The quaternion division ring contains an infinite number of elements $ u $ satisfying $ u^2=-1 $












2












$begingroup$



Show that the quaternion division ring contains an infinite number of elements $ u $ satisfying $ u^2=-1 $




I was trying to solve the above exercise on Page 133, Basic Algebra, Jacobson. Maybe it is convenient to consider $$ mathbb H=left{ left[begin{matrix} alpha & beta\-bar{beta} & bar{alpha} end{matrix} right]: alpha, betainmathbb C right}. $$



Of course,
$$ i= begin{bmatrix} sqrt{-1} & 0\ 0 & -sqrt{-1} end{bmatrix},quad j=begin{bmatrix} 0 & 1 \ -1 & 0 end{bmatrix},quad k=begin{bmatrix}
0 & sqrt{-1}\ sqrt{-1} & 0 end{bmatrix} $$
satisfy $ u^2=-1 $. But how to show that there are infinitely many elements in $ mathbb H $ satisfying the identity?






abstract-algebra ring-theory quaternions







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    $begingroup$
    "Maybe it is convenient....". It isn't.
    $endgroup$
    – Lord Shark the Unknown
    Jan 13 at 8:49
















2












$begingroup$



Show that the quaternion division ring contains an infinite number of elements $ u $ satisfying $ u^2=-1 $




I was trying to solve the above exercise on Page 133, Basic Algebra, Jacobson. Maybe it is convenient to consider $$ mathbb H=left{ left[begin{matrix} alpha & beta\-bar{beta} & bar{alpha} end{matrix} right]: alpha, betainmathbb C right}. $$



Of course,
$$ i= begin{bmatrix} sqrt{-1} & 0\ 0 & -sqrt{-1} end{bmatrix},quad j=begin{bmatrix} 0 & 1 \ -1 & 0 end{bmatrix},quad k=begin{bmatrix}
0 & sqrt{-1}\ sqrt{-1} & 0 end{bmatrix} $$
satisfy $ u^2=-1 $. But how to show that there are infinitely many elements in $ mathbb H $ satisfying the identity?






abstract-algebra ring-theory quaternions







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    "Maybe it is convenient....". It isn't.
    $endgroup$
    – Lord Shark the Unknown
    Jan 13 at 8:49














2












2








2





$begingroup$



Show that the quaternion division ring contains an infinite number of elements $ u $ satisfying $ u^2=-1 $




I was trying to solve the above exercise on Page 133, Basic Algebra, Jacobson. Maybe it is convenient to consider $$ mathbb H=left{ left[begin{matrix} alpha & beta\-bar{beta} & bar{alpha} end{matrix} right]: alpha, betainmathbb C right}. $$



Of course,
$$ i= begin{bmatrix} sqrt{-1} & 0\ 0 & -sqrt{-1} end{bmatrix},quad j=begin{bmatrix} 0 & 1 \ -1 & 0 end{bmatrix},quad k=begin{bmatrix}
0 & sqrt{-1}\ sqrt{-1} & 0 end{bmatrix} $$
satisfy $ u^2=-1 $. But how to show that there are infinitely many elements in $ mathbb H $ satisfying the identity?






abstract-algebra ring-theory quaternions







share|cite|improve this question









$endgroup$





Show that the quaternion division ring contains an infinite number of elements $ u $ satisfying $ u^2=-1 $




I was trying to solve the above exercise on Page 133, Basic Algebra, Jacobson. Maybe it is convenient to consider $$ mathbb H=left{ left[begin{matrix} alpha & beta\-bar{beta} & bar{alpha} end{matrix} right]: alpha, betainmathbb C right}. $$



Of course,
$$ i= begin{bmatrix} sqrt{-1} & 0\ 0 & -sqrt{-1} end{bmatrix},quad j=begin{bmatrix} 0 & 1 \ -1 & 0 end{bmatrix},quad k=begin{bmatrix}
0 & sqrt{-1}\ sqrt{-1} & 0 end{bmatrix} $$
satisfy $ u^2=-1 $. But how to show that there are infinitely many elements in $ mathbb H $ satisfying the identity?






abstract-algebra ring-theory quaternions




abstract-algebra ring-theory quaternions






share|cite|improve this question













share|cite|improve this question











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asked Jan 13 at 8:39









user549397user549397

1,5101418




1,5101418








  • 1




    $begingroup$
    "Maybe it is convenient....". It isn't.
    $endgroup$
    – Lord Shark the Unknown
    Jan 13 at 8:49














  • 1




    $begingroup$
    "Maybe it is convenient....". It isn't.
    $endgroup$
    – Lord Shark the Unknown
    Jan 13 at 8:49








1




1




$begingroup$
"Maybe it is convenient....". It isn't.
$endgroup$
– Lord Shark the Unknown
Jan 13 at 8:49




$begingroup$
"Maybe it is convenient....". It isn't.
$endgroup$
– Lord Shark the Unknown
Jan 13 at 8:49










2 Answers
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$begingroup$

The proof of this statement can be found on Wikipedia. Let the square root of $-1$ be $x=a+bi+cj+dk$, then $x^2=-1$ is equivalent to
$$a^2-b^2-c^2-d^2=-1qquad2ab=2ac=2ad=0$$
If $ane0$, $b=c=d=0$ and the first equation becomes $a^2=-1$, impossible since $ainmathbb R$. Hence $a=0$, at which point the second set of equations is automatically satisfied and the first equation becomes $b^2+c^2+d^2=1$. The quaternionic square roots of $-1$ thus form the unit sphere in $mathbb R^3$, so there are infinitely many solutions.






share|cite|improve this answer











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    $begingroup$

    Hint: Forget about the matrices for a moment. What is $(ai+bj)^2$?






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      2 Answers
      2






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      oldest

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      2 Answers
      2






      active

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      votes









      active

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      active

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      1












      $begingroup$

      The proof of this statement can be found on Wikipedia. Let the square root of $-1$ be $x=a+bi+cj+dk$, then $x^2=-1$ is equivalent to
      $$a^2-b^2-c^2-d^2=-1qquad2ab=2ac=2ad=0$$
      If $ane0$, $b=c=d=0$ and the first equation becomes $a^2=-1$, impossible since $ainmathbb R$. Hence $a=0$, at which point the second set of equations is automatically satisfied and the first equation becomes $b^2+c^2+d^2=1$. The quaternionic square roots of $-1$ thus form the unit sphere in $mathbb R^3$, so there are infinitely many solutions.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        The proof of this statement can be found on Wikipedia. Let the square root of $-1$ be $x=a+bi+cj+dk$, then $x^2=-1$ is equivalent to
        $$a^2-b^2-c^2-d^2=-1qquad2ab=2ac=2ad=0$$
        If $ane0$, $b=c=d=0$ and the first equation becomes $a^2=-1$, impossible since $ainmathbb R$. Hence $a=0$, at which point the second set of equations is automatically satisfied and the first equation becomes $b^2+c^2+d^2=1$. The quaternionic square roots of $-1$ thus form the unit sphere in $mathbb R^3$, so there are infinitely many solutions.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          The proof of this statement can be found on Wikipedia. Let the square root of $-1$ be $x=a+bi+cj+dk$, then $x^2=-1$ is equivalent to
          $$a^2-b^2-c^2-d^2=-1qquad2ab=2ac=2ad=0$$
          If $ane0$, $b=c=d=0$ and the first equation becomes $a^2=-1$, impossible since $ainmathbb R$. Hence $a=0$, at which point the second set of equations is automatically satisfied and the first equation becomes $b^2+c^2+d^2=1$. The quaternionic square roots of $-1$ thus form the unit sphere in $mathbb R^3$, so there are infinitely many solutions.






          share|cite|improve this answer











          $endgroup$



          The proof of this statement can be found on Wikipedia. Let the square root of $-1$ be $x=a+bi+cj+dk$, then $x^2=-1$ is equivalent to
          $$a^2-b^2-c^2-d^2=-1qquad2ab=2ac=2ad=0$$
          If $ane0$, $b=c=d=0$ and the first equation becomes $a^2=-1$, impossible since $ainmathbb R$. Hence $a=0$, at which point the second set of equations is automatically satisfied and the first equation becomes $b^2+c^2+d^2=1$. The quaternionic square roots of $-1$ thus form the unit sphere in $mathbb R^3$, so there are infinitely many solutions.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 13 at 8:59

























          answered Jan 13 at 8:49









          Parcly TaxelParcly Taxel

          41.8k1372101




          41.8k1372101























              4












              $begingroup$

              Hint: Forget about the matrices for a moment. What is $(ai+bj)^2$?






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                Hint: Forget about the matrices for a moment. What is $(ai+bj)^2$?






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  Hint: Forget about the matrices for a moment. What is $(ai+bj)^2$?






                  share|cite|improve this answer









                  $endgroup$



                  Hint: Forget about the matrices for a moment. What is $(ai+bj)^2$?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 13 at 8:45









                  ArthurArthur

                  115k7116198




                  115k7116198






























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