0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ Oh, ok then. Better to write it like this: $\mathbb{C}$. $\endgroup$ – Mark May 4 at 8:57
  • 1
    $\begingroup$ 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
2
$\begingroup$

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, $y = \frac{1}{2}((x + y) - (x -y)) \in (x-y, x +y))$.

$\endgroup$
  • 1
    $\begingroup$ There's a typo: in your last formula, the l.h.s. should be $y$. $\endgroup$ – Bernard May 4 at 9:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.