is the ideal $(x-y,x+y)$ same as $(x,y)$

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

• What makes you think that $x+y,x-y\in C$ implies that $y\in C$? – Mark May 4 at 8:55
• The ideal inside which ring? What is $C$? – punctured dusk May 4 at 8:55
• C is the complex field – rhombicosicodecahedron May 4 at 8:56
• Oh, ok then. Better to write it like this: $\mathbb{C}$. – Mark May 4 at 8:57
• Yes, it looks fine. It is indeed important that it is a field and not just a ring. – 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, $$y = \frac{1}{2}((x + y) - (x -y)) \in (x-y, x +y))$$.
• There's a typo: in your last formula, the l.h.s. should be $y$. – Bernard May 4 at 9:22