Is it correct to say the field of complex numbers is contained in the field of quaternions?












2












$begingroup$


I believe it is correct to refer to the complex numbers and their 'native algebra' as a field. See for example Linear Algebra and Matrix Theory, by Evar Nering. I assume the same may be said for the set of quaternions. Please correct me if that is wrong.



Based on that assumption, I ask: is the field of complex numbers a subset (sub-field?) of the quaternions in the same sense that the real numbers are 'contained' in the complex numbers?



Is there a term for such a subordination of number fields?



I admit that I am reaching well beyond my realm of familiarity, and that my question may be nonsensical to those who are more knowledgeable in these matters.






abstract-algebra complex-numbers definition quaternions







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$endgroup$












  • $begingroup$
    I'm guessing you actually meant to ask whether the complex field can be embedded in the (Hamilton) quaternions $;Bbb H;$ . Now, this last is a divison ring, or a non-commutative "integral domain" (very cautious with this as integral domains' definition usually carries commutativity as conditon), sometimes also named "as skew-field, whereas $;Bbb C;$ is commutative...so you need an embedding into the center of $;Bbb H;$ : what is it?
    $endgroup$
    – DonAntonio
    3 hours ago












  • $begingroup$
    Yes and no : $mathbb{C}$ is a field and $mathbb{H}$ is a skew-field, containing many copies of $mathbb{C}$ (but only one copy of $mathbb{R}$), $mathbb{H}$ is a non-commutative $mathbb{R}$-algebra with defining relations $j^2 = -1, zj = joverline{z}, z in mathbb{C}$ so it is not a $mathbb{C}$-algebra. In this sense it extends $mathbb{R}$, not $mathbb{C}$.
    $endgroup$
    – reuns
    3 hours ago


















2












$begingroup$


I believe it is correct to refer to the complex numbers and their 'native algebra' as a field. See for example Linear Algebra and Matrix Theory, by Evar Nering. I assume the same may be said for the set of quaternions. Please correct me if that is wrong.



Based on that assumption, I ask: is the field of complex numbers a subset (sub-field?) of the quaternions in the same sense that the real numbers are 'contained' in the complex numbers?



Is there a term for such a subordination of number fields?



I admit that I am reaching well beyond my realm of familiarity, and that my question may be nonsensical to those who are more knowledgeable in these matters.






abstract-algebra complex-numbers definition quaternions







share|cite|improve this question











$endgroup$












  • $begingroup$
    I'm guessing you actually meant to ask whether the complex field can be embedded in the (Hamilton) quaternions $;Bbb H;$ . Now, this last is a divison ring, or a non-commutative "integral domain" (very cautious with this as integral domains' definition usually carries commutativity as conditon), sometimes also named "as skew-field, whereas $;Bbb C;$ is commutative...so you need an embedding into the center of $;Bbb H;$ : what is it?
    $endgroup$
    – DonAntonio
    3 hours ago












  • $begingroup$
    Yes and no : $mathbb{C}$ is a field and $mathbb{H}$ is a skew-field, containing many copies of $mathbb{C}$ (but only one copy of $mathbb{R}$), $mathbb{H}$ is a non-commutative $mathbb{R}$-algebra with defining relations $j^2 = -1, zj = joverline{z}, z in mathbb{C}$ so it is not a $mathbb{C}$-algebra. In this sense it extends $mathbb{R}$, not $mathbb{C}$.
    $endgroup$
    – reuns
    3 hours ago
















2












2








2





$begingroup$


I believe it is correct to refer to the complex numbers and their 'native algebra' as a field. See for example Linear Algebra and Matrix Theory, by Evar Nering. I assume the same may be said for the set of quaternions. Please correct me if that is wrong.



Based on that assumption, I ask: is the field of complex numbers a subset (sub-field?) of the quaternions in the same sense that the real numbers are 'contained' in the complex numbers?



Is there a term for such a subordination of number fields?



I admit that I am reaching well beyond my realm of familiarity, and that my question may be nonsensical to those who are more knowledgeable in these matters.






abstract-algebra complex-numbers definition quaternions







share|cite|improve this question











$endgroup$




I believe it is correct to refer to the complex numbers and their 'native algebra' as a field. See for example Linear Algebra and Matrix Theory, by Evar Nering. I assume the same may be said for the set of quaternions. Please correct me if that is wrong.



Based on that assumption, I ask: is the field of complex numbers a subset (sub-field?) of the quaternions in the same sense that the real numbers are 'contained' in the complex numbers?



Is there a term for such a subordination of number fields?



I admit that I am reaching well beyond my realm of familiarity, and that my question may be nonsensical to those who are more knowledgeable in these matters.






abstract-algebra complex-numbers definition quaternions




abstract-algebra complex-numbers definition quaternions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago







Steven Hatton

















asked 3 hours ago









Steven HattonSteven Hatton

769315




769315












  • $begingroup$
    I'm guessing you actually meant to ask whether the complex field can be embedded in the (Hamilton) quaternions $;Bbb H;$ . Now, this last is a divison ring, or a non-commutative "integral domain" (very cautious with this as integral domains' definition usually carries commutativity as conditon), sometimes also named "as skew-field, whereas $;Bbb C;$ is commutative...so you need an embedding into the center of $;Bbb H;$ : what is it?
    $endgroup$
    – DonAntonio
    3 hours ago












  • $begingroup$
    Yes and no : $mathbb{C}$ is a field and $mathbb{H}$ is a skew-field, containing many copies of $mathbb{C}$ (but only one copy of $mathbb{R}$), $mathbb{H}$ is a non-commutative $mathbb{R}$-algebra with defining relations $j^2 = -1, zj = joverline{z}, z in mathbb{C}$ so it is not a $mathbb{C}$-algebra. In this sense it extends $mathbb{R}$, not $mathbb{C}$.
    $endgroup$
    – reuns
    3 hours ago




















  • $begingroup$
    I'm guessing you actually meant to ask whether the complex field can be embedded in the (Hamilton) quaternions $;Bbb H;$ . Now, this last is a divison ring, or a non-commutative "integral domain" (very cautious with this as integral domains' definition usually carries commutativity as conditon), sometimes also named "as skew-field, whereas $;Bbb C;$ is commutative...so you need an embedding into the center of $;Bbb H;$ : what is it?
    $endgroup$
    – DonAntonio
    3 hours ago












  • $begingroup$
    Yes and no : $mathbb{C}$ is a field and $mathbb{H}$ is a skew-field, containing many copies of $mathbb{C}$ (but only one copy of $mathbb{R}$), $mathbb{H}$ is a non-commutative $mathbb{R}$-algebra with defining relations $j^2 = -1, zj = joverline{z}, z in mathbb{C}$ so it is not a $mathbb{C}$-algebra. In this sense it extends $mathbb{R}$, not $mathbb{C}$.
    $endgroup$
    – reuns
    3 hours ago


















$begingroup$
I'm guessing you actually meant to ask whether the complex field can be embedded in the (Hamilton) quaternions $;Bbb H;$ . Now, this last is a divison ring, or a non-commutative "integral domain" (very cautious with this as integral domains' definition usually carries commutativity as conditon), sometimes also named "as skew-field, whereas $;Bbb C;$ is commutative...so you need an embedding into the center of $;Bbb H;$ : what is it?
$endgroup$
– DonAntonio
3 hours ago






$begingroup$
I'm guessing you actually meant to ask whether the complex field can be embedded in the (Hamilton) quaternions $;Bbb H;$ . Now, this last is a divison ring, or a non-commutative "integral domain" (very cautious with this as integral domains' definition usually carries commutativity as conditon), sometimes also named "as skew-field, whereas $;Bbb C;$ is commutative...so you need an embedding into the center of $;Bbb H;$ : what is it?
$endgroup$
– DonAntonio
3 hours ago














$begingroup$
Yes and no : $mathbb{C}$ is a field and $mathbb{H}$ is a skew-field, containing many copies of $mathbb{C}$ (but only one copy of $mathbb{R}$), $mathbb{H}$ is a non-commutative $mathbb{R}$-algebra with defining relations $j^2 = -1, zj = joverline{z}, z in mathbb{C}$ so it is not a $mathbb{C}$-algebra. In this sense it extends $mathbb{R}$, not $mathbb{C}$.
$endgroup$
– reuns
3 hours ago






$begingroup$
Yes and no : $mathbb{C}$ is a field and $mathbb{H}$ is a skew-field, containing many copies of $mathbb{C}$ (but only one copy of $mathbb{R}$), $mathbb{H}$ is a non-commutative $mathbb{R}$-algebra with defining relations $j^2 = -1, zj = joverline{z}, z in mathbb{C}$ so it is not a $mathbb{C}$-algebra. In this sense it extends $mathbb{R}$, not $mathbb{C}$.
$endgroup$
– reuns
3 hours ago












3 Answers
3






active

oldest

votes


















3












$begingroup$

The field of complex numbers is isomorphic to a subfield of the division ring of the quaternions. Also, the field of complex numbers is isomorphic to a subfield of the complex field.



Whether or not we have subsets here depends on the way the complex numbers and the quaternions are defined.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    The quaternion skew field ${Bbb H} = {Bbb C}times {Bbb C}$ has ${Bbb C}$ as a
    subfield by considering the ring monomorphism ${Bbb C}rightarrow {Bbb H}: zmapsto (z,0)$.






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      The ring of quaternions(it is not a field) contains infinitely many subrings that are fields and are isomorphic to the field of complex numbers.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        The field of complex numbers is isomorphic to a subfield of the division ring of the quaternions. Also, the field of complex numbers is isomorphic to a subfield of the complex field.



        Whether or not we have subsets here depends on the way the complex numbers and the quaternions are defined.






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          The field of complex numbers is isomorphic to a subfield of the division ring of the quaternions. Also, the field of complex numbers is isomorphic to a subfield of the complex field.



          Whether or not we have subsets here depends on the way the complex numbers and the quaternions are defined.






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            The field of complex numbers is isomorphic to a subfield of the division ring of the quaternions. Also, the field of complex numbers is isomorphic to a subfield of the complex field.



            Whether or not we have subsets here depends on the way the complex numbers and the quaternions are defined.






            share|cite|improve this answer









            $endgroup$



            The field of complex numbers is isomorphic to a subfield of the division ring of the quaternions. Also, the field of complex numbers is isomorphic to a subfield of the complex field.



            Whether or not we have subsets here depends on the way the complex numbers and the quaternions are defined.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 3 hours ago









            José Carlos SantosJosé Carlos Santos

            154k22123226




            154k22123226























                3












                $begingroup$

                The quaternion skew field ${Bbb H} = {Bbb C}times {Bbb C}$ has ${Bbb C}$ as a
                subfield by considering the ring monomorphism ${Bbb C}rightarrow {Bbb H}: zmapsto (z,0)$.






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  The quaternion skew field ${Bbb H} = {Bbb C}times {Bbb C}$ has ${Bbb C}$ as a
                  subfield by considering the ring monomorphism ${Bbb C}rightarrow {Bbb H}: zmapsto (z,0)$.






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    The quaternion skew field ${Bbb H} = {Bbb C}times {Bbb C}$ has ${Bbb C}$ as a
                    subfield by considering the ring monomorphism ${Bbb C}rightarrow {Bbb H}: zmapsto (z,0)$.






                    share|cite|improve this answer









                    $endgroup$



                    The quaternion skew field ${Bbb H} = {Bbb C}times {Bbb C}$ has ${Bbb C}$ as a
                    subfield by considering the ring monomorphism ${Bbb C}rightarrow {Bbb H}: zmapsto (z,0)$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 hours ago









                    WuestenfuxWuestenfux

                    4,0501411




                    4,0501411























                        2












                        $begingroup$

                        The ring of quaternions(it is not a field) contains infinitely many subrings that are fields and are isomorphic to the field of complex numbers.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          The ring of quaternions(it is not a field) contains infinitely many subrings that are fields and are isomorphic to the field of complex numbers.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            The ring of quaternions(it is not a field) contains infinitely many subrings that are fields and are isomorphic to the field of complex numbers.






                            share|cite|improve this answer









                            $endgroup$



                            The ring of quaternions(it is not a field) contains infinitely many subrings that are fields and are isomorphic to the field of complex numbers.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 3 hours ago









                            Emilio NovatiEmilio Novati

                            51.7k43473




                            51.7k43473






























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