What means $ℝ[i]$?












0












$begingroup$


From what i‘ve understood, a set, say $ℤ[i]$, denotes the set of all integers, in addition to i. ($i^2 = -1$) So for example, the complex number $1+i$ would be within that set.
In short, my question is, would then $ℝ[i]$ be the equivalent set as $ℂ$?






complex-numbers







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  • 3




    $begingroup$
    I see no reason it shouldn't be.
    $endgroup$
    – Arthur
    Dec 13 '18 at 21:33






  • 1




    $begingroup$
    yes, it is and it's exactly how you build $mathbb{C}$
    $endgroup$
    – qbert
    Dec 13 '18 at 21:34
















0












$begingroup$


From what i‘ve understood, a set, say $ℤ[i]$, denotes the set of all integers, in addition to i. ($i^2 = -1$) So for example, the complex number $1+i$ would be within that set.
In short, my question is, would then $ℝ[i]$ be the equivalent set as $ℂ$?






complex-numbers







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    I see no reason it shouldn't be.
    $endgroup$
    – Arthur
    Dec 13 '18 at 21:33






  • 1




    $begingroup$
    yes, it is and it's exactly how you build $mathbb{C}$
    $endgroup$
    – qbert
    Dec 13 '18 at 21:34














0












0








0





$begingroup$


From what i‘ve understood, a set, say $ℤ[i]$, denotes the set of all integers, in addition to i. ($i^2 = -1$) So for example, the complex number $1+i$ would be within that set.
In short, my question is, would then $ℝ[i]$ be the equivalent set as $ℂ$?






complex-numbers







share|cite|improve this question











$endgroup$




From what i‘ve understood, a set, say $ℤ[i]$, denotes the set of all integers, in addition to i. ($i^2 = -1$) So for example, the complex number $1+i$ would be within that set.
In short, my question is, would then $ℝ[i]$ be the equivalent set as $ℂ$?






complex-numbers




complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 21:49









Asaf Karagila♦

304k32434764




304k32434764










asked Dec 13 '18 at 21:32









Nils Phillip TalgöNils Phillip Talgö

559




559








  • 3




    $begingroup$
    I see no reason it shouldn't be.
    $endgroup$
    – Arthur
    Dec 13 '18 at 21:33






  • 1




    $begingroup$
    yes, it is and it's exactly how you build $mathbb{C}$
    $endgroup$
    – qbert
    Dec 13 '18 at 21:34














  • 3




    $begingroup$
    I see no reason it shouldn't be.
    $endgroup$
    – Arthur
    Dec 13 '18 at 21:33






  • 1




    $begingroup$
    yes, it is and it's exactly how you build $mathbb{C}$
    $endgroup$
    – qbert
    Dec 13 '18 at 21:34








3




3




$begingroup$
I see no reason it shouldn't be.
$endgroup$
– Arthur
Dec 13 '18 at 21:33




$begingroup$
I see no reason it shouldn't be.
$endgroup$
– Arthur
Dec 13 '18 at 21:33




1




1




$begingroup$
yes, it is and it's exactly how you build $mathbb{C}$
$endgroup$
– qbert
Dec 13 '18 at 21:34




$begingroup$
yes, it is and it's exactly how you build $mathbb{C}$
$endgroup$
– qbert
Dec 13 '18 at 21:34










1 Answer
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$begingroup$

In general, $S=R[a]$ is a ring extension
of $R$ by an element $a$ in a ring containing $R$, i.e., $S$ is the smallest ring which contains $R$ and $a$. Notice that all elements of $R[a]$ are necessarily in all rings containing $R$ and $a$, and it is itself a ring. Now apply this for $R=Bbb{Z}$ or $R=Bbb{R}$ in the ring $Bbb{C}$ and for $a=i$. This answers your question "what means $Bbb{R}[i]$?". Finally, $Bbb{R}[i]cong Bbb{C}$ as rings (and as fields).






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    $begingroup$

    In general, $S=R[a]$ is a ring extension
    of $R$ by an element $a$ in a ring containing $R$, i.e., $S$ is the smallest ring which contains $R$ and $a$. Notice that all elements of $R[a]$ are necessarily in all rings containing $R$ and $a$, and it is itself a ring. Now apply this for $R=Bbb{Z}$ or $R=Bbb{R}$ in the ring $Bbb{C}$ and for $a=i$. This answers your question "what means $Bbb{R}[i]$?". Finally, $Bbb{R}[i]cong Bbb{C}$ as rings (and as fields).






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      In general, $S=R[a]$ is a ring extension
      of $R$ by an element $a$ in a ring containing $R$, i.e., $S$ is the smallest ring which contains $R$ and $a$. Notice that all elements of $R[a]$ are necessarily in all rings containing $R$ and $a$, and it is itself a ring. Now apply this for $R=Bbb{Z}$ or $R=Bbb{R}$ in the ring $Bbb{C}$ and for $a=i$. This answers your question "what means $Bbb{R}[i]$?". Finally, $Bbb{R}[i]cong Bbb{C}$ as rings (and as fields).






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        In general, $S=R[a]$ is a ring extension
        of $R$ by an element $a$ in a ring containing $R$, i.e., $S$ is the smallest ring which contains $R$ and $a$. Notice that all elements of $R[a]$ are necessarily in all rings containing $R$ and $a$, and it is itself a ring. Now apply this for $R=Bbb{Z}$ or $R=Bbb{R}$ in the ring $Bbb{C}$ and for $a=i$. This answers your question "what means $Bbb{R}[i]$?". Finally, $Bbb{R}[i]cong Bbb{C}$ as rings (and as fields).






        share|cite|improve this answer











        $endgroup$



        In general, $S=R[a]$ is a ring extension
        of $R$ by an element $a$ in a ring containing $R$, i.e., $S$ is the smallest ring which contains $R$ and $a$. Notice that all elements of $R[a]$ are necessarily in all rings containing $R$ and $a$, and it is itself a ring. Now apply this for $R=Bbb{Z}$ or $R=Bbb{R}$ in the ring $Bbb{C}$ and for $a=i$. This answers your question "what means $Bbb{R}[i]$?". Finally, $Bbb{R}[i]cong Bbb{C}$ as rings (and as fields).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 13 '18 at 21:48

























        answered Dec 13 '18 at 21:43









        Dietrich BurdeDietrich Burde

        79.2k647103




        79.2k647103






























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