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.
abstract-algebra ring-theory
asked Sep 10 at 21:05
user101
566213
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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.
abstract-algebra ring-theory
asked Sep 10 at 21:05
user101
566213
add a comment |Â
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.
abstract-algebra ring-theory
asked Sep 10 at 21:05
user101
566213
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
abstract-algebra ring-theory
asked Sep 10 at 21:05
user101
566213
asked Sep 10 at 21:05
user101
566213
asked Sep 10 at 21:05
user101
566213
asked Sep 10 at 21:05
user101
566213
asked Sep 10 at 21:05
user101
566213
566213
add a comment |Â
add a comment |Â
1 Answer
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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 $f(x,y)in langle x^3-y^2rangle$.
answered Sep 10 at 21:24
SMM
2,03049
add a comment |Â
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 $f(x,y)in langle x^3-y^2rangle$.
answered Sep 10 at 21:24
SMM
2,03049
add a comment |Â
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 $f(x,y)in langle x^3-y^2rangle$.
answered Sep 10 at 21:24
SMM
2,03049
add a comment |Â
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 $f(x,y)in langle x^3-y^2rangle$.
answered Sep 10 at 21:24
SMM
2,03049
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 $f(x,y)in langle x^3-y^2rangle$.
answered Sep 10 at 21:24
SMM
2,03049
answered Sep 10 at 21:24
SMM
2,03049
answered Sep 10 at 21:24
SMM
2,03049
answered Sep 10 at 21:24
SMM
2,03049
2,03049
add a comment |Â
add a comment |Â
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