Kernel of polynomial ring homomorphism

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Consider the ring homomorphism $phi:mathbbC[x,y]rightarrow mathbbC[z]$ sending $xto z^2$ and $yto z^3$. Show that $ker(phi)=(x^3-y^2)$



I have shown the inclusion that $(x^3-y^2)subseteq ker(phi)$. But am stuck with the other side. I have tried using the divison algorithm but am unable to work it out.










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    Consider the ring homomorphism $phi:mathbbC[x,y]rightarrow mathbbC[z]$ sending $xto z^2$ and $yto z^3$. Show that $ker(phi)=(x^3-y^2)$



    I have shown the inclusion that $(x^3-y^2)subseteq ker(phi)$. But am stuck with the other side. I have tried using the divison algorithm but am unable to work it out.










    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Consider the ring homomorphism $phi:mathbbC[x,y]rightarrow mathbbC[z]$ sending $xto z^2$ and $yto z^3$. Show that $ker(phi)=(x^3-y^2)$



      I have shown the inclusion that $(x^3-y^2)subseteq ker(phi)$. But am stuck with the other side. I have tried using the divison algorithm but am unable to work it out.










      share|cite|improve this question













      Consider the ring homomorphism $phi:mathbbC[x,y]rightarrow mathbbC[z]$ sending $xto z^2$ and $yto z^3$. Show that $ker(phi)=(x^3-y^2)$



      I have shown the inclusion that $(x^3-y^2)subseteq ker(phi)$. But am stuck with the other side. I have tried using the divison algorithm but am unable to work it out.







      abstract-algebra ring-theory






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      asked Sep 10 at 21:05









      user101

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          For every $f(x,y)$ we can write $f(x,y)= (x^3-y^2)g(x,y)+ x^2a(y)+xb(y)+c(y)$. If $f(x,y)inkerphi$, we have $z^4a(z^3)+z^2b(z^3)+c(z^3)=0$. Note that powers of terms in $c(z^3)$ are equal $0$ modulo $3$, powers of terms in $z^2b(z^3)$ are equal $2$ modulo $3$, and powers of terms in $z^4a(z^3)$ are equal $1$ modulo $3$. This implies that there are no cancelations in $z^4a(z^3)+z^2b(z^3)+c(z^3)$, so coefficients of all polynomials $a$, $b$ and $c$ are $0$, i.e. $a=b=c=0$. Thus $f(x,y)=(x^3-y^2)g(x,y)$ and $f(x,y)in langle x^3-y^2rangle$.






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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            2
            down vote



            accepted










            For every $f(x,y)$ we can write $f(x,y)= (x^3-y^2)g(x,y)+ x^2a(y)+xb(y)+c(y)$. If $f(x,y)inkerphi$, we have $z^4a(z^3)+z^2b(z^3)+c(z^3)=0$. Note that powers of terms in $c(z^3)$ are equal $0$ modulo $3$, powers of terms in $z^2b(z^3)$ are equal $2$ modulo $3$, and powers of terms in $z^4a(z^3)$ are equal $1$ modulo $3$. This implies that there are no cancelations in $z^4a(z^3)+z^2b(z^3)+c(z^3)$, so coefficients of all polynomials $a$, $b$ and $c$ are $0$, i.e. $a=b=c=0$. Thus $f(x,y)=(x^3-y^2)g(x,y)$ and $f(x,y)in langle x^3-y^2rangle$.






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              up vote
              2
              down vote



              accepted










              For every $f(x,y)$ we can write $f(x,y)= (x^3-y^2)g(x,y)+ x^2a(y)+xb(y)+c(y)$. If $f(x,y)inkerphi$, we have $z^4a(z^3)+z^2b(z^3)+c(z^3)=0$. Note that powers of terms in $c(z^3)$ are equal $0$ modulo $3$, powers of terms in $z^2b(z^3)$ are equal $2$ modulo $3$, and powers of terms in $z^4a(z^3)$ are equal $1$ modulo $3$. This implies that there are no cancelations in $z^4a(z^3)+z^2b(z^3)+c(z^3)$, so coefficients of all polynomials $a$, $b$ and $c$ are $0$, i.e. $a=b=c=0$. Thus $f(x,y)=(x^3-y^2)g(x,y)$ and $f(x,y)in langle x^3-y^2rangle$.






              share|cite|improve this answer






















                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                For every $f(x,y)$ we can write $f(x,y)= (x^3-y^2)g(x,y)+ x^2a(y)+xb(y)+c(y)$. If $f(x,y)inkerphi$, we have $z^4a(z^3)+z^2b(z^3)+c(z^3)=0$. Note that powers of terms in $c(z^3)$ are equal $0$ modulo $3$, powers of terms in $z^2b(z^3)$ are equal $2$ modulo $3$, and powers of terms in $z^4a(z^3)$ are equal $1$ modulo $3$. This implies that there are no cancelations in $z^4a(z^3)+z^2b(z^3)+c(z^3)$, so coefficients of all polynomials $a$, $b$ and $c$ are $0$, i.e. $a=b=c=0$. Thus $f(x,y)=(x^3-y^2)g(x,y)$ and $f(x,y)in langle x^3-y^2rangle$.






                share|cite|improve this answer












                For every $f(x,y)$ we can write $f(x,y)= (x^3-y^2)g(x,y)+ x^2a(y)+xb(y)+c(y)$. If $f(x,y)inkerphi$, we have $z^4a(z^3)+z^2b(z^3)+c(z^3)=0$. Note that powers of terms in $c(z^3)$ are equal $0$ modulo $3$, powers of terms in $z^2b(z^3)$ are equal $2$ modulo $3$, and powers of terms in $z^4a(z^3)$ are equal $1$ modulo $3$. This implies that there are no cancelations in $z^4a(z^3)+z^2b(z^3)+c(z^3)$, so coefficients of all polynomials $a$, $b$ and $c$ are $0$, i.e. $a=b=c=0$. Thus $f(x,y)=(x^3-y^2)g(x,y)$ and $f(x,y)in langle x^3-y^2rangle$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 10 at 21:24









                SMM

                2,03049




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