Let $A = \mathbb{C}[X,Y]/(X^2Y + XY^2 - XY).$ Denote the images of $X,Y$ in $A$ as $\bar{X},\bar{Y}$ respectively. Let $\phi \colon A \to A$ be a ring homomorphism such that
- $\phi$ is an automorphism of $\mathbb{C}$;
- $\phi(X) = \bar{Y}$;
- $\phi(Y) = 1 - \bar{X} - \bar{Y}.$
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 $$\phi(X+Y+i^2) = \phi(X+Y-1) = -\bar{X}$$ and $$\phi(X+Y+i^2) = \phi(X) + \phi(Y) + \phi(i)^2 = 1-\bar{X}+\phi(i)^2.$$ So $\phi(i)^2 = -1,$ i.e. $\phi$ is either the identity or conjugation on $\mathbb{C}$. 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 $a\in\mathbb{C}.$
Any help will be much appreciated!