Is the ideal $(x-y,x+y)$ same as $(x,y)$, since $$x+y, x-y \in \mathbb{C},$$ so $ y \in \mathbb{C} $ because $\mathbb{C}$ is a field. And similarly $x $ in $\mathbb{C}$, so $ (x,y)\in (x-y,x+y)$ and the converse follows similarly.

This would show that $(x,y)$ is maximal in $\mathbb{C}[X,Y]$, which is what I originally wanted to prove.

Thank you

  • $\begingroup$ What makes you think that $x+y,x-y\in C$ implies that $y\in C$? $\endgroup$ – Mark May 4 at 8:55
  • $\begingroup$ The ideal inside which ring? What is $C$? $\endgroup$ – punctured dusk May 4 at 8:55
  • $\begingroup$ C is the complex field $\endgroup$ – rhombicosicodecahedron May 4 at 8:56
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    $\begingroup$ Oh, ok then. Better to write it like this: $\mathbb{C}$. $\endgroup$ – Mark May 4 at 8:57
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    $\begingroup$ Yes, it looks fine. It is indeed important that it is a field and not just a ring. $\endgroup$ – Mark May 4 at 9:13

I've failed in understanding about ideals in which ring is this topic, so my answer is about $\mathbb C[x, y]$.

The easiest way to prove that$(x-y, x+y) = (x, y)$ is to prove that $x, y \in (x-y, x+y)$ and $x - y, x + y \in (x, y)$. Both of these facts are trivial (for example, $y = \frac{1}{2}((x + y) - (x -y)) \in (x-y, x +y))$.

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    $\begingroup$ There's a typo: in your last formula, the l.h.s. should be $y$. $\endgroup$ – Bernard May 4 at 9:22

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