0

Let $r=r(x,y) \in \mathbb{C}[x,y]$ and define $e_{\alpha,\beta} : \mathbb{C}[x,y] \to \mathbb{C}$ by $e_{\alpha,\beta}(r(x,y)):=r(\alpha,\beta)$.

If I am not wrong, $e_{\alpha,\beta}$, the evaluation map, is a ring homomorphism (it is known to be a ring homomorphism for one variable; I guess that it is still a ring homomorphism for more than one variable?).

The kernel of $e_{\alpha,\beta}$ is the maximal ideal $(x-\alpha,y-\beta)$.

Now let $p=p(x,y),q=q(x,y) \in \mathbb{C}[x,y]$, with $n=\deg(f) \geq 2$ and $m=\deg(g) \geq 2$.

Can we find a commutative ring $R$ and a ring homomorphism $e: \mathbb{C}[x,y] \to R$ such that $(p,q)$ is the kernel of $e$?

(The special case where $R$ is an integral domain and $e$ is an epimorphism, in other words, $(p,q)$ is a prime ideal, was dealt with here).

Thank you very much!

user237522
  • 6,071
  • 3
  • 10
  • 21

1 Answers1

2

Yes, this is exactly what quotient rings are for. Very generally, if $A$ is any ring and $I$ is any (two-sided) ideal in $A$, then the quotient map $A\to A/I$ is a ring homomorphism whose kernel is $I$.

Eric Wofsey
  • 295,450
  • 24
  • 356
  • 562