrt{3}]$. You showed that $(beta)(theta) in Bbb{Z}[sqrt{3}]$. But how does this imply that $theta in Bbb{Z}[sqrt{3}]$?
            $endgroup$
            – Artus
            Mar 12 '15 at 14:05








          • 2




            $begingroup$
            That's a very good question! We can just think of the norm $N$ on $mathbb Z[sqrt 3]$ as being the norm on $mathbb Q[sqrt 3]$, restricted to $mathbb Z[sqrt 3]$.
            $endgroup$
            – Robert Cardona
            Mar 12 '15 at 15:13










          • $begingroup$
            why is |p - r |< 1/2 and |q- s| < 1/2
            $endgroup$
            – user10024395
            Apr 23 '15 at 9:33










          • $begingroup$
            @user136266, because we define $p, q$ to be the closest integers to $r, s$, respectively. Say $p = frac14$, the closest integer is $0$ and the difference is less than $frac12$, if $p = frac34$, then the closest integer is $1$. If $p = frac12$, then pick either $0$ or $1$ and the claim holds (less than or equal to).
            $endgroup$
            – Robert Cardona
            Apr 23 '15 at 14:21










          • $begingroup$
            @RobertCardona how is 1/4 and 1/2 integers? these are not integers. i think you meant r but even so why must r be only these 3 possibilities?If i have a, c = 1 and b,d = 0. Then r = 0 and so u can't find an integer p such that p-r < 1/2.
            $endgroup$
            – user10024395
            Apr 24 '15 at 1:45
















          9












          9








          9





          $begingroup$

          Define the norm on $mathbb Z[sqrt 3]$ to be $N(a + b sqrt 3) = vert a^2 - 3 b^2 vert$.



          Let $alpha, beta in mathbb Z[sqrt 3]$ with $beta neq 0$.



          Say $alpha = a + b sqrt 3$ and $beta = c + d sqrt 3$.



          Notice that
          begin{align*}
          fracalphabeta &= frac{a + b sqrt 3}{c + d sqrt 3} cdot frac{c - d sqrt 3}{c - d sqrt 3} \
          &= frac{ac - 3bd}{c^2 - 3d^2} + frac{-ad + bc}{c^2 - 3d^2} sqrt 3 \
          &= r + ssqrt 3
          end{align*}



          where $r = displaystyle frac{ac - 3bd}{c^2 - 3d^2}$ and $s = displaystyle frac{-ad + bc}{c^2 - 3d^2}$.



          Let $p$ be the closest integer to $r$ and let $q$ be the closest integer to $s$. Notice that $vert r - p vert leq 1/2$ and $vert s - q vert leq 1/2$.



          We want to show that $alpha = (p + qsqrt 3) beta + gamma$ for some $gamma in mathbb Z[sqrt 3]$ such that either $gamma = 0$ or $N(gamma) < N(beta)$. (We'll show the latter holds always.)



          Define $theta := (r - p) + (s - q)sqrt 3$ and define $gamma = beta cdot theta in mathbb Z[sqrt 3]$ and observe that
          begin{align*}
          gamma &= beta cdot theta\
          &= beta ( (r - p) + (s - q)sqrt 3)\
          &= beta (r + ssqrt 3) - beta(p + qsqrt 3) \
          &= beta cdotfracalphabeta - beta (p + qsqrt 3) \
          &= alpha - beta (p + qsqrt 3)
          end{align*}



          Hence we have $alpha = beta(p + qsqrt 3) + gamma$.



          Finally notice that
          begin{align*}
          N(gamma) &= N(beta cdot theta) \
          &= N(beta) cdot N(theta) \
          &= N(beta) cdot vert (r - p)^2 - 3 (s - q)^2 vert \
          &leq N(beta) cdot max{ (r - p)^2, 3(s - q)^2} \
          & leqfrac34 N(beta)\
          &< N(beta)
          end{align*}



          The key here was that $vert (r - p)^2 - 3 (s - q)^2 vert leq max{ (r - p)^2, 3(s - q)^2}$ since $(r - p)^2, 3(s - q)^2 geq 0$ and then we use that $(r - p)^2 leq 1/4$ and $3(s - q)^2 leq 3/4$.






          share|cite|improve this answer











          $endgroup$



          Define the norm on $mathbb Z[sqrt 3]$ to be $N(a + b sqrt 3) = vert a^2 - 3 b^2 vert$.



          Let $alpha, beta in mathbb Z[sqrt 3]$ with $beta neq 0$.



          Say $alpha = a + b sqrt 3$ and $beta = c + d sqrt 3$.



          Notice that
          begin{align*}
          fracalphabeta &= frac{a + b sqrt 3}{c + d sqrt 3} cdot frac{c - d sqrt 3}{c - d sqrt 3} \
          &= frac{ac - 3bd}{c^2 - 3d^2} + frac{-ad + bc}{c^2 - 3d^2} sqrt 3 \
          &= r + ssqrt 3
          end{align*}



          where $r = displaystyle frac{ac - 3bd}{c^2 - 3d^2}$ and $s = displaystyle frac{-ad + bc}{c^2 - 3d^2}$.



          Let $p$ be the closest integer to $r$ and let $q$ be the closest integer to $s$. Notice that $vert r - p vert leq 1/2$ and $vert s - q vert leq 1/2$.



          We want to show that $alpha = (p + qsqrt 3) beta + gamma$ for some $gamma in mathbb Z[sqrt 3]$ such that either $gamma = 0$ or $N(gamma) < N(beta)$. (We'll show the latter holds always.)



          Define $theta := (r - p) + (s - q)sqrt 3$ and define $gamma = beta cdot theta in mathbb Z[sqrt 3]$ and observe that
          begin{align*}
          gamma &= beta cdot theta\
          &= beta ( (r - p) + (s - q)sqrt 3)\
          &= beta (r + ssqrt 3) - beta(p + qsqrt 3) \
          &= beta cdotfracalphabeta - beta (p + qsqrt 3) \
          &= alpha - beta (p + qsqrt 3)
          end{align*}



          Hence we have $alpha = beta(p + qsqrt 3) + gamma$.



          Finally notice that
          begin{align*}
          N(gamma) &= N(beta cdot theta) \
          &= N(beta) cdot N(theta) \
          &= N(beta) cdot vert (r - p)^2 - 3 (s - q)^2 vert \
          &leq N(beta) cdot max{ (r - p)^2, 3(s - q)^2} \
          & leqfrac34 N(beta)\
          &< N(beta)
          end{align*}



          The key here was that $vert (r - p)^2 - 3 (s - q)^2 vert leq max{ (r - p)^2, 3(s - q)^2}$ since $(r - p)^2, 3(s - q)^2 geq 0$ and then we use that $(r - p)^2 leq 1/4$ and $3(s - q)^2 leq 3/4$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 16 at 7:10









          user549397

          1,5061418




          1,5061418










          answered Feb 18 '15 at 4:43









          Robert CardonaRobert Cardona

          5,29023499




          5,29023499








          • 1




            $begingroup$
            When you're proving that $N(gamma) < N(beta)$, you say that $N(theta) = |(p-r)^2 - 3(q-s)^2|$. However, $N$ was only defined for elements in $Bbb{Z}[sqrt{3}]$. You showed that $(beta)(theta) in Bbb{Z}[sqrt{3}]$. But how does this imply that $theta in Bbb{Z}[sqrt{3}]$?
            $endgroup$
            – Artus
            Mar 12 '15 at 14:05








          • 2




            $begingroup$
            That's a very good question! We can just think of the norm $N$ on $mathbb Z[sqrt 3]$ as being the norm on $mathbb Q[sqrt 3]$, restricted to $mathbb Z[sqrt 3]$.
            $endgroup$
            – Robert Cardona
            Mar 12 '15 at 15:13










          • $begingroup$
            why is |p - r |< 1/2 and |q- s| < 1/2
            $endgroup$
            – user10024395
            Apr 23 '15 at 9:33










          • $begingroup$
            @user136266, because we define $p, q$ to be the closest integers to $r, s$, respectively. Say $p = frac14$, the closest integer is $0$ and the difference is less than $frac12$, if $p = frac34$, then the closest integer is $1$. If $p = frac12$, then pick either $0$ or $1$ and the claim holds (less than or equal to).
            $endgroup$
            – Robert Cardona
            Apr 23 '15 at 14:21










          • $begingroup$
            @RobertCardona how is 1/4 and 1/2 integers? these are not integers. i think you meant r but even so why must r be only these 3 possibilities?If i have a, c = 1 and b,d = 0. Then r = 0 and so u can't find an integer p such that p-r < 1/2.
            $endgroup$
            – user10024395
            Apr 24 '15 at 1:45
















          • 1




            $begingroup$
            When you're proving that $N(gamma) < N(beta)$, you say that $N(theta) = |(p-r)^2 - 3(q-s)^2|$. However, $N$ was only defined for elements in $Bbb{Z}[sqrt{3}]$. You showed that $(beta)(theta) in Bbb{Z}[sqrt{3}]$. But how does this imply that $theta in Bbb{Z}[sqrt{3}]$?
            $endgroup$
            – Artus
            Mar 12 '15 at 14:05








          • 2




            $begingroup$
            That's a very good question! We can just think of the norm $N$ on $mathbb Z[sqrt 3]$ as being the norm on $mathbb Q[sqrt 3]$, restricted to $mathbb Z[sqrt 3]$.
            $endgroup$
            – Robert Cardona
            Mar 12 '15 at 15:13










          • $begingroup$
            why is |p - r |< 1/2 and |q- s| < 1/2
            $endgroup$
            – user10024395
            Apr 23 '15 at 9:33










          • $begingroup$
            @user136266, because we define $p, q$ to be the closest integers to $r, s$, respectively. Say $p = frac14$, the closest integer is $0$ and the difference is less than $frac12$, if $p = frac34$, then the closest integer is $1$. If $p = frac12$, then pick either $0$ or $1$ and the claim holds (less than or equal to).
            $endgroup$
            – Robert Cardona
            Apr 23 '15 at 14:21










          • $begingroup$
            @RobertCardona how is 1/4 and 1/2 integers? these are not integers. i think you meant r but even so why must r be only these 3 possibilities?If i have a, c = 1 and b,d = 0. Then r = 0 and so u can't find an integer p such that p-r < 1/2.
            $endgroup$
            – user10024395
            Apr 24 '15 at 1:45










          1




          1




          $begingroup$
          When you're proving that $N(gamma) < N(beta)$, you say that $N(theta) = |(p-r)^2 - 3(q-s)^2|$. However, $N$ was only defined for elements in $Bbb{Z}[sqrt{3}]$. You showed that $(beta)(theta) in Bbb{Z}[sqrt{3}]$. But how does this imply that $theta in Bbb{Z}[sqrt{3}]$?
          $endgroup$
          – Artus
          Mar 12 '15 at 14:05






          $begingroup$
          When you're proving that $N(gamma) < N(beta)$, you say that $N(theta) = |(p-r)^2 - 3(q-s)^2|$. However, $N$ was only defined for elements in $Bbb{Z}[sqrt{3}]$. You showed that $(beta)(theta) in Bbb{Z}[sqrt{3}]$. But how does this imply that $theta in Bbb{Z}[sqrt{3}]$?
          $endgroup$
          – Artus
          Mar 12 '15 at 14:05






          2




          2




          $begingroup$
          That's a very good question! We can just think of the norm $N$ on $mathbb Z[sqrt 3]$ as being the norm on $mathbb Q[sqrt 3]$, restricted to $mathbb Z[sqrt 3]$.
          $endgroup$
          – Robert Cardona
          Mar 12 '15 at 15:13




          $begingroup$
          That's a very good question! We can just think of the norm $N$ on $mathbb Z[sqrt 3]$ as being the norm on $mathbb Q[sqrt 3]$, restricted to $mathbb Z[sqrt 3]$.
          $endgroup$
          – Robert Cardona
          Mar 12 '15 at 15:13












          $begingroup$
          why is |p - r |< 1/2 and |q- s| < 1/2
          $endgroup$
          – user10024395
          Apr 23 '15 at 9:33




          $begingroup$
          why is |p - r |< 1/2 and |q- s| < 1/2
          $endgroup$
          – user10024395
          Apr 23 '15 at 9:33












          $begingroup$
          @user136266, because we define $p, q$ to be the closest integers to $r, s$, respectively. Say $p = frac14$, the closest integer is $0$ and the difference is less than $frac12$, if $p = frac34$, then the closest integer is $1$. If $p = frac12$, then pick either $0$ or $1$ and the claim holds (less than or equal to).
          $endgroup$
          – Robert Cardona
          Apr 23 '15 at 14:21




          $begingroup$
          @user136266, because we define $p, q$ to be the closest integers to $r, s$, respectively. Say $p = frac14$, the closest integer is $0$ and the difference is less than $frac12$, if $p = frac34$, then the closest integer is $1$. If $p = frac12$, then pick either $0$ or $1$ and the claim holds (less than or equal to).
          $endgroup$
          – Robert Cardona
          Apr 23 '15 at 14:21












          $begingroup$
          @RobertCardona how is 1/4 and 1/2 integers? these are not integers. i think you meant r but even so why must r be only these 3 possibilities?If i have a, c = 1 and b,d = 0. Then r = 0 and so u can't find an integer p such that p-r < 1/2.
          $endgroup$
          – user10024395
          Apr 24 '15 at 1:45






          $begingroup$
          @RobertCardona how is 1/4 and 1/2 integers? these are not integers. i think you meant r but even so why must r be only these 3 possibilities?If i have a, c = 1 and b,d = 0. Then r = 0 and so u can't find an integer p such that p-r < 1/2.
          $endgroup$
          – user10024395
          Apr 24 '15 at 1:45




















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