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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.

Rick Sanchez C-666
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1 Answers1

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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 $f\in I$.

Mark
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  • Cool trick, but i didn't understand the part: $f(x,y)+I=f(2,3)+I.$ – Rick Sanchez C-666 Aug 25 '22 at 00:40
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    Write explicitly $f(x,y)=\sum\limits_{i,j} a_{ij}x^iy^j$. As we have shown, in the quotient ring $x$ turns into $2$ and $y$ turns into $3$, so $f(x,y)+I=\sum\limits_{i,j}a_{ij}2^i3^j+I=f(2,3)+I$. – Mark Aug 25 '22 at 00:42
  • Thank you sir, that's a really nice trick that i'm gonna use from now on! – Rick Sanchez C-666 Aug 25 '22 at 00:44
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    @rick more generally in $R[y]$ we have $,(y!-!a,f(y)) = (y!-!a,f(a)),$ by $,f(y)\equiv f(a)\pmod{!y-a},,$ proved here. So $,(y!-!3,x!-!y!+!1) = (y!-!3,x!-!2).\ \ $ – Bill Dubuque Aug 25 '22 at 03:37