2

I am trying to prove that $\langle x-a, y-b, z-c\rangle$ is a maximal ideal in $K[x,y,z]$, where $K$ is a field.

My guess:

Define the homorphism from $K[x,y,z]\rightarrow K$ as $$f(x,y,z)\mapsto f(a,b,c)$$

This is a surjective homomorphism and the image is a field. So if we can show that the kernel is the ideal $\langle x-a, y-b, z-c\rangle$, then we are done. It is clear that $$\langle x-a, y-b, z-c\rangle\subseteq\text{Kernel}$$

How to prove the other inclusion ?

glS
  • 7,963
random_guy
  • 453

1 Answers1

1

For any $f\in K[x,y,z]$, by long division, we may write $f(x,y,z)=p(x,y,z)(x-a)+g(y,z)$ where $p$ and $g$ are polynomials. And then $g(y,z)=q(y,z)(y-b) + h(z)$, and finally $h(z)=r(z)(z-c)+k$ where $k$ is a constant. In total, this gives $$ f(x,y,z)=p(x,y,z)(x-a)+q(y,z)(y-b)+r(z)(z-c)+k $$ If $f$ is in the kernel of the evaluation homomorphism, we have $f(a,b,c)=0$, which forces $k=0$. (In fact, in general, the image of $f$ under the evaluation homomorphism is $k$.)

Arthur
  • 204,511