Quotient Space of Space of PolynomialsDoes a linear quotient map have sectionsPolynomial maps on indeterminate of vector space of polynomialsDetermine the quotient space $P$/$U_e$Are $P_>3$ in the vector space of all polynomials with coefficients?What are polynomials of degree less than $N$ in $x$?Doubts when calculating this space?Determine if a given subset is a subspace of a given vector spaceHow to find matrix representation for quotient map?Determinant matrix projective space?Homeomorphism between roots and coefficients of monic real polynomials

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Quotient Space of Space of Polynomials


Does a linear quotient map have sectionsPolynomial maps on indeterminate of vector space of polynomialsDetermine the quotient space $P$/$U_e$Are $P_>3$ in the vector space of all polynomials with coefficients?What are polynomials of degree less than $N$ in $x$?Doubts when calculating this space?Determine if a given subset is a subspace of a given vector spaceHow to find matrix representation for quotient map?Determinant matrix projective space?Homeomorphism between roots and coefficients of monic real polynomials













2












$begingroup$


Here's the question: Let $V = mathbbR[x]$ be the space of polynomials with real coefficients, and W the space of polynomials divisible by $x^2+1.$



Then the quotient space $V/W$ can be identified with the set $mathbbC$ of complex number, and the projection $Pcolon mathbbR[x]rightarrowmathbbC$ with the map $p(x)mapsto p(i)$ of evaluating a polynomial $p(x)$ at $x = i$.



I understand that two polynomials $p(x)$ and $p'(x)$ are equivalent modulo $W$ if and only if $p(x)-p'(x)$ is divisible by $x^2+1$. This means $p(x)$ and $p'(x)$ are equivalent if and only if $p(i)=p'(i)$ or $p(-i)=p'(-i)$. But why is the conclusion in bold font is true? Thanks!










share|cite|improve this question











$endgroup$
















    2












    $begingroup$


    Here's the question: Let $V = mathbbR[x]$ be the space of polynomials with real coefficients, and W the space of polynomials divisible by $x^2+1.$



    Then the quotient space $V/W$ can be identified with the set $mathbbC$ of complex number, and the projection $Pcolon mathbbR[x]rightarrowmathbbC$ with the map $p(x)mapsto p(i)$ of evaluating a polynomial $p(x)$ at $x = i$.



    I understand that two polynomials $p(x)$ and $p'(x)$ are equivalent modulo $W$ if and only if $p(x)-p'(x)$ is divisible by $x^2+1$. This means $p(x)$ and $p'(x)$ are equivalent if and only if $p(i)=p'(i)$ or $p(-i)=p'(-i)$. But why is the conclusion in bold font is true? Thanks!










    share|cite|improve this question











    $endgroup$














      2












      2








      2





      $begingroup$


      Here's the question: Let $V = mathbbR[x]$ be the space of polynomials with real coefficients, and W the space of polynomials divisible by $x^2+1.$



      Then the quotient space $V/W$ can be identified with the set $mathbbC$ of complex number, and the projection $Pcolon mathbbR[x]rightarrowmathbbC$ with the map $p(x)mapsto p(i)$ of evaluating a polynomial $p(x)$ at $x = i$.



      I understand that two polynomials $p(x)$ and $p'(x)$ are equivalent modulo $W$ if and only if $p(x)-p'(x)$ is divisible by $x^2+1$. This means $p(x)$ and $p'(x)$ are equivalent if and only if $p(i)=p'(i)$ or $p(-i)=p'(-i)$. But why is the conclusion in bold font is true? Thanks!










      share|cite|improve this question











      $endgroup$




      Here's the question: Let $V = mathbbR[x]$ be the space of polynomials with real coefficients, and W the space of polynomials divisible by $x^2+1.$



      Then the quotient space $V/W$ can be identified with the set $mathbbC$ of complex number, and the projection $Pcolon mathbbR[x]rightarrowmathbbC$ with the map $p(x)mapsto p(i)$ of evaluating a polynomial $p(x)$ at $x = i$.



      I understand that two polynomials $p(x)$ and $p'(x)$ are equivalent modulo $W$ if and only if $p(x)-p'(x)$ is divisible by $x^2+1$. This means $p(x)$ and $p'(x)$ are equivalent if and only if $p(i)=p'(i)$ or $p(-i)=p'(-i)$. But why is the conclusion in bold font is true? Thanks!







      linear-algebra






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 4 '16 at 9:53









      p Groups

      6,2631129




      6,2631129










      asked Jan 4 '16 at 0:23









      Huy TruongHuy Truong

      1177




      1177




















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          First, look at the space $mathbb R[x]/langle 1+x^2rangle$ which is the set of equivalence classes of polynomials with real coefficients, modulo $1+x^2$. For example $4+3x+2x^2+x^3 = (2+x)(1+x^2)+(2+2x)$:
          $$4+3x+2x^2+x^3 equiv 2+2x bmod 1+x^2$$
          For any polynomial $mathrm p(x)$ in $mathbb R[x]$, we divide by $1+x^2$ and identify it with its remainder. For each $mathrm p(x)$ in $mathbb R[x]$ we get a linear polynomial of the form $a+bx$ as the remainder, where $a,b in mathbb R$. We can use the notation $[a+bx]$ for the set of all polynomials which leave remainder $a+bx$ when divided by $1+x^2$. For example $4+3x+2x^2+x^3$ belongs to the equivalence class $[2+2x]$.



          The space $mathbb R[x]/langle 1+x^2rangle$ is given by all of these equivalence classes $[a+bx]$. Given a polynomial $mathrmp(x)$ in the set $[a+bx]$, we know that its members look like $mathrmq(x) cdot (1+x^2)+a+bx$ for some polynomial $mathrmq(x)$ in $mathbb R[x]$.



          We can see that $mathbb R[x]/langle 1+x^2rangle$ behaves like $mathbb C$. For example $(1+2mathrm i)(3+4mathrm i) =-5+10mathrmi$, while



          $$begineqnarray* \
          [1+2x]cdot [3+4x] &equiv& (1+2x)(3+4x) bmod 1+x^2 \ \
          &equiv& 3+10x+8x^2 bmod 1+x^2 \ \
          &equiv& 8(1+x^2) - 5 +10x bmod 1+x^2 \ \
          &equiv& -5+10x bmod 1+x^2
          endeqnarray*$$



          On the one hand, in $mathbb C$, we have $(1+2mathrm i)(3+4mathrm i) =-5+10mathrmi$. On the other, in $mathbb R[x]/langle 1+x^2rangle$, we have $[1+2x]cdot [3+4x] = [-5+10x]$. It seems that $[a+bx] mapsto a+mathrm i b$ could be a good candidate for an isomorphism $mathbb R[x]/langle 1+x^2rangle to mathbb C$. Let's go back and check our first example:
          $$4+3x+2x^2+x^3 in [2+2x]$$
          $$4 + 3(mathrmi) + 2(mathrmi)^2 + (mathrmi)^3 = 2+2mathrmi$$
          This makes perfect sense. All of the members of $[a+bx]$ look like
          $$mathrm q(x)cdot(1+x^2)+a+bx$$
          for some polynomial $mathrmq(x)$. Applying the rule $x mapsto mathrm i$ gives
          $$mathrm q(x)cdot(1+x^2)+a+bx longmapsto mathrm q(mathrm i)cdot(1+mathrm i^2)+a+bmathrm i = mathrmq(mathrm i) cdot 0 + a+bmathrm i = a+bmathrm i$$
          This show that the rule $mathrmp(x) mapsto mathrmp(mathrm i)$ takes all of the elements of $[a+bx]$ to the complex number $a+mathrmib$. It's not hard to check that this rule is in fact an isomorphism.






          share|cite|improve this answer











          $endgroup$




















            1












            $begingroup$

            One way to think of it is that (the ideal of) $x^2 + 1=0$ in the quotient space. So $x^2=-1$ in the quotient space. So $x$ acts as a square root of $-1$ in the quotient space.






            share|cite|improve this answer









            $endgroup$




















              1












              $begingroup$

              Let define the following ring homomorphism: $$varphi:left{beginarraycccmathbbR[X]&rightarrow&mathbbR[i]\P&mapsto&P(i)endarrayright..$$
              Using the euclidean division in $mathbbR[X]$, one has $textrmim(varphi)=mathbbR+imathbbR$ and $textrmker(varphi)=left(X^2+1right)$, therefore, one gets: $$mathbbR+imathbbRcongmathbbR[X]/left(X^2+1right).$$
              Hence the result.






              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












                $begingroup$

                First, look at the space $mathbb R[x]/langle 1+x^2rangle$ which is the set of equivalence classes of polynomials with real coefficients, modulo $1+x^2$. For example $4+3x+2x^2+x^3 = (2+x)(1+x^2)+(2+2x)$:
                $$4+3x+2x^2+x^3 equiv 2+2x bmod 1+x^2$$
                For any polynomial $mathrm p(x)$ in $mathbb R[x]$, we divide by $1+x^2$ and identify it with its remainder. For each $mathrm p(x)$ in $mathbb R[x]$ we get a linear polynomial of the form $a+bx$ as the remainder, where $a,b in mathbb R$. We can use the notation $[a+bx]$ for the set of all polynomials which leave remainder $a+bx$ when divided by $1+x^2$. For example $4+3x+2x^2+x^3$ belongs to the equivalence class $[2+2x]$.



                The space $mathbb R[x]/langle 1+x^2rangle$ is given by all of these equivalence classes $[a+bx]$. Given a polynomial $mathrmp(x)$ in the set $[a+bx]$, we know that its members look like $mathrmq(x) cdot (1+x^2)+a+bx$ for some polynomial $mathrmq(x)$ in $mathbb R[x]$.



                We can see that $mathbb R[x]/langle 1+x^2rangle$ behaves like $mathbb C$. For example $(1+2mathrm i)(3+4mathrm i) =-5+10mathrmi$, while



                $$begineqnarray* \
                [1+2x]cdot [3+4x] &equiv& (1+2x)(3+4x) bmod 1+x^2 \ \
                &equiv& 3+10x+8x^2 bmod 1+x^2 \ \
                &equiv& 8(1+x^2) - 5 +10x bmod 1+x^2 \ \
                &equiv& -5+10x bmod 1+x^2
                endeqnarray*$$



                On the one hand, in $mathbb C$, we have $(1+2mathrm i)(3+4mathrm i) =-5+10mathrmi$. On the other, in $mathbb R[x]/langle 1+x^2rangle$, we have $[1+2x]cdot [3+4x] = [-5+10x]$. It seems that $[a+bx] mapsto a+mathrm i b$ could be a good candidate for an isomorphism $mathbb R[x]/langle 1+x^2rangle to mathbb C$. Let's go back and check our first example:
                $$4+3x+2x^2+x^3 in [2+2x]$$
                $$4 + 3(mathrmi) + 2(mathrmi)^2 + (mathrmi)^3 = 2+2mathrmi$$
                This makes perfect sense. All of the members of $[a+bx]$ look like
                $$mathrm q(x)cdot(1+x^2)+a+bx$$
                for some polynomial $mathrmq(x)$. Applying the rule $x mapsto mathrm i$ gives
                $$mathrm q(x)cdot(1+x^2)+a+bx longmapsto mathrm q(mathrm i)cdot(1+mathrm i^2)+a+bmathrm i = mathrmq(mathrm i) cdot 0 + a+bmathrm i = a+bmathrm i$$
                This show that the rule $mathrmp(x) mapsto mathrmp(mathrm i)$ takes all of the elements of $[a+bx]$ to the complex number $a+mathrmib$. It's not hard to check that this rule is in fact an isomorphism.






                share|cite|improve this answer











                $endgroup$

















                  3












                  $begingroup$

                  First, look at the space $mathbb R[x]/langle 1+x^2rangle$ which is the set of equivalence classes of polynomials with real coefficients, modulo $1+x^2$. For example $4+3x+2x^2+x^3 = (2+x)(1+x^2)+(2+2x)$:
                  $$4+3x+2x^2+x^3 equiv 2+2x bmod 1+x^2$$
                  For any polynomial $mathrm p(x)$ in $mathbb R[x]$, we divide by $1+x^2$ and identify it with its remainder. For each $mathrm p(x)$ in $mathbb R[x]$ we get a linear polynomial of the form $a+bx$ as the remainder, where $a,b in mathbb R$. We can use the notation $[a+bx]$ for the set of all polynomials which leave remainder $a+bx$ when divided by $1+x^2$. For example $4+3x+2x^2+x^3$ belongs to the equivalence class $[2+2x]$.



                  The space $mathbb R[x]/langle 1+x^2rangle$ is given by all of these equivalence classes $[a+bx]$. Given a polynomial $mathrmp(x)$ in the set $[a+bx]$, we know that its members look like $mathrmq(x) cdot (1+x^2)+a+bx$ for some polynomial $mathrmq(x)$ in $mathbb R[x]$.



                  We can see that $mathbb R[x]/langle 1+x^2rangle$ behaves like $mathbb C$. For example $(1+2mathrm i)(3+4mathrm i) =-5+10mathrmi$, while



                  $$begineqnarray* \
                  [1+2x]cdot [3+4x] &equiv& (1+2x)(3+4x) bmod 1+x^2 \ \
                  &equiv& 3+10x+8x^2 bmod 1+x^2 \ \
                  &equiv& 8(1+x^2) - 5 +10x bmod 1+x^2 \ \
                  &equiv& -5+10x bmod 1+x^2
                  endeqnarray*$$



                  On the one hand, in $mathbb C$, we have $(1+2mathrm i)(3+4mathrm i) =-5+10mathrmi$. On the other, in $mathbb R[x]/langle 1+x^2rangle$, we have $[1+2x]cdot [3+4x] = [-5+10x]$. It seems that $[a+bx] mapsto a+mathrm i b$ could be a good candidate for an isomorphism $mathbb R[x]/langle 1+x^2rangle to mathbb C$. Let's go back and check our first example:
                  $$4+3x+2x^2+x^3 in [2+2x]$$
                  $$4 + 3(mathrmi) + 2(mathrmi)^2 + (mathrmi)^3 = 2+2mathrmi$$
                  This makes perfect sense. All of the members of $[a+bx]$ look like
                  $$mathrm q(x)cdot(1+x^2)+a+bx$$
                  for some polynomial $mathrmq(x)$. Applying the rule $x mapsto mathrm i$ gives
                  $$mathrm q(x)cdot(1+x^2)+a+bx longmapsto mathrm q(mathrm i)cdot(1+mathrm i^2)+a+bmathrm i = mathrmq(mathrm i) cdot 0 + a+bmathrm i = a+bmathrm i$$
                  This show that the rule $mathrmp(x) mapsto mathrmp(mathrm i)$ takes all of the elements of $[a+bx]$ to the complex number $a+mathrmib$. It's not hard to check that this rule is in fact an isomorphism.






                  share|cite|improve this answer











                  $endgroup$















                    3












                    3








                    3





                    $begingroup$

                    First, look at the space $mathbb R[x]/langle 1+x^2rangle$ which is the set of equivalence classes of polynomials with real coefficients, modulo $1+x^2$. For example $4+3x+2x^2+x^3 = (2+x)(1+x^2)+(2+2x)$:
                    $$4+3x+2x^2+x^3 equiv 2+2x bmod 1+x^2$$
                    For any polynomial $mathrm p(x)$ in $mathbb R[x]$, we divide by $1+x^2$ and identify it with its remainder. For each $mathrm p(x)$ in $mathbb R[x]$ we get a linear polynomial of the form $a+bx$ as the remainder, where $a,b in mathbb R$. We can use the notation $[a+bx]$ for the set of all polynomials which leave remainder $a+bx$ when divided by $1+x^2$. For example $4+3x+2x^2+x^3$ belongs to the equivalence class $[2+2x]$.



                    The space $mathbb R[x]/langle 1+x^2rangle$ is given by all of these equivalence classes $[a+bx]$. Given a polynomial $mathrmp(x)$ in the set $[a+bx]$, we know that its members look like $mathrmq(x) cdot (1+x^2)+a+bx$ for some polynomial $mathrmq(x)$ in $mathbb R[x]$.



                    We can see that $mathbb R[x]/langle 1+x^2rangle$ behaves like $mathbb C$. For example $(1+2mathrm i)(3+4mathrm i) =-5+10mathrmi$, while



                    $$begineqnarray* \
                    [1+2x]cdot [3+4x] &equiv& (1+2x)(3+4x) bmod 1+x^2 \ \
                    &equiv& 3+10x+8x^2 bmod 1+x^2 \ \
                    &equiv& 8(1+x^2) - 5 +10x bmod 1+x^2 \ \
                    &equiv& -5+10x bmod 1+x^2
                    endeqnarray*$$



                    On the one hand, in $mathbb C$, we have $(1+2mathrm i)(3+4mathrm i) =-5+10mathrmi$. On the other, in $mathbb R[x]/langle 1+x^2rangle$, we have $[1+2x]cdot [3+4x] = [-5+10x]$. It seems that $[a+bx] mapsto a+mathrm i b$ could be a good candidate for an isomorphism $mathbb R[x]/langle 1+x^2rangle to mathbb C$. Let's go back and check our first example:
                    $$4+3x+2x^2+x^3 in [2+2x]$$
                    $$4 + 3(mathrmi) + 2(mathrmi)^2 + (mathrmi)^3 = 2+2mathrmi$$
                    This makes perfect sense. All of the members of $[a+bx]$ look like
                    $$mathrm q(x)cdot(1+x^2)+a+bx$$
                    for some polynomial $mathrmq(x)$. Applying the rule $x mapsto mathrm i$ gives
                    $$mathrm q(x)cdot(1+x^2)+a+bx longmapsto mathrm q(mathrm i)cdot(1+mathrm i^2)+a+bmathrm i = mathrmq(mathrm i) cdot 0 + a+bmathrm i = a+bmathrm i$$
                    This show that the rule $mathrmp(x) mapsto mathrmp(mathrm i)$ takes all of the elements of $[a+bx]$ to the complex number $a+mathrmib$. It's not hard to check that this rule is in fact an isomorphism.






                    share|cite|improve this answer











                    $endgroup$



                    First, look at the space $mathbb R[x]/langle 1+x^2rangle$ which is the set of equivalence classes of polynomials with real coefficients, modulo $1+x^2$. For example $4+3x+2x^2+x^3 = (2+x)(1+x^2)+(2+2x)$:
                    $$4+3x+2x^2+x^3 equiv 2+2x bmod 1+x^2$$
                    For any polynomial $mathrm p(x)$ in $mathbb R[x]$, we divide by $1+x^2$ and identify it with its remainder. For each $mathrm p(x)$ in $mathbb R[x]$ we get a linear polynomial of the form $a+bx$ as the remainder, where $a,b in mathbb R$. We can use the notation $[a+bx]$ for the set of all polynomials which leave remainder $a+bx$ when divided by $1+x^2$. For example $4+3x+2x^2+x^3$ belongs to the equivalence class $[2+2x]$.



                    The space $mathbb R[x]/langle 1+x^2rangle$ is given by all of these equivalence classes $[a+bx]$. Given a polynomial $mathrmp(x)$ in the set $[a+bx]$, we know that its members look like $mathrmq(x) cdot (1+x^2)+a+bx$ for some polynomial $mathrmq(x)$ in $mathbb R[x]$.



                    We can see that $mathbb R[x]/langle 1+x^2rangle$ behaves like $mathbb C$. For example $(1+2mathrm i)(3+4mathrm i) =-5+10mathrmi$, while



                    $$begineqnarray* \
                    [1+2x]cdot [3+4x] &equiv& (1+2x)(3+4x) bmod 1+x^2 \ \
                    &equiv& 3+10x+8x^2 bmod 1+x^2 \ \
                    &equiv& 8(1+x^2) - 5 +10x bmod 1+x^2 \ \
                    &equiv& -5+10x bmod 1+x^2
                    endeqnarray*$$



                    On the one hand, in $mathbb C$, we have $(1+2mathrm i)(3+4mathrm i) =-5+10mathrmi$. On the other, in $mathbb R[x]/langle 1+x^2rangle$, we have $[1+2x]cdot [3+4x] = [-5+10x]$. It seems that $[a+bx] mapsto a+mathrm i b$ could be a good candidate for an isomorphism $mathbb R[x]/langle 1+x^2rangle to mathbb C$. Let's go back and check our first example:
                    $$4+3x+2x^2+x^3 in [2+2x]$$
                    $$4 + 3(mathrmi) + 2(mathrmi)^2 + (mathrmi)^3 = 2+2mathrmi$$
                    This makes perfect sense. All of the members of $[a+bx]$ look like
                    $$mathrm q(x)cdot(1+x^2)+a+bx$$
                    for some polynomial $mathrmq(x)$. Applying the rule $x mapsto mathrm i$ gives
                    $$mathrm q(x)cdot(1+x^2)+a+bx longmapsto mathrm q(mathrm i)cdot(1+mathrm i^2)+a+bmathrm i = mathrmq(mathrm i) cdot 0 + a+bmathrm i = a+bmathrm i$$
                    This show that the rule $mathrmp(x) mapsto mathrmp(mathrm i)$ takes all of the elements of $[a+bx]$ to the complex number $a+mathrmib$. It's not hard to check that this rule is in fact an isomorphism.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Mar 20 at 20:08

























                    answered Jan 4 '16 at 1:31









                    Fly by NightFly by Night

                    26.1k32978




                    26.1k32978





















                        1












                        $begingroup$

                        One way to think of it is that (the ideal of) $x^2 + 1=0$ in the quotient space. So $x^2=-1$ in the quotient space. So $x$ acts as a square root of $-1$ in the quotient space.






                        share|cite|improve this answer









                        $endgroup$

















                          1












                          $begingroup$

                          One way to think of it is that (the ideal of) $x^2 + 1=0$ in the quotient space. So $x^2=-1$ in the quotient space. So $x$ acts as a square root of $-1$ in the quotient space.






                          share|cite|improve this answer









                          $endgroup$















                            1












                            1








                            1





                            $begingroup$

                            One way to think of it is that (the ideal of) $x^2 + 1=0$ in the quotient space. So $x^2=-1$ in the quotient space. So $x$ acts as a square root of $-1$ in the quotient space.






                            share|cite|improve this answer









                            $endgroup$



                            One way to think of it is that (the ideal of) $x^2 + 1=0$ in the quotient space. So $x^2=-1$ in the quotient space. So $x$ acts as a square root of $-1$ in the quotient space.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 4 '16 at 0:36









                            paw88789paw88789

                            29.6k12351




                            29.6k12351





















                                1












                                $begingroup$

                                Let define the following ring homomorphism: $$varphi:left{beginarraycccmathbbR[X]&rightarrow&mathbbR[i]\P&mapsto&P(i)endarrayright..$$
                                Using the euclidean division in $mathbbR[X]$, one has $textrmim(varphi)=mathbbR+imathbbR$ and $textrmker(varphi)=left(X^2+1right)$, therefore, one gets: $$mathbbR+imathbbRcongmathbbR[X]/left(X^2+1right).$$
                                Hence the result.






                                share|cite|improve this answer









                                $endgroup$

















                                  1












                                  $begingroup$

                                  Let define the following ring homomorphism: $$varphi:left{beginarraycccmathbbR[X]&rightarrow&mathbbR[i]\P&mapsto&P(i)endarrayright..$$
                                  Using the euclidean division in $mathbbR[X]$, one has $textrmim(varphi)=mathbbR+imathbbR$ and $textrmker(varphi)=left(X^2+1right)$, therefore, one gets: $$mathbbR+imathbbRcongmathbbR[X]/left(X^2+1right).$$
                                  Hence the result.






                                  share|cite|improve this answer









                                  $endgroup$















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                                    Let define the following ring homomorphism: $$varphi:left{beginarraycccmathbbR[X]&rightarrow&mathbbR[i]\P&mapsto&P(i)endarrayright..$$
                                    Using the euclidean division in $mathbbR[X]$, one has $textrmim(varphi)=mathbbR+imathbbR$ and $textrmker(varphi)=left(X^2+1right)$, therefore, one gets: $$mathbbR+imathbbRcongmathbbR[X]/left(X^2+1right).$$
                                    Hence the result.






                                    share|cite|improve this answer









                                    $endgroup$



                                    Let define the following ring homomorphism: $$varphi:left{beginarraycccmathbbR[X]&rightarrow&mathbbR[i]\P&mapsto&P(i)endarrayright..$$
                                    Using the euclidean division in $mathbbR[X]$, one has $textrmim(varphi)=mathbbR+imathbbR$ and $textrmker(varphi)=left(X^2+1right)$, therefore, one gets: $$mathbbR+imathbbRcongmathbbR[X]/left(X^2+1right).$$
                                    Hence the result.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Jan 4 '16 at 0:38









                                    C. FalconC. Falcon

                                    15.3k41951




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