Show that there is only one endomorphism of $mathbbC[X,Y]/(X^2Y + XY^2 - XY)$ that is an automorphism on $mathbbC$
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Let $A = mathbbC[X,Y]/(X^2Y + XY^2 - XY).$ Denote the images of $X,Y$ in $A$ as $barX,barY$ respectively. Let $phi colon A to A$ be a ring homomorphism such that
- $phi$ is an automorphism of $mathbbC$;
- $phi(X) = barY$;
- $phi(Y) = 1 - barX - barY.$
The question asks me to show that there is only one possible choice for $phi.$ I'm not sure how I'm supposed to approach this.
Certainly, $phi$ is a well-defined homomorphism as it respects the quotients, i.e. $phi(X^2Y+XY^2-XY) = 0.$ Also, if we consider $phi(X+Y+i^2),$ we have that $$phi(X+Y+i^2) = phi(X+Y-1) = -barX$$ and $$phi(X+Y+i^2) = phi(X) + phi(Y) + phi(i)^2 = 1-barX+phi(i)^2.$$ So $phi(i)^2 = -1,$ i.e. $phi$ is either the identity or conjugation on $mathbbC$. However, how can I show that conjugation does not give us a ring homomorphism?
I'm not sure if it helps in this question, but the previous part asked me to list the prime ideals of $A,$ which I hope are $(X) ,(Y), (X+Y-1), (X,Y),$ and $(X-a,Y+a-1)$ for any $ainmathbbC.$
Any help will be much appreciated!
abstract-algebra ring-theory commutative-algebra
asked Aug 18 at 8:34
dhk628
1,031513
 |Â
show 1 more comment
up vote
0
down vote
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Let $A = mathbbC[X,Y]/(X^2Y + XY^2 - XY).$ Denote the images of $X,Y$ in $A$ as $barX,barY$ respectively. Let $phi colon A to A$ be a ring homomorphism such that
- $phi$ is an automorphism of $mathbbC$;
- $phi(X) = barY$;
- $phi(Y) = 1 - barX - barY.$
The question asks me to show that there is only one possible choice for $phi.$ I'm not sure how I'm supposed to approach this.
Certainly, $phi$ is a well-defined homomorphism as it respects the quotients, i.e. $phi(X^2Y+XY^2-XY) = 0.$ Also, if we consider $phi(X+Y+i^2),$ we have that $$phi(X+Y+i^2) = phi(X+Y-1) = -barX$$ and $$phi(X+Y+i^2) = phi(X) + phi(Y) + phi(i)^2 = 1-barX+phi(i)^2.$$ So $phi(i)^2 = -1,$ i.e. $phi$ is either the identity or conjugation on $mathbbC$. However, how can I show that conjugation does not give us a ring homomorphism?
I'm not sure if it helps in this question, but the previous part asked me to list the prime ideals of $A,$ which I hope are $(X) ,(Y), (X+Y-1), (X,Y),$ and $(X-a,Y+a-1)$ for any $ainmathbbC.$
Any help will be much appreciated!
abstract-algebra ring-theory commutative-algebra
asked Aug 18 at 8:34
dhk628
1,031513
You only proved that $phi(i)^2= -1$, not that $phi$ is the identity or the conjugation on $mathbbC$ : what about the real numbers ?
– Max
Aug 18 at 8:40
@Max right, forgot about those.. that makes matters a lot worse for me..
– dhk628
Aug 18 at 9:05
This corresponds to 3 lines cut out by the equation and indicates you should use CRT to look at lines first.
– user45765
Aug 18 at 12:12
1
Are you sure that this is all you have in the problem statement? The map $barvarphi:Ato A$ sending $zin mathbbC$ to $barz$, $barXmapsto barY$, and $barYmapsto 1-barX-barY$ does extend to a ring endomorphism of $A$. In fact, if you let $alpha$ be any field automorphism of $mathbbC$ that fixes $mathbbQ$ (and there are uncountably many of such $alpha$), then $alpha$ can be extended to a ring endomorphism of $A$ with $alpha(barX)=barY$ and $alpha(barY)=1-barX-barY$. Thus, your $varphi$ is certainly not unique with the required properties.
– Batominovski
Aug 18 at 20:31
On the other hand, if the problem statement says that $varphi$ is the identity on $mathbbC$, then yes, $varphi$ is unique, almost trivially.
– Batominovski
Aug 18 at 20:39
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $A = mathbbC[X,Y]/(X^2Y + XY^2 - XY).$ Denote the images of $X,Y$ in $A$ as $barX,barY$ respectively. Let $phi colon A to A$ be a ring homomorphism such that
- $phi$ is an automorphism of $mathbbC$;
- $phi(X) = barY$;
- $phi(Y) = 1 - barX - barY.$
The question asks me to show that there is only one possible choice for $phi.$ I'm not sure how I'm supposed to approach this.
Certainly, $phi$ is a well-defined homomorphism as it respects the quotients, i.e. $phi(X^2Y+XY^2-XY) = 0.$ Also, if we consider $phi(X+Y+i^2),$ we have that $$phi(X+Y+i^2) = phi(X+Y-1) = -barX$$ and $$phi(X+Y+i^2) = phi(X) + phi(Y) + phi(i)^2 = 1-barX+phi(i)^2.$$ So $phi(i)^2 = -1,$ i.e. $phi$ is either the identity or conjugation on $mathbbC$. However, how can I show that conjugation does not give us a ring homomorphism?
I'm not sure if it helps in this question, but the previous part asked me to list the prime ideals of $A,$ which I hope are $(X) ,(Y), (X+Y-1), (X,Y),$ and $(X-a,Y+a-1)$ for any $ainmathbbC.$
Any help will be much appreciated!
abstract-algebra ring-theory commutative-algebra
asked Aug 18 at 8:34
dhk628
1,031513
Let $A = mathbbC[X,Y]/(X^2Y + XY^2 - XY).$ Denote the images of $X,Y$ in $A$ as $barX,barY$ respectively. Let $phi colon A to A$ be a ring homomorphism such that
- $phi$ is an automorphism of $mathbbC$;
- $phi(X) = barY$;
- $phi(Y) = 1 - barX - barY.$
The question asks me to show that there is only one possible choice for $phi.$ I'm not sure how I'm supposed to approach this.
Certainly, $phi$ is a well-defined homomorphism as it respects the quotients, i.e. $phi(X^2Y+XY^2-XY) = 0.$ Also, if we consider $phi(X+Y+i^2),$ we have that $$phi(X+Y+i^2) = phi(X+Y-1) = -barX$$ and $$phi(X+Y+i^2) = phi(X) + phi(Y) + phi(i)^2 = 1-barX+phi(i)^2.$$ So $phi(i)^2 = -1,$ i.e. $phi$ is either the identity or conjugation on $mathbbC$. However, how can I show that conjugation does not give us a ring homomorphism?
I'm not sure if it helps in this question, but the previous part asked me to list the prime ideals of $A,$ which I hope are $(X) ,(Y), (X+Y-1), (X,Y),$ and $(X-a,Y+a-1)$ for any $ainmathbbC.$
Any help will be much appreciated!
abstract-algebra ring-theory commutative-algebra
asked Aug 18 at 8:34
dhk628
1,031513
asked Aug 18 at 8:34
dhk628
1,031513
asked Aug 18 at 8:34
dhk628
1,031513
asked Aug 18 at 8:34
dhk628
1,031513
1,031513
You only proved that $phi(i)^2= -1$, not that $phi$ is the identity or the conjugation on $mathbbC$ : what about the real numbers ?
– Max
Aug 18 at 8:40
@Max right, forgot about those.. that makes matters a lot worse for me..
– dhk628
Aug 18 at 9:05
This corresponds to 3 lines cut out by the equation and indicates you should use CRT to look at lines first.
– user45765
Aug 18 at 12:12
1
Are you sure that this is all you have in the problem statement? The map $barvarphi:Ato A$ sending $zin mathbbC$ to $barz$, $barXmapsto barY$, and $barYmapsto 1-barX-barY$ does extend to a ring endomorphism of $A$. In fact, if you let $alpha$ be any field automorphism of $mathbbC$ that fixes $mathbbQ$ (and there are uncountably many of such $alpha$), then $alpha$ can be extended to a ring endomorphism of $A$ with $alpha(barX)=barY$ and $alpha(barY)=1-barX-barY$. Thus, your $varphi$ is certainly not unique with the required properties.
– Batominovski
Aug 18 at 20:31
On the other hand, if the problem statement says that $varphi$ is the identity on $mathbbC$, then yes, $varphi$ is unique, almost trivially.
– Batominovski
Aug 18 at 20:39
 |Â
show 1 more comment
You only proved that $phi(i)^2= -1$, not that $phi$ is the identity or the conjugation on $mathbbC$ : what about the real numbers ?
– Max
Aug 18 at 8:40
@Max right, forgot about those.. that makes matters a lot worse for me..
– dhk628
Aug 18 at 9:05
This corresponds to 3 lines cut out by the equation and indicates you should use CRT to look at lines first.
– user45765
Aug 18 at 12:12
1
Are you sure that this is all you have in the problem statement? The map $barvarphi:Ato A$ sending $zin mathbbC$ to $barz$, $barXmapsto barY$, and $barYmapsto 1-barX-barY$ does extend to a ring endomorphism of $A$. In fact, if you let $alpha$ be any field automorphism of $mathbbC$ that fixes $mathbbQ$ (and there are uncountably many of such $alpha$), then $alpha$ can be extended to a ring endomorphism of $A$ with $alpha(barX)=barY$ and $alpha(barY)=1-barX-barY$. Thus, your $varphi$ is certainly not unique with the required properties.
– Batominovski
Aug 18 at 20:31
On the other hand, if the problem statement says that $varphi$ is the identity on $mathbbC$, then yes, $varphi$ is unique, almost trivially.
– Batominovski
Aug 18 at 20:39
You only proved that $phi(i)^2= -1$, not that $phi$ is the identity or the conjugation on $mathbbC$ : what about the real numbers ?
– Max
Aug 18 at 8:40
You only proved that $phi(i)^2= -1$, not that $phi$ is the identity or the conjugation on $mathbbC$ : what about the real numbers ?
– Max
Aug 18 at 8:40
@Max right, forgot about those.. that makes matters a lot worse for me..
– dhk628
Aug 18 at 9:05
@Max right, forgot about those.. that makes matters a lot worse for me..
– dhk628
Aug 18 at 9:05
This corresponds to 3 lines cut out by the equation and indicates you should use CRT to look at lines first.
– user45765
Aug 18 at 12:12
This corresponds to 3 lines cut out by the equation and indicates you should use CRT to look at lines first.
– user45765
Aug 18 at 12:12
1
1
Are you sure that this is all you have in the problem statement? The map $barvarphi:Ato A$ sending $zin mathbbC$ to $barz$, $barXmapsto barY$, and $barYmapsto 1-barX-barY$ does extend to a ring endomorphism of $A$. In fact, if you let $alpha$ be any field automorphism of $mathbbC$ that fixes $mathbbQ$ (and there are uncountably many of such $alpha$), then $alpha$ can be extended to a ring endomorphism of $A$ with $alpha(barX)=barY$ and $alpha(barY)=1-barX-barY$. Thus, your $varphi$ is certainly not unique with the required properties.
– Batominovski
Aug 18 at 20:31
Are you sure that this is all you have in the problem statement? The map $barvarphi:Ato A$ sending $zin mathbbC$ to $barz$, $barXmapsto barY$, and $barYmapsto 1-barX-barY$ does extend to a ring endomorphism of $A$. In fact, if you let $alpha$ be any field automorphism of $mathbbC$ that fixes $mathbbQ$ (and there are uncountably many of such $alpha$), then $alpha$ can be extended to a ring endomorphism of $A$ with $alpha(barX)=barY$ and $alpha(barY)=1-barX-barY$. Thus, your $varphi$ is certainly not unique with the required properties.
– Batominovski
Aug 18 at 20:31
On the other hand, if the problem statement says that $varphi$ is the identity on $mathbbC$, then yes, $varphi$ is unique, almost trivially.
– Batominovski
Aug 18 at 20:39
On the other hand, if the problem statement says that $varphi$ is the identity on $mathbbC$, then yes, $varphi$ is unique, almost trivially.
– Batominovski
Aug 18 at 20:39
 |Â
show 1 more comment
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You only proved that $phi(i)^2= -1$, not that $phi$ is the identity or the conjugation on $mathbbC$ : what about the real numbers ?
– Max
Aug 18 at 8:40
@Max right, forgot about those.. that makes matters a lot worse for me..
– dhk628
Aug 18 at 9:05
This corresponds to 3 lines cut out by the equation and indicates you should use CRT to look at lines first.
– user45765
Aug 18 at 12:12
1
Are you sure that this is all you have in the problem statement? The map $barvarphi:Ato A$ sending $zin mathbbC$ to $barz$, $barXmapsto barY$, and $barYmapsto 1-barX-barY$ does extend to a ring endomorphism of $A$. In fact, if you let $alpha$ be any field automorphism of $mathbbC$ that fixes $mathbbQ$ (and there are uncountably many of such $alpha$), then $alpha$ can be extended to a ring endomorphism of $A$ with $alpha(barX)=barY$ and $alpha(barY)=1-barX-barY$. Thus, your $varphi$ is certainly not unique with the required properties.
– Batominovski
Aug 18 at 20:31
On the other hand, if the problem statement says that $varphi$ is the identity on $mathbbC$, then yes, $varphi$ is unique, almost trivially.
– Batominovski
Aug 18 at 20:39