I'm trying to show that the ideal $I =\langle x-y+1,y-3\rangle$ of the ring $\mathbb C[x,y]$ is maximal.
My approach so far:
We define $\phi:\mathbb C[x,y]\to \mathbb C,$ with mapping
$$f(x,y)\mapsto f(2,3).$$
We can easily prove that $\phi$ is a surjective homomorphism $\big($i.e $f(x,y)=x-2+y-3+p(x,y)$, hence, $\phi(f)=p(x,y)$ $\big)$.
Its enough to show that ker($\phi$)=$I.$
$$f\in\langle x-y+1,y-3\rangle:$$
$$f(x,y)=(x-y+1)g(x,y)+(y-3)h(x,y)\Rightarrow f(2,3)=0\Rightarrow f\in\text{ker}(\phi)\Rightarrow$$
$$\Rightarrow \langle x-y+1,y-3\rangle\subseteq \text{ker}(\phi).$$
But if $f\in\text{ker}(\phi):$ how can i show that $f(x,y)=(x-y+1)g(x,y)+(y-3)h(x,y)?$
In order to claim that $\mathbb C[x,y]/\operatorname{ker}(\phi)\cong\operatorname{im}(\phi)=\mathbb C$ ($\to$ field $\iff I $ is maximal)
I'm a bit stuck.
Best Answer
Here is a nice trick which I really like. We can look what happens in the quotient $\mathbb{C}[x,y]/I$. We have $y-3\in I$, and so $y+I=3+I$. Also:
$x+I=x-(x-y+1)+I=(y-1)+I=(3-1)+I=2+I$
From here it easily follows that for each $f\in\mathbb{C}[x,y]$ we have $f(x,y)+I=f(2,3)+I$. In particular, if $f\in \text{ker}(\phi)$ then $f+I=f(2,3)+I=I$, and so $f\in I$.